Math $$$

SECTION 16
RINGS
Up to now we have been studying sets with a single binary operation defined
on them. For example, we have encountered groups such as (0, +) and
(0 – {O}, .). But of course there are times in real life when one considers both
addition and multiplication simultaneously, for instance on 0, or more
basically on lL. We will now consider an abstract notion designed to capture
the essence of such situations where two operations interact with each other.
What is the essence of the situation for addition and multiplication on lL,
for example? If we just look at addition, then we have an abelian group
(lL, + ). If we concentrate on multiplication, we have an associative commuta-
tive binary operation. There happens to be an identity element for ” but most
elements fail to have inverses. Finally, if we consider both operations at once,
then the most salient point is that they are connected by the distributive laws:
a(b + c)= ab + ac and (b + c)a= ba + ca.
During the nineteenth century, number theorists worked with systems
more inclusive than lL which satisfied the same properties with respect to +
and ‘. One motivation for their efforts was the hope that by considering such
systems, one might answer questions about lL that could not be answered by
thinking in terms of lL alone. Although this hope was only partially realized
(questions about lL can be hard!), a great deal was accomplished, and
moreover, the groundwork was laid for the development of an abstract theory
in the twentieth century.
The abstract concept which emerged is that of a ring. A ring has all the
properties of lL indicated above, except that in order to achieve greater
generality, one does not require that there exist an identity element for
multiplication, nor that multiplication be commutative. (A good deal of work
has also been done on systems for which multiplication fails even to be
associative, but we will not consider such nonassociative rings.)
153

154 Section 16. Rings
In writing down the axioms for a ring, we should perhaps use symbols
such as * and 0 to denote the two operations involved, to emphasize that
they do not have to be ordinary addition and multiplication of numbers.
However, we shall just use + and ” for simplicity. You are experienced
enough by this time to keep in mind that we are just talking about two binary
operations, even though we denote them by + and . and call them addition
and multiplication.
DEFINmON Suppose that R is a set and + and . are two binary operations
on R. Suppose further that:
i) (R, +) is an abelian group,
ii) . is associative, and
iii) the distributive laws hold, i.e.,
for all ‘,,’2,’3 in R.
Then R, together with the binary operations + and ” is called a ring. We
denote it by (R, +, .), or R for short.
The two distributive laws are referred to as the left and ,ight distributive
laws, respectively. Of course, if . happens to be commutative, then these two
laws say the same thing.
A ring for which . is commutative is called a commutative ring. In
general, the addition on a ring has already been assumed to be pretty nice,
and one gets more special rings by imposing more assumptions on the
multiplication.
The additive identity element of R, i.e., the identity element for (R, +), is
denoted by 0, or OR’ If there happens to be an identity element for’, then it is
an easy matter to see that there is only one such; it is called the multiplicative
identity element or the unity of R, and is denoted by 1, or 1 R’ A ring that
possesses a unity is called (what else?) a ring with unity.
Examples
1. (I, +, .) is a commutative ring with unity, as are (10, + , .) and
(IR, + , .). Here + and . denote ordinary addition and multiplication.
2. (21, +, .) is a commutative ring, but not a ring with unity.
3. Let R be the set of all real numbers that can be written in the form
a + b v’2 , where a, bEl. It is clear that the sum of two elements of R is in R,

Section 16. Rings 155
and if a + b Y2 and c + dY2 are in R, then their product
(ac + 2bd)+ (ad+ bc)Y2
is in R too. R is a commutative ring with unity under ordinary addition and
multiplication.
4. Let R = {O, 1,2, … ,n – I}, and let EB and 8 denote addition and
multiplication modulo n on R, that is,
aEBb= a+b and a8b= a·b,
where denotes remainders modulo n. Then (R, EB) is the familiar group
(In’ EB), and we claim that (R, EB, 8) is a commutative ring with unity, which
we will denote by (In’ EB, 8). In fact, the proofs of associativity for 8 and of
the distributive laws are very similar to the proof we gave for the associativity
of EB on In in Section 2 (Exercise 16.8). (In’ EB, 8) is commutative because
a8b= a·b = b’a =b8a.
The multiplicative identity element is 1.
The rings (In’ EB, 8) are interesting at this point because they already
begin to display behavior quite different from that of the prototype example
(l,+,·). For instance, look at (l6,EB,8). Here 283=0 although neither 2
nor 3 is 0. Things like that certainly don’t happen in l. To see something even
stranger, look at (ls, EB, 8). Here 23(=28282)=0, so a power of a nonzero
element can be 0.
We introduce some terminology in order to deal with such situations. An
element a E R is called a zero-divisor if there exists an element b =1= ° such that
either
ab=O or ba=O.
a is called nilpotent if there exists some positive integer n such that an = 0.
(Here an means a multiplied by itself n times.) Thus in l6′ 2 is a zero-divisor,
and so is 3 (and ° and 4). In ls’ 2 is nilpotent, as are 0, 4, and 6.
At the opposite extreme from these badly behaved elements are those
called units. Suppose R is a ring with unity. Then a E R is called a unit if there
exists an element b E R such that
ab=ba= 1.
Note right away that “unit” and “unity” are not the same thing! The unity is
a unit because 1·1 = 1, but there may be many units other than 1. For
instance, in ls, 1,3,5, and 7 are all units, since 12=32=52=72= 1. In lJO’ 1,
3, 7, and 9 are units because 12=92= I and 387=783= 1.
We shall see below that a unit can never be a zero-divisor. The general
situation for (In’ EB, 8) is that a Eln is a unit iff (a,n)= 1, and every element

156 Section 16. Rings
is either a unit or a zero-divisor (see Exercise 16.9). On the other hand, some
rings contain elements that are neither units nor zero-divisors. Can you give
an example?
It is easy to see that if a is a unit, then there is only one b such that
ab = ba = 1. We call b the multiplicative inverse of a, and denote it by a -I.
Examples (continued)
5. Let ~ ffi ~ be the set of all ordered pairs (a, b) of real numbers, with
addition and multiplication defined componentwise:
(a,b)+(c,d)=(a+c,b+d) and (a,b)·(c,d)=(ac,bd).
Then ~ffi~ is a commutative ring with unity. An element (a,b) in ~ffi~ is a
zero-divisor iff at least one of a, b is 0, and it is a unit iff neither of a, b is O.
For example, the equation
(0,3) ·(1,0)=(0,0)
shows that both (0,3) and (1,0) are zero-divisors, while
(5,6) . ( ~, ~) = (1,1)
shows that both (5,6) and (L ~) are units. In this ring it is true that an
element is a zero-divisor iff it is not a unit.
The ring IR ffi IR is called a direct sum. It is common practice, in ring
theory, to speak of “direct sums” rather than “direct products.” The additive
terminology is also commonly used in discussing abelian groups.
6. Generalizing the previous example, let R I’ R 2, ••• , Rn be rings. Then
their direct sum R 1 ffi R2 ffi … ffi ~ is the ring whose elements are all n-tuples
(‘I,’2″””n)’ with riERi, under componentwise addition and multiplication.
The direct sum is commutative iff each summand Ri is, and it has a
multiplicative identity iff each Ri has one.
7. Let R be the set of all real-valued functions defined on IR, under
addition and multiplication of functions. R is a commutative ring with unity.
An element fin R is a zero-divisor iff f(x) = 0 for at least one x E IR, and it is a
unit iff f(x) =1= 0 for all x E IR. Here, too, an element is a zero-divisor iff it is not
a unit.
Which elements of R are nilpotent?
8. Let MilR) denote the set of all 2 X 2 matrices with real entries, under
addition and multiplication of matrices. (Addition means adding correspond-
ing entries.) Then M2(1R) is a noncommutative ring with unity. For example,

Section 16. Rings 157
the left distributive law requires that
and this is easily checked.
Note that there exist elements A, B in MilR) such that AB = (g g) but
BA *(g g). For instance, we can take
A =(: g) and B=(g 0) I .
This example explains why, in the definition of zero-divisor, we said” ab = 0
or ba=O.”
Note also that this ring contains nonzero nilpotent elements. For exam-
ple,
0) o .
9. Let (G, + ) be any abelian group, and denote the identity element of G
by O. Define a multiplication on G by declaring a . b = 0 for all a, bEG. It is
then easy to check that (G, +, .) is a ring. We call this the ring on G with
trivial multiplication.
If, in particular, we start with G = {O}, the trivial group, then we get a ring
with one element. It is called the trivial ring, and is rather annoying. For
example, 0 satisfies the definition of a multiplicative identity element (O’x=
x·O=x for all x, right?), so 0 is 1. That should make you cringe, but there is
some comfort in noting that this anomaly can only occur in the trivial ring: If
R is a ring with unity and R has more than one element, then 0* I in R. In
order to prove this, we need some basic information.
We frequently denote multiplication in a ring by juxtaposition, writing ab
rather than a . b.
THEOREM 16.1 Let R be a ring, and let a,b be elements of R. Then:
i) a ·O=O·a=O;
ii) a( – b)=( – a)b= -(ab);
iii) (- a)( – b)= ab;
iv) m(ab)=(ma)b=a(mb) for any integer m;
v) mn(ab) = (ma)(nb) for any integers m and n.
PROOF. i) To show that a ·0=0 it is enough to show that
a ·0 + a ·0 = a ·0,

158 Section 16. Rings
for then we can add -(a ·0) to both sides. But
a·O + a ·O = a ·(0+0) = a ·0.
Similar reasoning works for O·a.
ii) To show that a(-b)= -(ab), it suffices to show that a(-b)+ab=O.
But
a( – b)+ ab= a( – b+ b)= a ·0=0
by (i). A similar argument, using right distributivity instead of left, shows that
(-a)b= -(ab).
iii) Replacing a by – a in the first equality of (ii), we obtain
(- a)( – b)= [ -( – a).]b= abo
iv) Exercise.
v) Exercise. 0
COROLLARY 16.1 Let R be a nontrivial ring with unity. Then 0+ 1 in R.
PROOF. Since R is nontrivial, we can pick a+O in R. Then if 0= 1 we get
a ·0= a . 1, that is, 0 = a, a contradiction. 0
COROLLARY 16.3 Let R be a ring with unity, and let u E R be a unit. Then u is
not a zero-divisor in R .
PROOF. We must show that if r is an element of R such that ur=O or ru=O,
then r=O. Now if ur=O, then
u-1(ur) = u – 1(O)=0,
that is,
r=O.
A similar argument works if ru = O. 0
The next corollary is technical, but useful.
COROLLARY 16.4 If band c are elements of a ring R , define b – c to mean
b+(-c). Then for any aER, we have
a(b-c)=”ab-ac, and (b-c)a=ba-ca.
PROOF.
a( b – c) = a( b + ( – c» = ab + a( – c) = ab + [ – (ac) ] = ab – ac.
Likewise for the second equality. 0
We mentioned above that one gets nicer and nicer rings by imposing
more and more assumptions on the multiplication. For example, we get rings

Section 16. Rings 159
that behave somewhat like (l, ED, 0) by adding the assumptions indicated in
the following
DEFINmON An integral domain (or just domain, for short) is a commutative
ring with unity in which 1 *0 and there are no nonzero zero-divisors.
Thus l, 0, and ~ are all domains. Another example is the ring in-
troduced in Example 3 above, consisting of all real numbers of the form
a+bV2, with a,bEl.
The following simple observation yields an alternative characterization of
integral domains.
11IEOREM 16.5 Let R be a ring and let a, b, c E R. Assume that a is not a
zero-divisor. Then if ab=ac, we have b=c.
PROOF. From ab=ac we get ab-ac=O, so a(b-c)=O. Since a is not a
zero-divisor, this means b – c = 0, so b = c. D
COROLLARY 16.6 Let R be a commutative ring with unity 1 *0. Then R is an
integral domain iff whenever a,b,cER satisfy ab=ac and a*O, we have
b=c.
PROOF. Assume that R is a domain. Let a,b,cER, a*O, and suppose that
ab=ac. Then, since a is not a zero-divisor, the theorem implies that b=c.
Conversely, suppose that R is not an integral domain. We will find
a,b,cER, a*O, such that
ab=ac but b*c.
In fact, we know that there is a nonzero zero-divisor a E R. Let bE R be such
that b*O and ab=O. Then we have
a ·b=a ·0, but b*O. D
Finally, we consider rings with unity in which every nonzero element has
a multiplicative inverse.
DEFINmONS R is called a division ring if R has a unity 1 *0 and every
nonzero element of R is a unit. A commutative division ring is called a field.
Another way of saying that R is a division ring is to say that the set
R – {O} forms a group under multiplication. Saying that R is a field amounts
to saying that this group is abelian.
Familiar examples of fields include 0, ~, and the complex numbers C. A
less familiar example is (lp, ED, 0), where p is a prime number. lp is a field
because each r E {l, 2, … ,p -l} satisfies (r,p) = I and is therefore a unit in lp
(see Exercise 16.9).

160 Section 16. Rings
Examples of division rings that are not fields are a little harder to come
by. In fact, a celebrated theorem of J. H. M. Wedderburn asserts that there
aren’t any finite examples: Every finite division ring is necessarily a field.
Proving this here would take us too far afield (so to speak), so we shall
content ourselves with a much simpler result. We shall see an infinite division
ring that is not a field in Section 17.
We obtain our easy substitute for Wedderburn’s Theorem by replacing
“division ring” by “integral domain.”
THEOREM 16.7 Every finite integral domain is a field.
PROOF. Let R be a finite domain. Then R is commutative, and 1 *0 in R. We
must show that if rER, r*O, then r has a multiplicative inverse in R.
Since R is finite we can list its elements as rl>r2, ••• ,’n’ Consider the
elements
Since r*O, these are all distinct by Corollary 16.6. Since they are all in Rand
R has only n elements altogether, they must account for all the elements in R.
In particular, one of them is 1, so
for some i, and r is a unit. 0
This proof will look familiar to you if you worked Exercise 3.15.
Of course, there exist infinite domains that are not fields-oZ, for in-
stance. On the other hand, every field, finite or infinite, is a domain, because
units are not zero-divisors. Thus the notions of “domain” and “field” coincide
for finite rings, but, in general, “field” is stronger.
EXERCISES
16.1 Let R be a ring with unity 1 R’ Show that ( – 1 R)a = – a for all a E R.
16.2 a) If r.s=2(r+s) and rDs=rs, is (IR,.,D) a ring?
b) If r.s=2rs and rDs=rs, is (IR-{O},.,D) a ring?
c) If r.s=rs and rDs=rs, is (IR+,.,D) a ring?
16.3 Show that the set of all real numbers of the form a + b V2 , where a, bE Q,
forms a field under ordinary addition and multiplication.
16.4 Consider (0,., D), where. is the addition given by a. b = a + b -1, and ° is
the multiplication given by aDb = a + b – abo Is (0,., D) a field?

Section 16. Rings 161
16.5 (A construction of the complex numbers.) Let F be the set of all 2 x 2 matrices
of the form
(-~ ~),
where a, bE IR. Show that F forms a field under addition and multiplication
of matrices.
Remarks. Note that if we think of
(-~ ~)
as representing the complex number a + bi, then addition and multiplication
in F correspond to the usual operations on complex numbers. For example,
(-~ ~)( -~ ~)=( -:~=~~ :~t~),
which corresponds to (ac- bd)+(ad+ bc)i. Thus this exercise shows you how
to construct a field having all the desired properties of C, starting only with IR.
16.6 Let Fbe a field. For a,bEF, b’l=O, define alb to mean ab- I. Show that:
(.)(a)(c) ac ( .. )a c_(ad+bc) Ib’li = bd; 11 b+li- bd
16.7 Let F be a field, let a,b E F, and assume a’l=O. Show that the equation
ax+b=O
can be solved for x in F; that is, there is x E F which makes the equation true.
16.8 Prove the following for (Zn’ Ell, 8):
a) associativity for 8;
b) the distributive laws.
16.9 a) Show that an element a E(Zn’ Ell, 8) is a unit iff (a,n)= 1.
b) Show that every element of Zn is either a unit or a zero-divisor.
c) Which elements of Zn are nilpotent?
16.10 Prove parts (iv) and (v) of Theorem 16.1.
16.11 Find all units, zero-divisors, and nilpotent elements in the following rings:
a) ZEllZ;
b) Z3EllZ3;
c) Z4 Ell Z6′
16.12 a) Show that the trivial ring is the only ring in which 0 is not a zero-divisor.
b) Show that in any ring except the trivial ring, every nilpotent element is a
zero-divisor.
16.13 a) Show that if R is a ring with unity, then the multiplicative identity element
in R is unique.

162 Section 16. Rings
b) Show that if R is a ring with unity and a E R is a unit, then the multiplicative
inverse of a is unique.
16.14 An element r in a ring R is called idempotent if r2 = r. Find all the idempotent
elements in the ring of real-valued functions on IR under addition and
multiplication of functions.
16.15 (See Exercise 16.14.) Let R be a nontrivial ring with unity. Let rER be
idempotent. Show that:
a) 1 – r is also idempotent, and
b) either r or 1 – r is a zero-divisor.
16.16 Let R be a ring with unity. R is called Boolean if every element of R is
idempotent. Show that if R is Boolean then:
a) 2r = 0 for every r E R (in other words, r = – r);
b) R is commutative.
[Hint jor (a): Consider (r + r)2.J
16.17 Let X be a set and let R be the set of all subsets of X.
a) Show that (R, Co. , n) is a commutative ring with unity, where Co. denotes the
operation of symmetric difference.
b) Show that (R, Co., n) is Boolean (see Exercise 16.16).
16.18 Let R be a ring with unity, and assume that R has no nonzero zero-divisors.
Let a, b E R, and assume that ab = 1. Show that ba = 1 too, and therefore a and
b are units.
16.19 Let R be a ring with unity, and let a E R. Assume that there is a unique bE R
such that ab = 1. Show that ba = 1, and therefore a is a unit.
16.20 a) Let S be the set of all real-valued functions on IR. Is (S, +, 0) a ring? (Here
o denotes composition of functions.)
b) Let R be the set of all real-valued functions on IR that are homomorphisms
of the additive group (IR, +). Is (R, +, 0) a ring?
16.21 Let G be the infinite direct product Z X Z X Z X Z X •.. , where there is one copy
of Z for each positive integer, and the operation on each copy is ordinary
addition. If ‘PI and ‘P2 are homomorphisms from G into itself, define ‘PI + ‘P2 by
(‘PI + ‘P2)(g) = ‘PI(g) + ‘P2(g)
for all gE G.
a) Show that ‘PI + ‘P2 is a homomorphism, and that the set of all homomor-
phisms from G into itself forms a ring R with unity under the operations
+ and o.
b) Show that there exist elements ‘P and I/; in R such that ‘PI/; = 1 but I/;’P=F 1
(here 1 denotes the unity of R).
16.22 Let (R, +, . ) be a ring, and let S be a set. Let R S denote the set of all functions
from S to R. Show that R S forms a ring under addition and multiplication of
functions.

Section 16. Rings 163
16.23 Let R be a ring with unity, and let U denote the set of units in R. Show that
U is a group under the multiplication in R.
16.24 a) Let 1[ i] denote the set of all comple~ numbers of the form a + bi, where
a,b E1. Show that l[i] is a commutative ring with unity under ordinary
addition and multiplication of complex numbers. l[i] is called the ring of
Gaussian integers.
b) For r= a + bi E l[i], define the norm N(r) of r by N(r) = a2 + b2• Show that
if r,sE1[i], then N(rs)=N(r)N(s).
c) Show that r=a+bi is a unit in l[i] iff N(r) = l. Using this information, find
all the units in 1[ i].
d) (See Exercise 16.23.) The group of units of l[i] is isomorphic to a familiar
group. Which one?
16.25 Let R denote the ring of all real numbers of the form a + b VI , where a, bEl.
For r=a+bVI ER, define N(r) by N(r)=a2 -2b2•
a) Show that if r,sER then N(rs)=N(r)N(s).
b) Show that r is a unit in R iff N(r) = ± l.
c) Show that there are infinitely many units in R.
16.26 Give an example of a finite noncommutative ring.
16.27 Give an example of a noncommutative ring with no multiplicative identity.
16.28 Let R be an integral domain. If there exists a positive integer n such that
n . I = 0, then the smallest such integer is called the characteristic of R. If no
such n exists, then we say that R has characteristic O.
a) Show that if R has characteristic n, then n ·r=O for every rER.
b) Show that if R has characteristic n > 0, then n is a prime number.
c) For each prime number p, give an example of a field of characteristic p. Give
an example of a field of characteristic O.
16.29 Must every ring with a prime number of elements be commutative? Either prove
that it must, or give a counterexample.
16.30 Let R be a finite nontrivial ring with no nonzero zero-divisors. Show that R
is a division ring.
16.31 Prove that if F is a finite field then there exist a prime number p and a positive integer
j such that IFI = pi. [Suggestion: Use the result of ExerC:lse 16.28(b).]
16.32 a) Suppose R is a ring with unity and for all x andy inR we have (xyi =~y. Prove
that R is commutative.
b) Give an example to show that the result of part (a) may fail if R does not have a
unity.

SECTION 17
SUBRINGS, IDEALS,
AND QUOTIENT RINGS
In this section we shall develop analogues, for rings, of some of the concepts
we encountered in dealing with groups. As for groups, the purpose of doing
this is to develop ways of talking about the internal structure of a given ring
and the relationships between different rings.
We begin with the analogue of “subgroup.”
DEFINmON Let (R, +, . ) be a ring. A subset S of R is called a subring of R if
the elements of S form a ring under + and ‘.
In particular, the definition requires that (S, +) be a subgroup of (R, +).
Thus if S is a subring of R, then we know that the additive identity element 0
of R is in S, and that S is closed under addition and under additive inverses.
The relationship between Rand S with respect to multiplication need not be
so clean. For example, if R has a multiplicative identity 1, then 1 need not be
in S, and it is even possible that S may have an identity element different
from that of R.
Examples
1. Let (l., +, .) be the integers under ordinary addition and multiplica-
tion, and let 2l. be the set of even integers. Then it is easy to see that 2l. is a
subring of l.. Although l. is a ring with unity, 2l. has no unity.
2. Consider the ring REB R. Let S denote the set of all pairs of the form
(r,O), where rER Then S is a subring of REBR, and REBR has unity (1,1),
164

Section 17. Subrings, Ideals, and Quotient Rings 165
which is not in S. In this case, S has its own unity, namely (1,0). Note that
(1,0) is not a unity for IREBIR, but only for S.
As for subgroups, the definition of “subring” can be recast III more
compact form.
THEOREM 17.1 Let (R, +, .) be a ring, and let S be a subset of R. Then S is a
subring of R iff the following two conditions are satisfied:
i) (S, +) is a subgroup of (R, +); and
ii) S is closed under multiplication, that is, if rpr2ES then r(r2ES.
PROOF. It is clear that if S is a subring of R, then (i) and (ii) hold. Conversely,
assume that (i) and (ii) hold for S. Then + and . are both binary operations
on S, and (S, +) is an abelian group, so we need only check that . is
associative on S and that the distributive laws hold in (S, +, .). But associa-
tivity and distributivity hold for all elements of R, hence for those of S. 0
Condition (i) can be reduced further by using your favorite subgroup
criteria. For example, by using the result of Exercise 5.24 we obtain the
following.
COROLLARY 17.2 Let (R, +,.) be a ring and let S be a nonempty subset of R.
Then S is a subring of R iff the following two conditions hold:
i) for every r(,r2 E S, we have r( – r 2 E S;
ii) for every r(,r2 ES, we have r(r2 ES.
Examples (continued)
3. Let’s find all the subrings of (Z, +, .). We know that any subring must
be a subgroup of (Z, +) and hence must be additively a cyclic subgroup of
the form mZ, for some m. We have only to figure out which of these
subgroups constitute subrings. The extra requirement that mZ must satisfy to
be a subring is closure under multiplication. But clearly if i, j are integers,
then (im)(jm) = (ijm)m E mZ. Thus every subgroup of (Z, + ) is a subring of
(Z, +, .).
Similar reasoning shows that every subgroup of (Zn’ EB) is a subring of
(In’ EB, 0).
4. Let R be the set of all real-valued functions defined on IR under
pointwise addition and multiplication of functions. Let S be the subset of R
consisting of all the continuous functions. Then S is a subring of R, for if j,g
are continuous functions so is j- g and so is jg. (We have used Corollary
17.2.)

166 Section 17. Subrings, Ideals, and Quotient Rings
5. Let R be as in the previous example and let
S={jERIJ(O)=O},
so that S consists of those functions that take the value 0 at x = O. Then S is a
subring of R, for if J,g E S then J(O) = 0 and g(O) = 0, so
(1- g)(O) = J(O) – g(O) =0,
which shows that J – g E S, and
Jg(O) = J(O)g(O) = 0,
which shows that Jg E S.
Observe that if T= {f E R IJ(O) = I}, then T is not a subring of R, because
if J(O) = g(O) = 1 then (1- g)(O) = 1-1 =0.
6. Let MilR) be the ring of all 2 X 2 matrices with real entries. Let S
consist of all matrices of the form (~ ~). Then S is a subring of MilR), for if
(a b) and (e f) are in S then so is (a b)_(e f) and so is (a b)(e f).
Od Oh ‘ Od Oh’ OdOh
On the other hand, if T consists of all matrices of the form (: ~), then T
is not closed under multiplication, so T is not a subring of MilR).
7. Let M 2(C) be the ring of all 2 X 2 matrices with comp~ex entries. Let
1=(6 0) J=(i 1 ‘ 0
0) K= ( 0
– i ‘ -1 6)’ and L=JK=(~ i) O’
Let IHI be the following subset of MiC):
IHI = {a1 + bJ + cK + dLla,b,c,d EIR}.
(Note: The product of a constant and a matrix is obtained by multiplying
each entry of the matrix by the constant.) IHI is a subring of M 2(C), for it is
clear that IHI is an additive subgroup, and closure of IHI under multiplication
follows from distributivity in MiC) and the fact that {± 1, ±J, ± K, ± L} is
closed under multiplication. [{ ± 1, ± J, ± K, ± L}, under multiplication, is
Q8′]
IHI is called the ring of quatemions, and “IHI” is for Hamilton, the man who
discovered this ring. IHI is a noncommutative ring, for the same reason that Qs
is a nonabelian group: JK-:I=KJ. IHI has an identity 1*(g g), and it is easy to
see that H is in fact a division ring. For if
a1+bJ+cK+dL=( a+bi
-c+di
C+di)EIHI
a-bi

Section 17. Subrings, Ideals, and Quotient Rings 167
and at least one of a,b,c,d is not zero, we have
( a+bi
-c+di
c + di ) – I I ( a – bi – c – di )
a-bi = a2 +b2 +c2 +d2 c-di a+bi
I
—– (aI – bJ – cK – dL) E IHI.
a 2 + b2 + c2 + d 2
IHI is the example of a noncommutative division ring that we promised you in
Section 16.
Hamilton discovered the quaternions in 1843, after he had spent ten or
fifteen years seeking a generalization of C that could be used in connection
with geometric and physical problems in 3-space. One reason why it took him
so long was that he started out looking for a commutative generalization;
coming up with a noncommutative one was, at the time, a revolutionary step.
The definition of IHI via MiC) was not possible until 1858, when Cayley
introduced matrices. Hamilton thought of IHI while taking a stroll on the
evening of October 16, 1843; it occurred to him as a set of elements of the
form
a+ bi+ cj+ dk,
where a,b,c,dEIR and
i2 =f=k2 =ijk= -I.
Although he made other distinguished contributions to science, Hamilton
considered the discovery of the quaternions to be the crowning achievement
of his life. He spent twenty years studying them, and wrote several huge
volumes about them.
In dealing with groups, we found that some subgroups were better than
others. For example, in attempting to construct the quotient group modulo a
subgroup H, we saw that it was crucial for H to be normal. We encounter a
similar situation when we try to construct quotient rings.
Let (R, +,.) be a ring and let S be a subring of R. We know that (S, +)
is a subgroup of (R, + ), and in fact there is no problem about (S, + ) being a
normal subgroup of (R, +), because (R, +) is abelian. Thus we already know
how to construct a quotient group (R/ S, +), the elements of which are the
co sets of Sin R, with addition defined by
(S+ a)+(S+b)= S+(a+b).
We would like to endow this quotient group with a multiplication, arising
naturally from the given multiplication in R, in such a way that the quotient
becomes a ring. As usual with these things, there is only one reasonable

168 Section 17. Subrings, Ideals, and Quotient Rings
attempt; we want to define
(S+a)(S+b)=S+ab.
What we have to check is that this is a well-defined operation, that is, if
S+a= S+a’ and S+b= S+b’,
then
S + ab = S + a’ b’,
so that the product doesn’t depend on which representatives we use to define
it.
So we assume a – a’ E Sand b – b’ E S, and we wish to show that
ab – a’ b’ E S. It is clear that if this is to work for all possible choices of
a,a’,b,b’, then S has to be rather special. For instance, if we take aES, a’=O,
b arbitrary, and b’ = b, then
a-a’ES and b-b’ES,
so we want
ab-ObES, that is, abES.
In other words, we require that if a E Sand b is any element of R, then
ab E S. Similar reasoning shows that we also require that if bE S and a E R is
arbitrary, then ab E S.
Now we claim that these two conditions are enough to make our
multiplication work out. For suppose S satisfies both conditions. Assume
a-a’ES and b-b’ES. Then
ab- a’b’ =(a- a’)b+ a'(b- b’).
By the conditions on S, (a – a’)b E Sand a'( b – b’) E S, so
(a- a’)b+ a'(b- b’) E S
since S is an additive subgroup. Thus ab – a’ b’ E S, and our multiplication is
well defined.
Subrings that have the special properties required to make multiplication
of the additive cosets well defined are called ideals.
DEFINl110N A subring S of a ring R is called an ideal of R if for every s E S
and rER we have rsES and srES.
We have seen that if S is a subring of R, then the natUral attempt at
introducing a multiplication on R/ S will succeed iff S is an ideal. It is now
easy to complete the proof of

Section 17. Subrings, Ideals, and Quotient Rings 169
THEOREM 17.3 Let (R, +, .) be a ring, and let S be an ideal of R. Then the set
R/ S of cosets of the additive subgroup (S, +) is a ring under the operations
(S+ a)+(S+ b)= S+(a+b),
(S+a)(S+ b)= S+ab.
PROOF. We know that R/ S is a group under the indicated addition, and this
group is abelian because (R, +) is. We have seen above that the indicated
multiplication yields a binary operation on R/ S, so all that we have to check
is that this multiplication is associative and that the distributive laws hold.
Associativity requires that
[(S+ a)(S+ b) J(S+ e) =(S+ a)[ (S+ b)(S+ e)]
for all a,b,e E R. This amounts to
(S+ab)(S+e)=(S+a)(S+be), that is,
S+ (ab)e = S+ a(be),
which is true by associativity of multiplication in R. Similarly, R/ S inherits
distributivity from R; we leave the details of this to the reader. 0
R/ S is called the quotient ring (or factor ring) of R by S.
There is a certain redundancy in our definition of “ideal,” in that the
condition that rs E Sand sr E S for every r E Rand s E S already implies part
of the condition that S be a sUbring. The next result provides a neater
characterization of ideals.
THEOREM 17.4 Let R be a ring and let S be a subset of R. Then S is an ideal
of R iff the following two conditions hold:
i) S is an additive subgroup of R (equivalently, S is nonempty and closed under
subtraction);
ii) For every r E Rand s E S, we have rs E Sand sr E S.
PROOF. Exercise.
We will usually denote an ideal by I, rather than S, from now on.
Examples (continued)
8. As we have seen, the subrings of (l, +, .) are precisely the additive
subgroups ml. In this case [likewise for (In’ EB, 0)], every subring is also an
ideal, because if we multiply an element of ml by any integer we get an
element of ml.
9. l is a subring of (Q, +, .), but not an ideal. For example, 1 Eland
4 E Q, but 1· 4 EtC l.

170 Section 17. Subrings, Ideals, and Quotient Rings
Actually, there is something more general going on in this example. If R
is a ring with unity 1, then the only ideal of R that contains 1 is R itself.
(Prove it!)
Incidentally, R will always be an ideal of R, no matter what ring R is. We
will call R the improper ideal; all other ideals are called proper. The ideal {O}
is called trivial.
10. Refer back to Examples 4 and 5, which present two examples of
subrings of the ring R of all real-valued functions on R. The subring S in
Example 4 is not an ideal, because if j E Sand g E R thenjg need not be in S,
since it need not be continuous. (Can you give an example?) The subring S in
Example 5 is an ideal, however, because if j(O) = 0 and g is any element of R,
then
jg(O) = j(O)g(O) =O·g(O) =0,
sojgES. Likewise gfES.
11. Let I be the subring of REBR consisting of all pairs of the form (r,O).
Then I is an ideal, because (r,O)(a,b)=(ra,O)EI for any r,a,bER, and
likewise (a,b)(r,O)EI. Note that I has a unity, and yet I is a proper ideal of
R. Why does this not contradict our observation in Example 9 about ideals
that contain the unity?
12. Let R be a commutative ring with unity, and let a E R. Let
aR={arlrER},
so that aR is the set of all multiples of a in R. Then aR is an ideal of R. First
of all, if ar1,ar2EaR then arl-ar2=a(rl-r~EaR, so aR is an additive
subgroup of R. Secondly, if ar E aR and t E R, then
t(ar) = (ar)t = a(rt) EaR,
since R is commutative, and this verifies that aR is an ideal.
We call aR the principal ideal generated by a. Note that a EaR, since
a = a . 1. Also observe that aR.= Ra = {ralr E R}.
13. Let R = (Z, + , .) and let I be the ideal nZ for some positive integer n.
Then (R / I, +, .) has n elements, namely, 1+ 0,1 + 1,1 + 2, … ,1 + (n – 1). We
saw in Section 13 that (R/ I, +) is isomorphic, as a group, to (ZII’ $). In
Section 18 we shall introduce a notion of isomorphism for rings, and it will
tum out that (Z/ nZ, +, .) is isomorphic, as a ring, to (ZII’ $, 0).
Suppose now that R is a ring and I is an ideal of R. Then R/ I is a ring,
and we can sensibly (and profitably) ask which properties in R translate into
familiar properties for the elements of R/ I. For example, if a E R, when is
I + a a zero-divisor in R / I?

Section 17. Subrings, Ideals, and Quotient Rings 171
I + a is a zero-divisor iff there is some 1+ b ‘f= 1+ 0 such that either
(I+a)(I+b)=I+O or (I+b)(I+a)=I+O. This boils down to there being
some b f1. I such that either ab E I or ba E I. In particular, I + a is a nontrivial
(i.e., nonzero) zero-divisor in R/ I iff a f1. I and there is some b f1. I such that
abEl or baEI.
From this it is clear what conditions we need on I to rule out nontrivial
zero-divisors in R/ I.
DEFINmON Let R be a ring, I an ideal in R. Then I is prime if whenever
a, bE Rand ab E I, then at least one of a or b is in I.
Example Let p be a prime in 71.. Then p7L is a prime ideal, because if ab is
divisible by p, then one of a or b must be divisible by p.
TIlEOREM 17.5 R/ I has no nontrivial zero-divisors iff I is prime.
The proof is immediate from the above discussion. Specializing to the
case where R is a commutative ring with unity, we get:
COROLLARY 17.6 Let R be a commutative ring with unity, I an ideal in R.
Then R/ I is an integral domain iff I is a proper prime ideal.
PROOF. R/ I is a commutative ring with unity (see Exercise 17.14). Thus R/ I
is an integral domain iff it is nontrivial and has no nontrivial zero-divisors,
that is, iff I is proper and prime. 0
When is R/ I a field?
DEFINmON. An ideal I of R is called maximal if I is proper and there is no
proper ideal J 7. I.
Thus I is maximal iff it is proper and cannot be extended to a larger
proper ideal.
TIlEOREM 17.7 Let R be a commutative ring with unity. If I is an ideal in R,
then R / I is a field iff I is maximal.
PROOF. R/ I is a commutative ring with unity. Thus it is a field iff it is
nontrivial and each of its nonzero elements is a unit.
Now R/ I is nontrivial iff I is proper. And every nonzero element of R/ I
is a unit ~ for every a E R – I, there is bE R such that (I + a)(I + b) = 1+1,
in other words,
for every a f1. I, there is b such that ab – I E I. [17.1]
Thus R/ I is a field iff I is proper and [17.1] holds. To conclude the proof, we
will show that a proper ideal I is maximal iff [17.1] holds.

172 Section 17. Subrings, Ideals, and Quotient Rings
First suppose that I is maximal, and take a f/. I. Then the set
l={ar+xlrER andxEI}
is an ideal that properly includes I, hence IS not proper. Therefore, aro+ xo= 1
for some ro and x o’ and this yields
aro-l = -xoE/,
which means that [17.1] holds. Conversely, if [17.1] holds, then let I’ be an
ideal such that I’ ~ I, with the aim of showing that I’ = R. Take a E I’ – I,
and take b such that ab – 1= y E I. Then a E I’ andy E I’, so ab – y E I’. Thus
IEI’, so I’ = R, and I is maximal. 0
COROLLARY 17.8 Let R be a commutative ring with unity. Then every maxi-
mal ideal of R is prime.
PROOF. If I is maximal, then Rj I is a field, hence an integral domain. Thus I
is prime by Corollary 17.6. 0
This last result can fail when we try to weaken the assumptions. For
instance, let R be the ring on (Z2′ EEl) with trivial multiplication. Then R is a
commutative ring without unity, and {O} is a maximal ideal which is not
prime.
On the other hand, prime ideals need not be maximal, even if R is a
commutative ring with unity. According to our definitions, in fact, any ring is
a prime ideal of itself which is not maximal. If this seems like cheating, note
that {O} is a prime ideal in Z which is not maximal. And for an example of a
nontrivial proper prime ideal which is not maximal, take the subset [of zez
consisting of all pairs (a,O). Note that
{(a,2b)la,bEZ}
is a proper ideal which is strictly larger than [.
Some conditions under which proper prime ideals are maximal are
indicated in Exercises 17.29-17.31.
Here are a few more examples of maximal ideals.
Examples
1. We have observed that every ideal in Z has the form nl for some n,
and that if n is a prime p, then pI is a prime ideal. If n > 2 is not prime, then
nl is clearly not a prime ideal, so the only other prime ideals in I are {O}
and I.

Section 17. Subrings, Ideals, and Quotient Rings 173
What are the maximal ideals? Since I is a commutative ring with unity,
any maximal ideal must be prime; and since {o} and I are obviously not
maximal, the only possible maximal ideals are of the form pl. In fact every
p I is maximal. For suppose J ;;. p I and J is an ideal. If x E J – p I then
(x,p) = I, so there are a,b such that ax+ bp = 1. Since ax EJ and bp EJ, I EJ,
so J=l.
Thus for I the maximal ideals are the nontrivial proper prime ideals.
2. The ideal I={(a ,2b)la,bEI} is maximal in lEal. For if J is an ideal
properly including I, we have (m ,2n + I)EJ for some m,n EZ. Since (m,2n)
EI, subtracting shows that (0, I)EJ. Also, (1,O)EI r:;;;,J. Adding, we see that
(1, I)EJ, so J=lfBI .
3. Let us return to Example 5 above. R is the ring of all real-valued
functions on JR, under pointwise addition and multiplication, and I is the
ideal consisting of all 1 such that 1(0) = 0. We want to show that I is maximal.
Suppose J is an ideal that is strictly larger than I, and take g E J – I. Then
g(O)*O, so there is hER such that
I
h(O) = g(O) .
We have g(O)h(O) = I, so if 11 denotes the multiplicative identity in R, the
value of the function gh – 11 at ° is 0. Hence
gh-11 EI r:;;;,J,
so since gh EJ, 11 EJ. Thus J = R, and I is maximal.
The same argument demonstrates that for any r E IR, I, = U E R 11(r) = O}
is a maximal ideal in R.
4. Let S be the subring of MiR) introduced in Example 6 above, that is,
S={(~ !)la,b,dER}. We contend that I={(~ ~)la,bER} is a maximal
ideal of S. First of all, it is clear that I is closed under subtraction in S, and if
(~ ~)ES and (~ ~)EI, we have
(~ {)( ~
(~ ~)(~
~)=( ~
{)=( ~
~)EI and
a1+bh)EI
° ‘
so I is an ideal. To see that I is maximal, let J be an ideal of S that properly
includes I. Then we have (~ !) EJ for some a,b,d, with d=foO, and therefore
(~ l~d)(~ !)=(~ nEJ.

17 4 Section 17. Subrings, Ideals, and Quotient Rings
Since (~ ~) E /, we get
so J = S, as desired.
EXERCISES
17.1 Which of the following subsets of M2(1R) are subrings?
a) S=all matrices of the form (~ !)
b) S = all matrices of the form (: ~)
c) S = GL(2, IR)
d) S = all matrices of the form (: !).
17.2 Let R be the ring of real-valued functions on the real line, under pointwise
operations. Which of the following subsets S of Rare subrings? Which are
ideals?
a) S={jERlf(l)=O}
b) S= {jE R If(1)=O or f(2)=0}
c) S={jERlf(3)=f(4)}.
17.3 Let R be the ring with trivial multiplication on some abelian group G. What
are the ideals of R?
17.4 Let S, T be subrings of R. Under what conditions is S uTa subring of R?
17.5 Let R be a finite ring with, say, n elements. Let S be a subring of R that has
m elements. Show that m divides n.
17.6 Find a maximal ideal in
17.7 Find all the maximal ideals in lln.
17.8 Let R =2ll be the ring of even integers, under ordinary addition and
multiplication, and consider the subset S=4ll. Show that S is an ideal of R.
Is S maximal? Is it prime?
17.9 Let R={qEQlq=ajb, a,bEll and b is odd}. Show that R has a unique
maximal ideal.
17.10 Let X be a nonempty set and let R be the ring (P(X), 6, n).
a) Show that if Y s: X, then P(Y) is an ideal in R and has a unity different from
that of R.
b) Find a maximal ideal in R.

Section 17. Subrings, Ideals, and Quotient Rings 175
17.11 Prove Theorem 17.4.
17.12 Let R be a ring with unity. Show that a nonempty subset S of R is an ideal iff
the following two conditions are satisfied:
i) SI +S2ES for every Sl>s2ES;
ii) rs and sr are in S for every sE S, rER.
17.13 Show that if I is an ideal of R, then the distributive laws hold in Rj I.
17.14 Let R be a ring and I an ideal of R. Show that
a) if R is commutative, so is Rj I;
b) if R has a unity, so does Rj I.
17.15 Show that if S is a subring of R and I is an ideal such that I !;;;; S, then S j I is
a subring of Rj I. Show that if S is an ideal of R, then Sj I is an ideal of Rj I.
17.16 Show that M 2(R) has no ideals other than the trivial ideal and the improper
ideal.
17.17 Let S be the ring of all matrices of the form (~ :), with a,b,dER. Find a
maximal ideal of S other than the one found in the text, and show that every
proper ideal of S is contained in one of these two maximal ideals.
17.18 Let I(i] be the ring of Gaussian integers (see Exercise 16.24). Let I be the
principal ideal generated by the element 2 + 2i. How many elements are there
in I[i]j I?
17.19 Let R be a commutative ring with unity I ~O.
a) Show that R is a domain iff {OJ is a prime ideal in R.
b) Show that R is a field iff {OJ is a maximal ideal in R.
17.20 Let R be a commutative ring with unity, and let a E R. Show that aR = Riff
a is a unit.
17.21 a) Let R be a ring. Define the center of R to consist of all r E R such that rx = xr
for every x E R. Show that the center of R is a subring of R.
b)Must the center of a ring be an ideal?
17.22 a) Let R be a commutative ring and X a subset of R. Define
Ann(X) = {rERlrx=O for every xEX}.
Ann(X) is called the annihilator of X. Show that Ann(X) is an ideal.
b) In (112′ EB, 0), find Ann({2}).
c) If R were not commutative, then the set Ann(X) defined above would be
called the left annihilator of X. Show that in this case, if X is itself an ideal,
then Ann(X) is still an ideal.
17.23 a) Let R I,R2, … ,Rn be rings with unity. Show that every ideal of
RI EBR2 EB’ .. EBRn is of the form II EBI2 EB· .. EBim where I; is an ideal of
R; for each i.

176 Section 17. Subrings, Ideals, and Quotient Rings
b) Show that this result may fail if the R;’s are not rings with unity.
17.24 Let R b R2, ••• ,Rn be rings with unity. What do the maximal ideals of
Rl EBR2 EB· .. EBRn look like?
17.15 a) Let R be a ring and let I andJ be ideals of R. Show that I nJ is an ideal
of R.
b) Suppose that I and J are prime. Must I nJ be prime?
17.16 a) Give an example of a ring R, an ideal I of R, and an ideal J of I such that
J is not an ideal of R.
b) Show that J must be an ideal of R if it is a prime ideal of l.
17.27 Let R be a commutative ring.
a) Show that the set of nilpotent elements in R forms an ideal.
b) . Show that the quotient of R by this ideal has no nonzero nilpotent elements.
17.28 Let R be a commutative ring, and let I be an ideal of R. Define the radical, VI ,
of I to consist of all r E R such that some power of r is in l.
a) Show that VI is an ideal of R.
b) I is called semiprime if I = VI . Show that I is semiprime iff R/ I has no
nontrivial nilpotent elements.
17.29 Let R be a finite commutative ring with unity. Show that every proper prime
ideal of R is maximal.
17.30 Let R be an integral domain. We call R a principal ideal domain (PID) if every
ideal of R is of the form aR for some a E R. Show that in a PID every nontrivial
proper prime ideal is maximal.
17.31 Let R be a Boolean ring (see Exercise 16.16). Show that every proper prime
ideal of R is maximal.
17.32 Prove that every nontrivial finite subring of a division ring is a division ring.
17.33 a) Let I and J be ideals of R. Define their sum 1+ J by
I+J= {x+ylxEI,YEJ}.
Show that 1+ J is an ideal.
b) Find 6l+ 14l in (l, +, .).
17.34 Let I and J be ideals of R. Define their product IJ to be the set of all finite sums
of the form XIYl + … +xnYn, where each x;EI and eachy;EJ. Show that IJ
is an ideal, and that IJ ~ In J.
17.35 Let I, J, K be ideals of R. Assume that IJ ~ K and K is prime. Show that at
least one of I or J is contained in K.
17.36 Let R be a commutative ring with unity, and let I and J be ideals of R such
that I+J=R. Show that IJ=InJ.
17.37 Let I, J, K be ideals of R. Show that I(J+K)=IJ+IK.

SECTION 18
RING HOMOMORPHISMS
We have seen that group homomorphisms enable us to relate different groups
to each other. Ring homomorphisms do the same thing for rings.
In the context of rings, a homomorphism must be “sensible” with respect
to both operations:
DEFINITION Let Rand S be rings, and let cp: R~S be a function. Then cp is
called a (ring) homomorphism if for every a, b E R we have
i) cp(a+ b)=cp(a)+cp(b) and
ii) cp(ab)=cp(a)cp(b).
Thus a ring homomorphism is in particular a group homomorphism from
(R, +) into (S, +). As such, it has many familiar properties. For instance,
CP(OR)=OS’ and cp(na)=ncp(a) for every aER and nEZ.
A ring homomorphism cp must also preserve products, but in general this
doesn’t have as many consequences as the corresponding fact for sums,
because (R – {OJ, .) and (S – {OJ, .) need not be groups. It is entirely possi-
ble, for instance, that Rand S are both rings with unity, and yet cp(l R) * Is.
Because of this, it is possible that u E R is a unit but cp( u) E S is not.
Examples Let cp: R~R be given by cp(r) = 0 for all r E R. Then cp is a ring
homomorphism, cp(IR)#’ IR’ and cp(r) is never a unit, although every r#’O is a
unit in R.
For a slightly less trivial example, let
cp: R~REBR
be given by cp(r)=(r,O) for every rER. Clearly cp is a ring homomorphism,
but cp(lR)=(l,O), which is not the multiplicative identity in REBR. Again cp
maps all the units in R to nonunits.
177

178 Section 18. Ring Homomorphisms
Many times people who are working solely in the context of rings with
unity tighten up their definition of homomorphism so as to insist that
‘P(lR) = Is· This is nice in that it avoids maps like the ‘P’s in our examples, but
in our general context our definition is probably better, and we will stick to it.
We record some basic properties of ring homomorphisms for reference.
11fEOREM 18.1 Let l, and let Cp:(Z,+,·)~(Zn,$,0) be given by cp(x)=x, the
remainder of x (mod n). Then cp is an onto homomorphism, and ker(cp)=nZ,
so
(Z, +,. )/nZ~(Zn’ $, 0).
The distinct subrings of (Zn’ $, 0) are (dl), … ,(dk ), where dl, … ,dk are
the positive divisors of n (Corollary 5.6). The distinct subrings of (Z, +, .)
that contain nZ are dI Z,d2Z, … ,dk Z, so the correspondence between the
subrings is very transparent in this case.

182 Section 18. Ring Homomorphisms
2. Let cP: lLffilL~lL be given by cp[(m,n)] = n. Then cP is onto and ker(cp) is
the ideal {(m,O)lm ElL). We have
(lLffilL) /ker( cp) ~ lL.
The ideals of lLffilL that contain ker(cp) correspond in an obvious way to the
ideals of lL.
3. Let R be the ring of real-valued functions on Ill, under pointwise
operations. Fix some real number r in mind, and define a mapping CPr: R~1Il
by
CPr(J)=f(r).
CPr is an onto homomorphism, and ker(cpr)=Ir={jERlf(r)=O}. Since
R/ker( CPr) ~ III
and R is a field, Theorem 17.7 tells us that Ir is maximal in R for every r. We
verified this by more heavy-handed means in Example 3 on p. 173.
The proofs of the second and third isomorphism theorems offer no
surprises, so we leave them as easy exercises.
THEOREM 18.6 (Second isomorphism theorem for rings) Let S be a suoring of
R and let I be an ideal. Then S n I is an ideal of S, and
S/(S n 1)~(S+ 1)/ I.
Here S+I is the subring {s+xlsES, xEI} of R.
THEOREM 18.7 (Third isomorphism theorem for rings) Let I and J be ideals of
R and suppose I r:;,J. Then J / I is an ideal of R/ I and
R/I
J/I~R/J.
With these theorems stated, we have now remodeled all our basic
machinery so that it is suitable for use in ring theory. We will proceed to
some new results, each of which involves ring homomorphisms.
We first want to show that every field F has a subfield that is isomorphic
either to Q or to (lLp, ffi, 0) for some prime p. In fact, more is true: F has
exactly one such subfield. We are interested in this result because it provides
a useful classification of fields. The nature of the solutions of equations
involving elements of F can be heavily influenced by whether it is Q, on the
one hand, or some lLp, on the other, which is isomorphic to a subfield of F.
THEOREM 18.8 Let F be a field. Then the intersection of all the sub fields of F is
itself a sub field of F, and is isomorphic either to Ql or to Zp for some prime p. This
is the only subfield of F that is isomorphic either to Ql or to some Zp.

Section 18. Ring Homomorphisms 183
PROOF. Let K be the intersection of all the subfields of F. As in Exercise 5.14 we
see that K is a subfield of F. K is contained in every subfield of F, and the
multiplicative identity element e of Fis in K. (We use “e” instead of “1” or “II’
for clarity in what follows.) We consider two cases depending on the order o(e)
of e in the group (F, +).
Case 1: If o(e) is infinite then S = {me(netl I m, nEZ, n:f:. O} !;; K. It is easy to
verify that S is a subfield of F, so we have K !;;; S and therefore K = S. The
mapping rp: Q -+K given by rp (mIn) = me (netl is a well-defined isomorphism.
Case 2: If o(e) = n E Z+ then it follows from Exercise 16.28 that n is a prime
number p. If S = {OF> e, 2e, … , (p – 1) e} then S!;;; K. S is a subring of F isomorphic
to the field Zp , so S is a subfield of F and thus K!;;; S. So K = S and K == Zp.
To prove the uniqueness assertion, suppose L is a subfield of F and If/: L -+ Q
is an isomorphism. Then If/(e) = 1, so since Q = {m I n I m, nEZ, n:f:. O},
L = {me (netll m, nEZ, n:f:. O} = K
(because o(e) is infinite in (F, +». Likewise, if L == Zp for some prime p then L =
{OF> e, 2e, … , (P-l) e} = K 0
The subfield of F that is isomorphic to Q or some l.p is called the prime
subfield of F. If F is itself either Q or some l.p’ then F is its own prime
sub field, and Q and the l./s are accordingly called prime fields.
If the prime subfield of F is isomorphic to Q, equivalently if 1 F has
infinite order in (F, +), we say that F is of characteristic O. Thus Q, R, and C
are all of characteristic O. If the prime subfield of F is isomorphic to l.p’ we
say that F is of characteristic p. It is clear, for example, that every finite field
must be of characteristic p for some prime p.
Our next two results are examples of what are called embedding theorems,
in that they state that a ring of some kind can be embedded in a ring of some
other kind. To say that R can be embedded into S is simply to say that there
is an embedding (monomorphism) from R into S, that is, R is isomorphic to a
subring of S. The usual purpose of an embedding theorem is to remedy some
defect of a given ring by embedding the ring in a bigger ring, where things are
better. For example, we know that some rings lack multiplicative identities.
But:
TIIEOREM 18.9 Let R be a ring. Then R can be embedded in a ring with unity.
PROOF. Let S={(r,n)lrER, nEl.}, with operations defined by
(r,n) +(s,m) =(r+s,n+ m)
and
(r,n)(s,m)= (rs+ ns+ mr,nm).

184 Section 18. Ring Homomorphisms
It is clear that S forms an abelian group under addition, and a routine
calculation shows that multiplication is associative and the distributive laws
hold. Thus S is a ring.
We map R~S by x) to mean thaty – x E F+. Show that, for all x,y,z E F:

190 Section 18. Ring Homomorphisms
a) exactly one of the following holds:
xy;
b) if xO andy >0, then x+y>O and xy>O;
d) if x 0, then xz yz.

SECTION 19
POL YNOMIALS
We have all been familiar with polynomials since our days in high school
algebra. We are by now accustomed to thinking of them as functions of the
formf(x)=ao+a\x+'” +anxn, with x being a variable and the a;’s being
real constants. In this section we will take another look at them, from the
point of view of abstract algebra.
We ‘Shall denote variables by upper-case letters: X, Y, … . If R is a ring,
then by a polynomial in X with coefficients from R we mean an infinite formal
symbol
Qo+a\X+a2X 2 +a3X 3 + … ,
where each ajER and there is some n such that aj=O for all i>n. The a/s are
called the coefficients of the polynomial, and aj is called the coefficient of xj.
If an*O and aj=O for all i>n, then we usually write the above polynomial
more simply as
but we still regard it as having a coefficient aj for every j > O. This approach is
often very convenient. For instance, it makes it easy for us to say what we
mean by two polynomials being equal. If
and
are polynomials with coefficients from R, then we say that they are equal, and
we write f(X) = g(X), if aj = hj for every i. Notice that we don’t have to worry
about one of f(X) or g(X) having more coefficients than the other (for
instance, extra O-coefficients) because we regard them both as having a
coefficient for each power of X.
191

192 Section 19. Polynomials
Observe that a polynomial f(X) with coefficients from R gives us a
function on R, obtained by plugging elements of R in for X and interpreting
addition and multiplication as the operations in R. However, our point of
view at the moment is that f(X) is a formal expression, and not the function
this expression induces on R. This distinction is a significant one, because it is
possible for two different polynomials to induce the same function on R. For
example, if R is (.l2,E9,O), then both 0+OX+OX 2+”. and 0+lX+IX2
give us the function that maps every element of R to O.
In denoting polynomials, we usually omit terms with coefficient 0,
wherever they occur. For instance, ao+OX + a2X2 can be written as ao+
a2X 2• On the other hand, if R is a ring with unity, then we usually do not
bother to write I R when it occurs as a coefficient. Thus if we are considering
polynomials with coefficients from .l2′ then we write X + X2 in place of
IX+ lX2.
We denote the set of all polynomials in X with coefficients from R by
R[X]. The elements of R[X] are also sometimes called polynomials over R.
We turn R[X] into a ring by introducing the natural addition and
multiplication. If
and
then we addf(X) and g(X) by adding corresponding coefficients:
f(X) + g(X) = (ao+ bo)+ (a l + bl)X + (a2+b2)X2+ ….
We multiply just as we did in high school algebra: f(X)g(X) = co+ c\X +
C2X 2+ … , where for each n,
cn = aobn + a\bn_\ + a2bn-2 + … + an_\b\ + anbO’
Thus the product is obtained by multiplying everything out, using the rule
aXj·bXj = abXi+j, and then collecting terms that involve the same power of
X. The product is indeed a polynomial over R, because there exists n such
that cj=O for every i>n. Specifically, if aj=O for all i>m and bj=O for all
i > I, then we can take n = m + I. For with this choice of n, each term ajbj in
the expression for ck , k>n, must have i+ j>m+ I, so either i>m or j>l, and
therefore either aj = 0 or bj = 0, whence ajbj = O.
Observe that, under these definitions, a polynomial such as ao+a\X +
a2X 2 is actually the sum of ao’ a1X, and a2X 2 in R[X]. By the same token, if
R is a ring with unity, so that X ER[X], then a2X 2 is the product of a2X and
X.
It is not very difficult (but neither is it very exciting) to verify that R[X]
does form a ring under the given operations. We shall forego getting into this,

Section 19. Polynomials 193
and instead take up the more interesting question of which properties of R
carry over to the new ring R[X].
Some things can be seen at once. For instance, if R has unity 1, then
R[X] has unity 1+OX+OX2+ … , which we also denote by l. If R is
commutative, then so is R[X], for if
j(X)=ao+a\X+a2X 2 +… and
then the coefficient of Xn inj(X)g(X) is
aobn+a\bn_\+··· +an_\b\+anbO’
and the coefficient of xn in g(X)j(X) is
boan+b\an_\+'” +bn_\a\+bnao’
These are the same, by the commutativity of multiplication in R.
What if R is a domain? Must R[X] be one too? The answer is yes,
because if j(X) and g(X) are as in the last paragraph, and neither one is the
zero polynomial
0+OX+OX 2 + … ,
then we can let n be such that an*O and a;=O for all i>n, and let m be such
that bm*O and b;=O for all i>m. It then follows that the coefficient of xn+m
in j(X)g(X) is anbm, which is not ° since an,bm are nonzero elements of a
domain. Thusj(X)g(X) is not the zero polynomial, and it follows that R[X]
is a domain.
Arguments such as the one we have just given can be expressed more
succinctly if we allow ourselves some additional terminology. If j(X)ER[X]
is not the zero polynomial, then we can write
j(X)=ao+a\X+··· +anXn, where an *0.
The integer n is called the degree of f(x), and an is called the leading coeffi-
cient. We denote the degree off(x) by deg (j), or deg (f(X). Notice that deg (j)
= ° ifff(x) is a nonzero constant polynominal, that is, ifff(x) is the polynomi-
nal ao. for some nonzero ao E R. The zero polynominal is not assigned a degree.t
The argument we gave above can now be expressed more precisely by
saying that if R is a domain, andj(X),g(X)ER[X] have leading coefficients
an and bm, respectively, then j(X)g(X) has leading coefficient anbm. As a
consequence, we’ have
“tNo confusion should result from using the same symbol to denote both “0 E R and the constant
polynomial “0. In fact, R is isomorphic to the subring of R[X] consisting of all the constant
polynomials. and we often think of R itself as a subring of R[X].

194 Section 19. Polynomials
THEOREM 19.1 (Degree rule) If R is a domain and J(X),g(X) are nonzero
elements of R[X], then
deg(f(X)g(X) = deg(f(X) + deg(g(X).
The assumption that R is a domain is crucial here. For instance, in the
nondomain 1.6, we have (2X)(3X + 1)=2X, or, even worse, (2X)(3X +3)=0.
We have seen above that a number of properties will always be passed on
from R to R[X). One property that will obviously not be passed on is that of
being a field. For if F is a field, then the element X is clearly not a unit in
F[X].
Nevertheless, it does seem reasonable to expect that assuming F to be a
field will have a beneficial impact on F[X], beyond that of making it a
domain. Much of what can be said about F[X] depends on the following
analogue of the division algorithm for 1..
THEOREM 19.2 (Division algorithm for F[ X]) Let F be a field, and let
J(X),g(X)EF[X]. If g(X) *0, then there exist q(X),r(X)EF[X] such that
J(X) = q(X)g(X) + r(X)
and either r(X)=O or deg(r)< deg(g). PROOF. If J(X) = 0, or if J(X) '1= ° and deg(f) < deg( g), we write J(X)=O·g(X)+ J(X), and we are done. We now proceed by induction on deg(f). If deg(f) = 0, then by the above we are done unless deg(g) = 0. But in this case both J(X) and g(X) are constant polynomials-say, J(X) = ao, g(X) = bo-and thus we can write J(X) = (aobo-l)g(X) + 0. Note that bO- 1 exists because F is a field and bo'l=O. Now assume the result has been proved for deg(f) 0, so J(X) has at most n
roots and cannot have all the elements of S as roots. 0
COROLLARY 19.6 Let F be an infinite field, S an infinite subset of F. Suppose
J(X),g(X) E F[X] and J(s) = g(s) for every s E S. Then J(X) = g(X), that is,
J(X) and g(X) are the same polynomial.
PROOF. Apply Corollary 19.5 to the polynomiaIJ(X)- g(X). 0
Examples
1. Two polynomials in R[X] induce the same function on R iff they are
the same polynomial.
2. Corollary 19.4 may fail if F is not a field. For instance, let R = Z2 Ea Z2′
and letJ(X)=X2+XER[X]. Then deg(f)=2, but each of the four elements
of R is a root of J(X).
Corollary 19.4 (and, with it, 19.5 and 19.6) does hold true for any domain
R, however. See Exercise 19.16.
3. Corollary 19.4 says that if deg(f)=n, thenJ(X) has at most n roots in
F. It may have fewer; indeed, it may have none. For example, the polynomial
X2+ 1 in R[X] has no roots in R
The problem of trying to determine the roots of a polynomial in R[X] is
one that we all encountered in our earliest experience with polynomials. One
of the standard methods for dealing with this problem is, of course, to write
the given polynomial as a product of simpler factors, and then to find the
roots of all the factors. You may remember that in R[X], every nonconstant
polynomial can be written as a product of factors that either have degree 1, or
have degree 2 and are irreducible in the sense that they cannot be factored
any further (Exercise 19.5).
DEFINITION Let F be a field. A nonconstant polynomialJ(X)EF[X] is called
irreducible in F[ X] (or irreducible over F) if J cannot be written as the product
of two nonconstant polynomials in F[X].
In general, F[X] may contain irreducible polynomials of degree higher
than 2. It is still true, however, that every nonconstant polynomial can be
written as a product of irreducible factors.

Section 19. Polynomials 197
THEOREM 19.7 Let F be a field, and let j(X) be a nonconstant polynomial in
F[X]. Then there exist irreducible polynomials jJ(X)”” ,Jk(X) in F[X] such
that j(X) = jJ(X)jzk orj>l.
Since p does not divide akbl but does divide all the other terms, p does not
divide Ck+I’ and Gauss’ Lemma is proved. 0
We put it to work:
LEMMA 19.10 Let f(X)EZ[X]. If f(X) can be written as the product of two
nonconstant polynomials in O[X], thenf(X) can be written as the product of
two nonconstant polynomials in Z[X].
PROOF. As in the discussion preceding the statement of Gauss’ Lemma,
suppose g,h EO[X] are nonconstant polynomials such that
f(X) = g(X)h(X),
and write g(X)=(I/ c)g·(X), h(X)=(l/ d)h·(X), with g·,h· EZ[X]. We can
now write
g·(X) = content (g.) ‘g\(X) and h·(X)=content(h·) ·h\(X),
with g\ and h\ primitive. Therefore,
f(X)= content(g.~~ontent(h.) g\(X)h\(X),
and since g\ and h\ have the same degrees as g and h, we will be done if we
can show that the constant factor on the right-hand side is in Z. Now
cd·f(X) = content( g·)content( h·)g \(X)h\(X),
so, taking contents on both sides,
Icdl’ content(j) =content( g·)content(h·)content(g\h\).
By Gauss’ Lemma, content (g\h\)= I, so we obtain
(f)
– content( g·)content( h·)
content – Icdl
Since content(f) E Z, we have what we want. 0
We now reap the first benefit of the preceding discussion by stating a
criterion for irreducibility in O[X] due to Ferdinand Eisenstein (1823-1852).

200 Section 19. Polynomials
Eisenstein was a prize pupil of Gauss, and his result is very much in the spirit
of Gauss’ Lemma.
lHEOREM 19.11 (The Eisenstein Criterion) Let f(X) = ao + a\X + … + anxn E
l[X), and suppose p is a prime such that
plao,pla\, … ,plan_\,p.{an, and p2.{ao.
Thenf(X) is irreducible in Q[X).
PROOF. If not, then by Lemma 19.10 there are nonconstant g(X),h(X) El[X)
such that f(X) = g(X)h(X). Say
with 1 1, deg(h);;> 1. We thus obtain
j(X) = g(X)h(X) (see Exercise 19.13), and since deg(h=deg(f) we must have
deg(g)= deg(g) >1, deg(h)=deg(h);;> 1. Thus AX) is reducible in 7l.p [X],
contrary to our assumption. 0
Example The polynomial X 3 + 2X + 20 is irreducible in Q[ X], because if we
reduce its coefficients mod 3, we get X 3 + 2X + 2 E 7l.3[X], which has the same
degree and is irreducible by Theorem 19.8. Note that if we reduce the
coefficients mod 2, we get X 3 E 7l.2[X], which is reducible. Thus if we reduce
mod some prime and get a reducible polynomial, it does not follow that the
original polynomial is reducible.
EXERCISES
19.1 Letf(X)=ao+ a1X + … +a,X’ EZ[X]. Suppose mlnEQ, with (m,n)= 1.
Show that if min is a root of f(X), then mlao and nla,.
19.2 Determine which of the following are irreducible in Q[X].
a) X 3+X+36
b) 2X3_8X2-6X+20

202 Section 19. Polynomials
c) 2X4+3X3+ 15X +6
d) X4+2X3+X2+X+l
e) XS+14X 2+4X+6
f) X4+X3+X2+X + 1 [This is (X s-l)/(X-l). Recall how we dealt with
X4+ 1.]
g)x”-3XZ+6X+l
h) X -XZ+ 1
19.3 Write each polynomial as a product of irreducible polynomials over the given
field.
a) 2X3+X2+2, over I3
b) X 3+3X 2+X+4, over Is
c) X2+5, over I7
d) X4+X3+2X2+X+2, over I3
e) X5+X2-X-l, over I2
19.4 a) Suppose thatf(x) E JR[X] and c = a + bi E C is a root ofj{X). Show that the
complex conjugate c = a – bi of c is also a root ofj{X).
b) The Fundamental Theorem of Algebra asserts that every nonconstantj{X) E C[X]
has a factorization in C[X] of the form
Assuming this, show that every nonconstantf(X) E JR[X] can be factored in JR[X]
as a product of irreducible polynomials of degree at most 2.
19.5 Suppose F is a field and a(), ai, … , an E F. Let
j{X) = a”x’ + … + atX + ao
and
Prove thatj{X) is irreducible over F iff g(X) is irreducible over F.
19.6 Letj(X) = aX2+ bX + c E IR[X], a,..,O. Show thatj(X) is irreducible in R[X]
iff b2 – 4ac < O. 19.7 Let a E I +. Show that X4 + a is reducible in Q[X] iff a = 4b4 for some integer b. 19.8 Letj(X),g(X) be nonzero polynomials in I[X]. Show that content(fg) = content(f)· content( g). 19.9 Show that the polynomials q(X) and r(X) in Theorem 19.2 are uniquely determined by j(X) and g(X). Section 19. Polynomials 203 19.10 Let F be a field. a) What are the units in F[Xl? b) Show that if cEF, c~O, then for anyj(X)EF[Xl,j(X) and cj(X) generate the same principal ideal in F[Xl. 19.11 Let R be a ring. Verify the following for R[Xl: a) the left distributive law; b) associativity of multiplication. 19.12 Let R be a commutative ring, rER,j(X),g(X)ER[Xl. Let h(X)= j(X) + g(X) and k(X)= j(X)g(X). Show that h(r) = j(r) + g(r) and k(r) = j(r)g(r). Thus the mapping (fir: R[Xl~R given by (fIr(J(X» = j(r) is a homomorphism. It is called an evaluation homomorphism. 19.13 LetR and Sbe rings and let cp : R ~ Sbe a homomorphism. Forf(X) E R[X], letf"(X) E S [X] denote the polynomial obtained by replacing each coefficient aj off(X) by cp(aj). Show that the mapping R[X] -+ S[X] given by f ~ f" is a homomorphism. 19.14 Give another proof of Eisenstein's Criterion, by considering the homomorphism Z[Xl~Zp[Xl obtained by reducing all the coefficients of polynomials in Z[Xl modp. 19.15 Let F be a field, let bb ... ,bn +) be n+ 1 distinct elements of F, and let c), ... ,cn +) be n + 1 elements of F (not necessarily distinct). a) Find a polynomialj(X)EF[Xl such thatj(X)=O or deg(f) I and X – a does not divide g(X).
a) Show that m is uniquely determined by f and a. That is, if we also have
f(X)=(X – a)’h(X), where X-a does not divide h(X), then r= m.
b) If m:> 2, we say that a is a multiple root of f . Show that a is a multiple root
of f iff 1′(a)=O. (See Exercise 19.17.)
19.19 Let F be a finite field with q elements.
a) Show that a q – I = I for every a*O in F.
b) Letf(X)EF[X]. Show that there exists a polynomialr(X)EF[X] such that
either r =0 or deg(r) deg(f).
So let S={deg(g)lgEI and g*O}. Since S is a nonempty set of non-
negative integers, it has a smallest element, n. Choose J(X) in I such that
deg(f) = n. It is clear that (f(X»!: I, and to establish the reverse inclusion we
use the division algorithm for F[X].
Let g(X)EI, and write g(X)= J(X)q(X) + r(X), where either r(X) =0 or
deg(r) 1. Then there is an extension field K of F that contains a root of f(X).
PROOF. We can assume thatf(X) is irreducible in F[X), because if it isn’t we
can work with one of its irreducible factors.
Under this assumption, F[X]/U(X» is a field, in which f(X)=O. If
f(X)=ao+aIX+··· +anXn, this means that
iio+iiIX+·.· + iinxn =0. [20.1)
We have an embedding ‘P:F~F[X]/(f(X», given by ‘P(a) = ii, for every
a E F. ‘P is clearly a homomorphism, and it is one-to-one since if aI’ a2 E F and

208 Section 20. From Polynomials to Fields
al=a2, thenf(X) divides al -a2 in F[X]. This means that al-a2=0, since
deg(f) > 1, so a l = a2•
Thus { ala E F} is a subfield of F[X]/(f(X», isomorphic to F. If we
replace each element a by a, we obtain a field K that extends F, in which
ao+aIX+· · · +anXn=O, by [20.1]. 0
Theorem 20.3 is the key to an in-depth analysis of the roots of polynomi-
als, because it tells us we can always get our hands on a complete set of roots
to work with:
COROLLARY 20.4 Let f(X) E F[X] have degree n> 1. Then there is a field
K d F such that in K[X] we can write f(X) = a(X – c1)(X – c~· .. (X – cn)’
PROOF. By induction on n.For n= 1 we have f(X)=aX+b, with a,bEF,
a*O, so we can writef(X)=a(X-(-b/a» in F[X]. Now suppose deg(f)=
m and the result is proved for polynomials of degree m – 1. By Theorem 20.3,
we can let KI d F be a field containing a root c1 of f(X). In K1[X], we can
write f(X) = (X – cl)g(X), where g(X) has degree m – 1 and is therefore
subject to the inductive hypothesis. Let K d KI be such that in K[X] we have
g{X) = a{X – C2)’ .. (X – cn)’
Then K is an extension of F, and we have
f{X) = a{X – cl)·· . (X – cn)
in K[X]. 0
Of course, some of c1,,, , ,cn may already lie in F, and c1″ “,cn need not
be distinct.
Example Let f(X) = X4- X 3_ X + 1 EIs[X]. Then 1 is a root of f(X), and
long division yields
f{X)={X -1){X3- 1)
in Is[X]. Since 1 is again a root of X 3 – 1, we obtain
f{X) = (X -1){X-I){X2+ X + I).
The factor X2+ X + 1 has no roots in Is, but we know there is a root C in
some extension K of Is. We divide X2+ X + 1 by X – c in K[X]:
X +(c+l)
X+l
cX
(c+ l)X + 1
(c + l)X – c( C + 1)
c( C + I) + 1 = c 2 + C + 1 = 0.

Section 20. From Polynomials to Fields 209
Thus, in K[X),
f(X)= (X -1)(X -1)(X – c)(X – (- c-l».
EXEROSES
20.1 Letp be a prime. Show that lp[Xl!(X 2 + 1) is a field iff the equation x 2=-1
has no solution (modp).
20.2 Is O[X]/«X – 1)2) a domain?
20.3 Is O[XJ!(X 3 +2X +2) a field? How about IR[XJ!(X 3 +2X +2)?
20.4 LetK= {O, 1,X,X + I} be the four-element field constructed in Example 1 on
pp. 206-207. Write ~ + X + 1 as a product of factors of degree 1 in K[X].
2O.S The elements of the field K in Exercise 20.4 can each be written uniquely in
the form a+bX, with a,bEl2• Find a general rule for writing the product
(a + bX)( c + dX) in this form.
20.6 a) Let f(X) be irreducible in F[X], and let K be the field obtained from
F[XJ!(f(X» by replacing a by a, for each a EF. Show that if deg(f) = n,
then every element of K has a unique representation in the form
ao+alX + … +an_IXn-t,
with ajEF.
b) Show that if we start with F = lp and deg(f) = n, then the field K in part
(a) has p n elements.
20.7 Use the result of Exercise 20.6 to construct a field with m elements, for m=
a) 8;
b) 9;
c) 27;
d) 25;
e) 125.
20.8 Let F be a field. What are the prime ideals in F[X]?
20.9 Let n;;. 1, letp be a prime, and letf(X)=XP· – X Elp[X]. By Corollary 20.4,
let K be an extension of lp such that in K[X] we have
f(X)=(X – c l )”, (X -cp.).
a) Show that in this case CI>””cp • are all distinct. (Use Exercise 19.18.)
b) Show that {CI,’ .. ,cp .} is a subfield of K, and hence that there exists a field
with p n elements.
20.10 IfF is a field and.f{X), g(X) are elements of F[X] then a polynomial h(X) E F[X] is
called a greatest common divisor (g.c.d) of.f{X) and g(X) if
i) h(X) divides both.f{X) and g(X), and
ii) every k(X) in F[X] that divides both.f{X) and g(X) divides h(X).

210 Section 20. From Polynomials to Fields
a) By applying Theorem 20.1 to the ideal
/= {a(X)f{X) + b(X)g(X) I a(X), b(X) E FIXJ},
prove that any two polynomialsf{.x), g(.x) in FIX] have a g.c.d. that can be written
as a(X}f{.x)+b(.x)g(.x) for some a(.x), b(.x) E FIX].
b) Prove that if at least one of f{.x) or g(X) is not the zero polynomial then any two
g.c.d.’s off{.x) andg(.x) differ by a constant factor.
20.11 Let F be a field, let a E F, and let p be a prime. Let fiX) = ~ -a. Prove that f (.x) is
irreducible in FIX] if and only iff{X) has no roots in F.
20.12 Let F be a field of prime characteristic p, let a E F, and let n E Z+. Prove that X pn –
a is irreducible in FIX] if and only if ~ – a has no roots in F.

SECTION 21
UNIQUE FACTORIZATION
DOMAINS
We have called a positive integer p a prime if p~ I and the only positive
divisors of p are I and p itself. This notion of “prime” generalizes in a natural
way to allow negative primes: an integer n is called prime if n~O, n~ ± I,
and the only divisors of n are ± I, ± n. Thus 2, – 2, 3, – 3,5, – 5,7,
– 7, II, – II… are all primes in lL..
In these terms, the Fundamental Theorem of Arithmetic may be taken as
the statement that every integer n which is neither zero nor ± I can be
factored into a product of primes, and that if
n=PIP2″‘Pr and n=QIQ2″‘qs
are two such factorizations, then r = s and, after rearranging the q;’s if
necessary, we have Pj = ± qj. For example, we have – 60 = (- 2)(2)(3)(5) =
( – 3)(2)( – 5)( – 2), and if we rearrange the second factorization it becomes
( – 2)(2)( – 3)( – 5).
The Fundamental Theorem of Arithmetic seems obvious to most of us,
because we all grew up with it in school. For this reason, it is easy to get
lulled into thinking that the theorem doesn’t require any proof. One purpose
of this section is to convince you that it does require proof by showing you
some integral domains for which the analogue of the Fundamental Theorem
is false. With such examples in hand, it becomes interesting to try to
enunciate some conditions on a domain which will guarantee that the
analogue of the Fundamental Theorem does hold.
The necessity for such an investigation is underscored by the fact that
some notable mathematicians have fallen into the trap of assuming that the
Fundamental Theorem holds in cases where in fact it does not. Efforts to
salvage the results which they based on these faulty assumptions have been
largely responsible for the development of ring theory as we know it today.
t This section can be omitted without loss of continuity.
211

212 Section 21. Unique Factorization Domains
Before we recount a specific instance, let us establish some vocabulary, so
that we can say things a bit more precisely.
Let D be a domain. (We concentrate on domains because domains are
the natural generalization of 7L.) An element dE D is called irreducible if d is
neither 0 D nor a unit, and whenever a, bED and d = ab, then either a or b is a
unit.
Examples
1. Since the units in 7L are ± 1, the irreducible elements of 7L are precisely
the primes.
2. If F is a field, the units in F[X) are the nonzero elements of F. Thus
the irreducible elements of F[X) are the nonconstant polynomials that are
irreducible according to the meaning of the word in Section 19.
3. We claim that a nonconstant polynomial J(X) E7L[X) is irreducible in
7L[X) iff J(X) is primitive and cannot be written as the product of two
nonconstant polynomials in 7L[X).
First of all, if J is irreducible then J must be primitive, for otherwise we
could write J(X)=p·g(X) for some prime p and nonconstant g(X), and
neither p nor g(X) is a unit in 7L[X]. Likewise, J cannot be written as the
product of two nonconstant polynomials.
Conversely, suppose that J has the two indicated properties. Then if
J(X)=g(X)h(X) in 7L[X), either g or h must be a constant; since J is
primitive, this constant must be ± 1, so it is a unit in 7L[X). Thus J is
irreducible.
It is easy to see that a constant is irreducible in 7L[X) iff it is a prime in 7L,
and therefore the irreducible elements of 7L[X) are the primes of 7L and the
nonconstant irreducibles described above.
In light of Example 2, our use of the term “irreducible element” seems
very natural; but in terms of Example 1, “prime element” might seem more
appropriate. In general, the word “prime” is used in connection with a
concept that is sometimes stronger than irreducibility. If D is a domain, and
x,y ED, then we say that x dividesy, and we write xly, if there is some zED
such that y = xz. An element d ED is called prime if d is neither 0 D nor a unit
and whenever a,bED and dlab, then dla or dlb.
Prime elements are always irreducible, in any integral domain:
THEOREM 21.1 Let D be a domain, and let d E D be prime. Then d is
irreducible.

Section 21. Unique Factorization Domains 213
PROOF. Suppose d=ab; we must show that either a or b is a unit. Since d is
prime, we know that d divides either a or b. If dla, then we have a = de for
some e, and so
d=(de)b.
Since d=f’O and D is a domain, we conclude that 1= eb and b is a unit.
Similarly, if dlb then a is a unit. 0
For (Z, +, .), the notions of “prime” and “irreducible” coincide (Theo-
rem 4.3) and in general the truth of the implication “irreducible ~ prime” for
a domain D is very intimately connected with the truth of an analogue of the
Fundamental Theorem for D (see Exercise 21.9). We will soon see examples
of irreducibles that are not prime.
Two elements a and b of a domain D are said to be associates if a = bu
for some unit u. For example, any element a is an associate of itself, because
a = a . 1. Similarly, if a and b are associates, then b and a are associates,
because a = bu implies b = au – ‘. The relation
aRb iff a and b are associates
is in fact an equivalence relation; transitivity is easily verified.
We now have the words we need in order to say precisely what it means
for an analogue of the Fundamental Theorem to hold for D.
DEFINITION Let D be an integral domain. Then D is called a unique factoriza-
tion domain (UFD) if:
i) Every element d ED that is neither 0 nor a unit can be factored as the
product of a finite number of irreducible elements; and
ii) If d=P,P2·· ·Pr and d=Q,q2··· Qs are two such factorizations, then r=s,
and there is a permutation f of {1, 2, … , s} such that Pi and Qf(i) are
associates, for each iE{1,2, … ,s}.
Note that this is a direct generalization of the statement of the Funda-
mental Theorem that we gave on p. 211, because in Z, two elements a and b
are associates iff a = ± b.
Now for a bit of history. We all know that there are triples x, y, z of
nonzero integers such that x2 + y2 = z2; for instance, 32 + 42 = 52. In 1637,
Pierre de Fermat claimed to have discovered a “truly remarkable proof’
that if n > 2 then there do not exist nonzero integers such that xn + yn = zn.
This result is referred to as Fermat’s Last Theorem, although Fermat
never revealed a proof and there is a great deal of doubt as to whether he
ever had one. Mathematicians sought a proof for over 350 years, until a
proof was finally published by Andrew Wiles in 1995.

214 Section 21. Unique Factorization Domains
The problem is very easily reduced to showing that x n + y n = Z n is
impossible (for nonzero x,y,z) if n is 4 or an odd prime. (It is fun to do this
for yourself, and we won’t spoil it by showing you how.) The case n=4 can
be disposed of by elementary means, and we know that Fermat did prove this
case himself. In any event, the general problem boils down to proving that
x P + yP = zP is impossible if p is an odd prime and xy z =1= O.
A proof for p = 3 was published in 1770 by Leonhard Euler, the most
prolific mathematician of all time. Euler’s proof involved using numbers of
the form a+bv=:3, where a,bElL and v=:3 =V3i. At one point in his
argument, he made a claim about these numbers which was apparently based
on the tacit assumption that they obey unique factorization. His claim was
correct, but the tacit assumption behind it was not, and his proof remained
incomplete until the missing justification was supplied by Legendre some time
later.
A proof for p = 5 was given by Legendre, and independently by Dirichlet,
around 1825. The casep=7 was handled by Lame in 1840. The first general
-and by far the most significant-attack on the problem was made by E.
Kummer in 1843. Kummer’s basic idea was to consider numbers of the form
where a; E lL and ~P is a complex number =1= I such that ~: = 1. (For example,
~P = cos(2’17″/ p) + i sin(2?T / p ).”t) These numbers form a subring of C, which we
denote by lL[~p]. Using them, it is possible to factor x P + yP completely, and
the equation x P + yP = zP becomes
Assuming that lL[~p] is a UFD, Kummer used this form of the equation to
prove that x P + yP = zP is impossible if xyz =1=0.
Kummer presented his proof to Dirichlet (a more established mathemati-
cian), who pointed out that Kummer had neglected to verify the assumption
that factorization into irreducibles is unique in lL[~p]’ (Kummer was later to
point out a similar flaw in an attempt by Lame.) In 1847, Cauchy (after
having made the same mistake himself) pointed out that factorization is not
unique in lL[~23]’ Thus Fermat’s Last Theorem remained unproved.
1’That [cos(2’IT /p)+ isin(2’IT /p)Y’ = 1 follows from De Moivre’s Theorem:
[cos8+ isin8t = cosk8 + isink8.
De Moivre’s Theorem can be proved by induction on k, using the fact that
[cos a + i sin aJ[cos {:I + isin{:l] = (cos a cos{:l- sin a sin{:l) + i(sina cos {:I + cos a sin,B)
= cos(a + {:I) + isin(a+ {:I).

Section 21. Unique Factorization Domains 215
Undaunted, Kummer set about trying to modify Z[~p] so as to restore
the uniqueness of factorization. He introduced what he called ideal
numbers, and the theory he developed was a precursor of the modern theory
of ideals. Kummer succeeded in proving Fermat’s Last Theorem for certain
primes which he called regular.
The complete proof that Wiles published in 1995 employed modern and
sophisticated ideas far beyond those known to Kummer (and far beyond the
scope of this book). But Kummer’s method had a lasting impact on the
development of algebra. Attempts by number theorists to exploit the
properties of specific systems more inclusive than Z led naturally to a
study of such systems in general, hence to the emergence of an abstract
theory of rings .
After we have developed some more information about UFDs, we will
discuss in detail the application of another extended number system (the ring
of Gaussian integers) to the proof of Fermat’s classic “two squares” theorem.
But first we want to show you an explicit example of the failure of unique
factorization in a fairly simple domain.
In fact, let us use the set of all complex numbers of the form a + b V-3 ,
with a, bEl, referred to above. If we denote this set by l[V-3], then
l[v=3] is a subring of C containing 1, hence an integral domain. In
l[v=3] we have the following two factorizations of the number 4:
4=2·2=(1 + V-3 )(1- V-3).
We are going to show that 2, 1 + V-3 , and 1- V-3 are all irreducible in
l[v=3] and that no two of them are associates. This will establish that
l[v=3] is not a UFD.
For a=a+bV-3 , define the norm N(a) of a to be N(a)=a 2 +3b2• For
every a El[v=3 ], N(a) is a nonnegative integer, and N(a)=O iff a=O. An
easy calculation shows that if a, {3 E l[ v=3 ], then
N(<<{3) = N(a)N( {3). Using norms, we can readily determine all the units in l[V-3]. For if a is a unit, then a{3 = 1 for some {3, whence N(a)N({3)=N(I)= 1, which implies that N(a) = 1, since N(a) and N({3) are both nonnegative integers. Conversely, if N(a) = 1 and, say, a=a+bV-3, then a2 +3b2 = 1, which implies that a = ± 1 and b =0. Thus a = ± 1, and a is a unit. Thus we see that a is a unit iff N(a) = 1, and the only units in l[V-3] are ± 1. Now we claim that 2 is irreducible in l[ V-3 ]. Clearly, 2 is not 0, and it is not a unit since 2* ± 1. We must show that if 2= a{3, then either a or {3 must be a unit, that is, either N( a) or N( {3) must be 1. Now if 2 = a{3, then 216 Section 21. Unique Factorization Domains N(2)=N(a)N(f3) 4=N(a)N(f3), so all we have to show is that neither N(a) nor N(f3) can be 2. But the norm of any element of Z [-v=3 ] has the form a2 + 3b2, and it is clear that this never gives us 2. The same argument shows that both 1 + -v=3 and 1 - v=-3 are irre- ducible, because their norms are both 4. No two of our three irreducible elements are associates, because the only units are ± 1. Thus Z[ -v=3 ] is not a UFO. Notice a couple of other interesting things. The equation 2·2=(1 + -v=3 )(1- -v=3) shows that 2 divides the product (I + -v=3 )(1- -v=3), but it obviously does not divide either factor. Thus 2 is irreducible, but not prime. The same can be said for I + -v=3 and I - -v=3 . Observe that if a EZ[-v=3 ] and N(a) is a prime integer, then a is irreducible. For if a=f3y, then N(a)=N(f3)N(y), so one of N(f3), N(y) is 1. The converse is false; for example, 2 is irreducible, but N(2) = 4 is not a prime integer. By considering norms, it can be shown that every element of Z[-v=3] which is neither ° nor a unit can be factored into a product of irreducibles (Exercise 21.5), and thus it is only the uniqueness of factorization that fails, not the existence. On the other hand, it is not true that every element *0, ± 1 can be written as a product of prime elements. For instance, 2 is not a prime, and if then N(2) = 4 = N( al)N( a2)' .. N( ak)' so some lX; has norm 4 and the others are units, not primes. We have not yet exhibited a prime in Z[-v=3]. An example is given in Exercise 21.22. Under what conditions might it be impossible to factor an element into irreducible factors? Suppose D is a domain, a E D is neither ° nor a unit, and a cannot be written as the product of a finite number of irreducible elements. In particular, then, a is not itself irreducible, so we can write a = alb l, where neither al nor bl is a unit. Furthermore, our assumptions on a imply that a l and b l cannot both be written as products of irreducible elements. Say b l cannot be so written. We have a = albl; al,bl are not units; b l is not a product of irreducible elements. [21.1] Section 21. Unique Factorization Domains 217 Now bl has the same properties that a had when we started, so we can write b l = a2b2, where neither a2 nor b2 is a unit, and, say, b2 cannot be written as a product of irreducible elements. Looking at b2, we write b2 = a3b3, where neither a3 nor b3 is a unit, and b3 cannot be written as a product of irreducible elements. Clearly we can continue like this indefinitely by induction; we have bn=an+lbn+I' with an+l,bn+1 not units and bn+1 not a product of irreducible elements. We have the following equations: a = alb l = a 1a2b2 = a la2a3b3 = ala2a3a4b4 = .... What conditions on D could possibly rule this out? Think what the recurring situation [21.1] means. If we have a= alb l, then aEblD, the principal ideal of D generated bl, and this implies that aD ~ biD, since biD is an ideal. Is b l D ~ aD too? If so, then in particular b l E aD, so b l = ac for some c. Then bl =ac=alblc. Since b l =F 0 and D is a domain, this yields I = a I c, and a I is a unit. Contradiction! Thus aD f biD. Since b2 was obtained from b l just as b l was obtained from a, we get biD f b2D by an identical argument. And, in general, bnD f bn+ID, so we have a strictly increasing chain of ideals: aD f biD f b2D f b3D f b4D f .... The union of this chain, aD U biD U b2D U .'., is easily seen to be an ideal. If this ideal were principal, then the fact that our chain of ideals is strictly increasing would be contradicted. For if aDubIDub2DUb3DU'" =dD, then dE bjD for some i, hence dD ~ biD, and the chain would stop at biD. Thus one way to ensure that elements factor into irreducibles is to assume that every ideal in D is principal. An integral domain with the property that every ideal is principal is called a principal ideal domain (PID). For example, 71.. is a PID, and so is F[X], for any field F. Our discussion proves TIfEOREM 21.2 Let D be a PID, and let a E D. If a is neither 0 nor a unit, then a can be written as the product of finitely many irreducible elements. This theorem constitutes half of a proof that every PID is a UFD. The other half deals with the uniqueness of the factorization, and for this we take a hint from 71... The key to the uniqueness of factorization for 71.. (Exercise 4.28) was the fact that if p were irreducible and p divided a product ab, then p had to divide either a or b. In our current terminology this says that every irreducible element was prime. We thus rest our hope for unique factorization in PID's on 218 Section 21. Unique Factorization Domains THEOREM 21.3 Every irreducible element of a PID is prime. We will establish Theorem 21.3 by proving the following sharper resu~t, which also generalizes Theorem 20.2. THEOREM 21.4 Let D be a PID, let d ED, and assume that d is neither OD nor a unit. Then the following are equivalent: i) d is prime ii) d is irreducible iii) dD is a maximal ideal iv) dD is a prime ideal. PROOF. The proof is not much longer than the statement of the theorem; we will show that (i) ~ (ii) ~ (iii) ~ (iv) ~ (i). (i) ~ (ii): See Theorem 21.1. (ii) ~ (iii): Since d is not a unit, It£. dD, so dD is a proper ideal. We must show that if dD ~ bD, then bD = D. Now if dD ~ bD, then in particular dE bD, so we have d= be for some e. Since d is irreducible, either b or e must be a unit. If c is a unit, then b = de -, and bE dD, so bD c;;" dD, a contradic- tion. Thus b is a unit, and bD = D. (iii) ~ (iv): See Corollary 17.8. (iv) ~ (i): Suppose dD is prime, and suppose d divides a product be. Then bCEdD, so either bEdD or eEiiD, that is, either dlb or die, as desired. 0 THEOREM 21.5 Every PID is a UFD. PROOF. By Theorem 21.2, it suffices to show that if P,P2"'Pr=q,qz"'qs [21.2] for irreducible elements Pi' qj' then r = s and there is a permutation f of {l,2,oo.,s} such that Pi and qJ(i) are associates for each iE{1,2,oo.,s}. Now, from the above equation, p,lq,q2' .. qs' so by Theorem 21.3 and induction,p, divides some qj' By renaming the q's if necessary, we can assume j = 1. Thus q, = ap, for some a, and since q, and p, are irreducible, this means that a is a unit, so q, andp, are associates. We have P,Pz" 'Pr=ap,qz'" qs' and since we are in a domain, this becomes (a - ~2)P3' .. Pr= qZq3' .. qs' [21.31 We now proceed by induction on r. If r= 1, then s must be 1, else Equation [21.3] would show that the irreducible element q2 is a unit, which is nonsense. Thus if r= 1, then the original equation [21.2] had one irreducible on each side, and p, = q,. Section 21. Unique Factorization Domains 219 Assuming the result for r = n, suppose that r = n + 1. Then by [21.3], s> 1,
and by the inductive hypothesis, r – I = s – I and there is a permutation g of
{2, … ,s} such that Pi and qg(i) are associates for 2 < i 1. Thus Condition (i) in
the definition of Euclidean domain is satisfied.
Now for Condition (ii). Let a,f3 E l..[i], 13=1=0. We seek to ·find y and p in
1..[i] such that a = yf3 + p, and either p = 0 or v(p) < v( /3). Since 13=1= 0, a / 13 is a complex number, so there are real numbers x and y such that a = (x + yi)f3. 222 Section 21. Unique Factorization Domains Our idea is to use for 'I an element of l[i] that is close to x + yi. Specifically, choose integers a and b such that Ix - al <;; i and I y - bl '" i. Then a =(a+ bi){J+ [(x-a) +(y - b)i] {J, so if we take y=a+bi and p=[(x-a)+(y-b)i){J, we have a=y{J+p. Now if p=O, we are done. Otherwise v[(x-a)+(y-b)i) is defined, and we have as desired. v(p) = v[(x- a) +(y - b)i]v( {J) = [ (x - a)2 + (y - b)2] v( {J) < [ ( ~ r + ( ~ r] v( {J) I = 2" v( {J) < v( {J), Unique factorization in land lli) provide a very appealing proof of a classic theorem of Fermat. Fermat observed that some primes can be ex- pressed in the form a2 +b2 (with a,bEl), while others cannot. For instance, 5=22+ 12, 13=32+22, 17=42+ 12,29=52+22,37=62+ 12; but none of 3,7, 11, 19,23, or 31 can be written as the sum of two squares. It is also true that 5, 13, 17,29, and 37 are all congruent to 1 (mod 4), while 3, 7, 11, 19,23, and 31 are all congruent to 3. Fermat's Two Squares Theorem asserts that, in general, a positive odd prime p is the sum of two squares iff p = 1 (mod 4). Fermat claimed to have proved this result in a letter he wrote in 1640. In accordance with the practice of the time, he never published a proof, and the first published proof was given by Euler in 1754. It is said that Euler worked on and off for seven years to find a proof. THEOREM 21.8 (Fermat) Let p > 0 be an odd prime. Then there exist integers a
and b such thatp=a2 +b2 iff p=1 (mod 4). If p=1 (mod4), thenp can be
written as a2 + b2 in only one way (we do not count things like b2 + a2 or
(- ai+( – bi as different ways.)
PROOF. Suppose p = a2 + b2• Then p = a2 + b2 (mod 4), and each of a2,b2 is
congruent (mod4) to one of 02, 12, 22, or 32; that is, each of a2,b2 is =0 or 1
(mod 4). Since p is odd, one of a2,b2 is = 0 and the other is = 1 (mod 4).
Hence p = 1 (mod 4).
That was the easy half. To finish the proof, we must show that if p = I
(mod4), then p can be written (uniquely) as a2 + b2• The idea of our proof is
to view a2 + b2 as being (a + bi)( a – bi) in l[ i), and to shQW that p can be

Section 21. Unique Factorization Domains 223
written as (a + bi)( a – bi) for some integers a, b.
Our first step is to show that p is at least not irreducible in 1[ i]. We know
that if p were irreducible, then since l[i] is a PlD, p would be prime in l[i].
We are going to show that this is not the case by showing that there exists an
integer m such that
(m – i)(m+ i) = kp
for some integer k. Thus pl(m – i)(m + i), but clearly p divides neither (m – i)
nor (m + i) in 1[i], so p is not prime.
We find m by observing that having (m – i)(m + i) = kp for some k is the
same as having m2 + 1 = 0 (mod p), that is, m2 = – 1 (mod p). We know
something that is = -1 (mod p), namely (p -I)! (see Wilson’s Theorem,
Exercise 10.19). Thus to fmd a suitable m it suffices to show that (p – I)! = m2
for some m. Now we know that p = 4n + 1 for some n> 0, hence
(p -I)! = (1)(2)· .. (2n)(2n + 1)(2n +2)· .. (4n -1)(4n)
=(1)(2)·· · (2n)( -2n)( -(2n-I»··· (-2)( -l)(mod p)
=( -li”122232 ••• (2nf(modp)
=[(1)(2)(3)··· (2n)]2(modp).
Thus if we take m=(l)(2)(3)· ·· (2n), we have the required m, and we see that
p is not prime, hence not irreducible, in 1[ i].
The rest is easy. Write
p =(a+ bi)(c+ di),
where neither a + bi nor c + di is a unit in 1[i]. Then
v(p) = v(a + bi)v(c + di),
that is,
p2 = (a2 + b2)( c2 + d2).
Since neither a + bi nor c + di is a unit, neither a2 + b2 nor c2 + d 2 is
(Exercise 16.24), so by unique factorization in 1 ,
and we have shown that p can be written as the sum of two squares. (Note
that fromp=(a+bi)(c+di) it follows that c=a and d= -b.)
All that remains is to establish the essential uniqueness of the representa-
tion. We leave this to you as Exercise 21.19.
EXERCISES
21.1 Let D be a domain. Show that it is not possible to express a unit of D as a
product of prime elements.

224 Section 21. Unique Factorization Domains
21.2 Let D be a domain and let a, bED. Under what conditions is it true that
Da=Db?
21.3 Let R be a UFD, r E R. Show that r is irreducible in R iff it is irreducible in
R[X].
21.4 Complete the proof that Z[X] is a UFD by showing that every element which
is neither ° nor a unit can be written as a product of finitely many irreducible
elements.
21.5 Show that every element *0, ± I in Z[v=3] can be expressed as the product
of finitely many irreducible elements.
21.6 Show that the following domains are not UFDs.
a) D= {a+bViO la,bEZ}. (Consider N(a+bViO )=a2 -lOb2, and find
two distinct factorizations of 6.)
b) D= {a+ bY=5 la,bEZy. (Consider N(a+ bY=5 )=a2 +5b2, and find
three distinct factorizations of 21.)
21.7 Let D be a Euclidean domain, with v: D – {OD }~Z+ U {O}.
a) Show’ that if d ED, then d is a unit iff v( d) = vel D)’
b) Show that if v is a constant function then D is a field.
c) Show that, in general, if two nonzero elements a and b of a Euclidean
domain are associates, then v(a)=v(b).
21.8 Show that the following domains are Euclidean with the given function v.
a) D= {a+bY2la,bEZ}, with v(a+bY2 )=la2 -2b21
b) D= {a+bv=2la,bEZ}, with v(a+bv=2 )=a2 +2b2
21.9 Let D be a domain with the property that every element of D which is Ineither
o nor a unit can be written as the product of a finite number of irreducibles.
Show that D is a UFD iff every irreducible element of D is prime.
21.10 Let F be a field, and let X and Y be variables. Define the polynomial ring
F[X, Y] in two variables over F by
F[X, Y]=F[X][Y].
Thus F[X, Y] is the ring of polynomials in Y, over F[X].
a) Show that every element of F[X, Y] is a finite sum of terms of the form
aXiyj, where a E F and i,j are nonnegative integers.
b) Show that F[X, Y] is not a PID.
21.11 Let D be a domain, and let a, bED. An element dE D is called a greatest
common divisor (g.c.d.) of a and b if
i) dla and dlb; and
ii) if c is any element such that cia and clb, then cld.
Show that if D is a PID, then any two elements of D have a g.c.d. which can
be written in the form xa + yb, for some x,y ED. (Suggestion: Consider the
ideal {xa+yblx,y ED}.)

Section 21. Unique Factorization Domains 225
21.12 In the context of 7L, the definition in Exercise 21.11 assigns two g.c.d.’s to each
pair of nonzero elements a, b. For instance, the g.c.d.’s of 12 and 15 are 3 and
– 3. (Compare Exercise 4.26.) Show that, in general, if D is a domain, and
dE D is a g.c.d. of a and b, then all the g.c.d.’s of a and b are precisely the
associates of d.
21.13 Show that the elements 4 and 2(1 +V-3) in 7L[v=3] have no g.c.d.
21.14 Use the Euclidean algorithm (that is, repeated application of the division
algorithm) in O[X] to find a g.c.d. of the elements 2X3 + 9X2 + 12X + 5 and
2X s+5X4 +8X +20.
21.15 Let a E 7L[i]. Show that a is a prime element in 7L[i] iff either v(a) is a prime
integer or a is an associate of some prime integer p such thatp==:3 (mod4).
21.16 Write each of the following elements of 7L[i] as a product of primes (see Exercise
21.15).
a) 1+ 3i
b)7+8i
c) 99+27i
21.17 a) Prove that if a,bE7L andp is a prime integer,p==:3 (mod4), such that
pl(a2 + b2), then p21(a2 + b2).
(Use the fact that p is prime in 7L[i].)
b) Prove that an integer n;;. 2 is the sum of two (integer) squares iff in the prime
factorization of n (in 7L), every prime p ==: 3 (mod4) occurs to an even power.
21.18 a) Use the Euclidean algorithm in 7L[i] to find a g.c.d. for 53 + 9i and 1+ 7 i.
(See the proof that 7L[i] is a Euclidean domain, and use the fact that for
complex numbers a + bi and c + di,
a+bi = (a+bi)(c-di) )
c+di c2 +d2 •
b) The proof that I[i] is Euclidean provides an upper bound on the number
of steps it will take the Euclidean algorithm to produce a g.c.d. What is this
bound?
21.19 Prove the uniqueness statement of Theorem 21.7.
21.20 Give an example of two domains D and D’ such that D is a UFD, D’ is a
homomorphic image of D, and D’ is not a UFD.
21.21 Let R be a PID and I an ideal of R.
a) Show that every ideal of R/ I is principal. Must R/ I be a PID?
b) Show that RII has only finitely many ideals if I is nontrivial.
21.22 Show that v=3 is prime in 7L[ v=3 ].
21.23 Let
-1+v=3
~3 = COS(21T /3) + i sin(21T /3) = – ~ + i(V3 ) /2 = 2

226 Section 21. Unique Factorization Domains
Let
Z[t31 = {a + bt3+ ctlla,b,cEZ},
and observe that tl= 1. Thus Z[t31 is the domain considered by Kummer in
his work on Fermat’s Last Theorem, for p = 3. This exercise will result in a proof
that Z[t31 is a Euclidean domain. [A proof of Fermat’s Last Theorem forp=3,
using Z[t31, can be found in Hardy and Wright, An Introduction to the Theory
of Numbers, Chapter XIII (Oxford: Clarendon Press, 1960).]
a) Show that Z[t3] = {a+ bt3la,bEZ}, and that every element of Z[t3] has a
unique representation in the form a + bh
b) Let us write just t instead of h for simplicity. Define
v(a+ bn= (a + bO( a+ bt 2).
Show that v(a + bn = a2 + b2 – ab, and that v is multiplicative, that is,
v(afJ) = v(a)v(fJ).
c) Show that v maps Zm- {OJ into Z+, and that Z[n with the function v,
is a Euclidean domain.
21.24 a) Show that Z[v=3 ]C;Z[t3]’ (See the preceding exercise.)
b) Determine the relationship between Z[v=3 ] and Z[t31 by describing the
elements of Z[tJ] – Z[v=3 ] in terms of integers and ‘\f=3 .
1l.lS See Exercise 21.23.
a) Show that if a EZ[t3], then a is a unit iff v(a)= 1.
b) Show that if a EZ[t3] and v(a) is a prime integer, then a is irreducible.
Conclude that 1- t3 is irreducible.
c) Show that 2 is prime in Z[t3]. (Hint: Show that if 21(a2 + b2 -ab), then both
a and b are even.) There are many more primes in Z[t3]; see Hardy and
Wright, Chapter XV.
21.l6 Let D be a Euclidean domain. Show that the q and r in Condition (ii) for a
Euclidean domain will be unique for every choice of a and b iff v has the
property that v(a+b)<:max{v(a),v(b)} for all nonzero a and b such that a+b*O. SECTION 22 EXTENSIONS OF FIELDS In this section we begin an in-depth study of fields. The work begun here will culminate in the presentation of Galois theory in Section 25. The developments in this and the succeeding sections represent a beautiful interplay between the ideas we have developed so far. Aside from this, we will see a number of striking applications, some in Sections 25 and 26 and some even sooner in Section 23. If E and F are fields we write F ~ E to indicate that E is an extension of F, i.e. F is a subfield of E. In this context we are of course using the symbol "~" to mean more than just "is a subset of'; here the symbol means that (F, +) is a subgroup of (E, +) and (F - {OJ, -) is a subgroup of(E - {OJ, .). If F ~ E and a E E we will often study a by considering the smallest subfield of E containing F u {a}. Notation. If F ~ E and a E E then F(a) denotes the intersection of all subfields of E containing F u {a}. (The notation "F(a)" is read "F adjoin a".) It is easy to see that F(a) is a subfield of E containing F u {a} and IS contained in every subfield of E that contains F u {a}. THEOREM 22.1 F(a) "" {g(a) / h(a) I g(x), heX) E F[X] and h(a) *- OJ. PROOF. If we denote the set on the right side of the equation by S then clearly every element g(a) / h(a) of S is in F(a) since F(a) is a sub field of E containing F u {a}. On the other hand, S is a subfield of E containing F u {a}, so F(a) is a subset of S. Thus F(a) = S. 0 There is one very important situation in which the description of F(a) in Theorem 22.1 can be simplified considerably. This is the situation in which the element a happens to be a root of some nonzero polynomial in F[X]. 227 228 Section 22. Extensions of Fields DEFINITIONS If F ~ E and a E E then a is algebraic over F if there exists a nonzero polynomial j{X) E F[X] such thatj{a) == O. We say that E is an algebraic extension of F (or that E is algebraic over F) if every element of E is algebraic over F. If a is algebraic over Fthen the set of polynomials 1= (f(X) E F[X] I I(a) = O} is a nontrivial ideal in F[X]. By Theorem 20.1 there must existfiX) E I such that 1= (f(X), the prinicipal ideal generated by fiX). It will be convenient to have several equivalent ways to describe those polynomials that generate I. THEOREM 22.2 If F ~ E, a E E is algebraic over F, and I = (f(X) E F[X] I j{a) == O} then the following statements about an element j{X) E I are all equivalent to each other: i) 1= (t(X) ii)j{X) has minimal degree among all the nonzero polynomials in I iii)j{X) is irreducible. PROOF. That (i) implies (ii) is clear, since if 1= (f(X) then fiX) divides every g(X) E I, so if g(X) is nonzero then deg(f) :s deg(g) by the degree rule. To see that (ii) implies (iii) note that if fiX) is reducible and fiX) = g(X)h(X) with g(X) and heX) each of degree less than deg(f), then 0 = fia) = g(a)h(a) in E, so either g(a) = 0 or h(a) = O. Thus either g(X) E lor heX) E I, and thus (ii) cannot be true. Finally, (iii) implies (i) because iffiX) E I is irreducible and g(X) is any generator for I then fiX) = g(X)h(X) for some heX), so since fiX) is irreducible heX) must be a nonzero constant. ThusfiX) is a generator for I. 0 A polynomial is called monic if its leading coefficient is 1. We can obtain a monic generator for the ideal I by starting with any generator and multiplying by an element of F. I has only one monic generator, because any two generators divide each other and therefore differ by a constant factor. DEFINITIONS Let F ~ E and let a E E be algebraic over F. The unique monic irreducible polynomial in F[X] having a as a root is called the irreducible polynomial of a over F. We denote it by irr(a/F)' The degree of irr(alF) is called the degree of a over F and denoted by deg(alF). It follows from Theorem 22.2 that if j(){) is any nonzero polynomial in F[X] such thatj{a) = 0 then deg(a/F) :s deg(j{X), and we have deg(alF) = deg(j{X) iff j{X) is irreducible in F[X]. Examples Section 22. Extensions of Fields 229 1. Let F be any field and let a E F. Then a is algebraic over F, irr(alF) is X - a, and deg(alF) = I. Conversely, if F S;;;; E and a E E is such that deg(alF) = I, then a is a root of some polynomial of degree I in F[X] and thus a E F because F is a field . 2. Consider JR s;;;; (X) = X
”
-I
= Xp-I + Xp-2 + . .. + X + 1 p X -I
‘
so (p is algebraic over Q. We claim that irr«(IQ) = p(X), and therefore deg«(IQ)
= p – I.
To see this it suffices to show that P(X) is irreducuble over Q, which we do
by applying Eisenstein’s Criterion to p(X + I):
(X + IY -1
(X +I)-1
X p + pX ,,-1 + … + pX
X
where for each r E {I, … , p – I} the coefficient of Xfrr in the numerator is the
binomial coefficient I( P� )1 and is thus an integer divisible by p since p is prime. r. p r .
By Eisenstein’s Criterion, iX) is called the pth cyclotomic polynomial. Its roots in the complex plane,
together with the number 1, are p equally spaced points on the circle of radius 1
around the origin. (The word “cyclotomic” means “circle -cutting”.) As we will see
later, the fact that deg«(IQ) = p – 1 is the key to determining for which primes pit
is possible to construct a regular p-gon using straightedge and compass
4. If F is a field then by Theorem 22.1 the quotient field of the domain F[X] is
F(X). The element X in F(X) is not algebraic over F, because if ao, ai, … , an E F
and anXn + an_IXn-1 + . . . + alX + ao is 0 in F[X] then all the coefficients aj are 0 in
F because this is what it means for anXn + an_IXn-1 + … + alX + ao to be 0 in F[X].
In general, any element that is not algebraic over F is called transcendental
over F.

230 Section 22. Extensions of Fields
We can now simplify our description of F(a) in the case where a is algebraic
over F.
THEOREM 22.3 Let F <;;;; E and let a E E be algebraic over F. Then every element of F(a) has a unique representation in the formj(a), wherej(X) E F[X] is either 0 or of degree less than deg(a/F). PROOF. Let K = {g(a) I g(X) E F[X]}. We first show that F(a) = K. Every element of K is in F(a) since F(a) is a subfield of E containing Fu{a}. To establish the reverse inclusion it suffices to show that K is a field (for then K is a subfield of E containing Fu {a} ). We have an evaluation homomorphism cpa : F[X] -+ K such that cpa(g(X» = g(a) for all g(X) E F[X] (Exercise 19.12). The kernel of cpa is {g(X) I g(a) = O},and this is the principal ideal! = (irr(a/F). We thus have an isomorphism qJa : F[X]/!-+ K, so to show that K is a field it suffices, by Theorem 17.7, to show that! is a maximal ideal. But this is so by Theorem 20.2, since irr(a/F) is irreducible over F. We now know that F(a) = K. We also know (Exercise 20.6) that each element of Flx]/! is the coset j(X) represented by a unique polynomial that is either 0 or has degree less than deg(a/F). Since 'Pa ([(X) = f(a), we see that each element of K = F(a) isf(a) for a unique such polynomialf(.X).o We have also proved THEOREM 22.4 Suppose F <;;;; E and a E E is algebraic over F. Let! = (irr(a/F) and denote each element! +j(X) of F[X]/I by j(X). Then there is an isomorphism qJ" : F[X]/!-+ F(a) such that 'Pa ([(X» = f(a) for allf(X) E F[X]. In particular 'Pa (c) = c for all c E F and 'Pa(X) = a. In Section 20 we saw that if we start with an irreducible polynomialf(X) over F then we can fmd a field extending F and containing a root of f(X) by forming F[X]/([(X». Theorem 22.4 shows that this is the only way to adjoin a root for f(X) to F. COROLLARY 22.5 Suppose fiX) is irreducible over F and al and a2 are roots of fiX) in extension fields El and E2 of F. Then there is an isomorphism Section 22. Extensions of Fields 231 cp : F(al) ---+ F(a2) such that q:>(c) = c for all c E F and q:>(a l ) = az.
PROOF Note that irr(a/F) = irr(a2 /F) since both are obtained by multiplyingJ{X)
by the multiplicative inverse of its leading coefficient. Thus by Theorem 22.4 q:> =
((J,a 0 q:>a -I has the desired properties. 0
2 1
Our purpose in considering F(a) is to use it to draw conclusions about the
element a. To do this we will need to use some ideas that will be familiar to you if
you have studied linear algebra.
In general, if F S; E then a subset S of E is said to be linearly independent
over F if whenever SI~ S2, “‘, Sn are finitely many distinct elements of Sand CI, Cz,
.•. , Cn are elements of F such that CIS) + C282 + … + CnSn = a in E, then CI = C2 = …
= c” = 0 in F. On the other hand, a subset S of E is said to span E over F if for
every element bEE there exist finitely many elements s], S2, … , s” in Sand Cj, cz,
… , Cn in F such that CISI + C2S2 + … + CnSn = b, i.e. b is a linear combination of SI,
sz, … , Sn with coefficients c, E F. S is called a basis for E over F if S spans E over
F and is linearly independent over F.
Example Theorem 22.3 shows that if a E E is algebraic over F and deg(a/F) = n
then the set {I, a, a2, •.• , an-I} is a basis for F(a) over F.
As a specific example, consider Q <;:;::: Q(lJ5) <;:;::: R Since lJ5 is a root of X 3 - 5, which is irreducible over Q by Eisenstein's Criterion, deg(lJ5/Q) = 3. Thus {I, 51/3, 52/3} is a basis for QOJS) over Q. THEOREM 22.6 Suppose F t;;;; E. If {SI' Sz, ... , sn} spans E over F then no subset of E that is I inearly independent over F can have more than n elements. PROOF. Suppose for a contradiction that {b l, b2, ... , bn+ I} is a set of n + I elements linearly independent over F. Since {SI' S2, ... , sn} spans E over F there exist CI, Cz, ... , en in F such that Since {bl, b2, ... , hn+d is linearly independent, b l * 0 and therefore at least one c,* O. By renumbering if necessary we can assume CI * 0, and therefore we can express SI as a linear combination of b], Sz, S3, ... , S". It follows that since {SI' 82, ... , Sn} spans E over F, so does {bl, S2, S3, ... , sn}. Therefore there are d], d2, ... , dn in F such that If d2 = ... = dn = 0 this contradicts the independence of {bl, b2, ... , btl+ l }, so by renumbering if necessary we can assume that d2 * 0 and therefore we can express 232 Section 22. Extensions of Fields S2 as a linear combination of bl, b20 83, ... , Sn. Thus since {bl, 82, ... , 8n } spans E over F, so does {bl, b20 83, ... , sn}. Continuing in this way, we conclude that {bl, ... , bn } spans E over F. In particular, bn+1 is a linear combination of b l , ... , bn , and this contradicts the linear independence of {b], ... , bn+I }. 0 COROLLARY 22.7 If F ~ E and S is a finite basis for E over F then every basis for E over F has the same number of elements as S. PROOF. If B is any basis for E over F then, since S spans E over F and B is linearly independent over F, Theorem 22.6 shows that the number of elements in B is no larger than the number in S. But then reversing the roles of Sand B shows that the number of elements in S is no larger than the number in B. 0 DEFINITION If F ~ E, we say that E is a finite extension of F (or that E is finite over F) if there exists a finite basis for E over F. The number of elements in such a basis is called the degree (or dimension) of E over F. We denote this degree by [E: FJ. Note that [E : FJ is well defined by Corollary 22.7. Examples 1. If E = F then {I} is a basis for E over F, so [E : F1 = 1. Conversely, if [E : F1 = I and {b} is a basis for E over F then since {b} spans E over F we can in particular write 1 = cb for some c E F. Thus b = c-l E F, so since every element of E can be written in the form db with d E F, E = F. 2. If F c;;;, E and a E E is algebraic over F then F(a) is a fmite extension of F and [F(a) : F1 = deg(a/F). For we have observed that {I, a, if, ... , an-- l } is a basis for F(a) over F, where n = deg(a/F). 3. Consider F c;;;, F(X), where F(X) is the quotient field of F[X]. Since the subset {1,X,X2,X3, ... } is linearly independent over F, theorem 22.6 shows that F(X) is not a finite extension of F. We noted previously that F(X) is not an algebraic extension of F, and therefore the fact that F(X) is not finite over F also follows from THEOREM 22.S Every finite extension is algebraic. PROOF. Suppose E is a finite extension of F, with [E: F1 = n. We want to show that every a E E is algebraic over F. If d" = d' for two distinct positive integers m and k then a is a root of X'" - x", Section 22. Extensions of Fields 233 so a is algebraic over F. If am1' d' for all m * k then {a, cl, ... , an+1} is a set of more than n elements of E and cannot be linearly independent over F by Theorem 22.6. Therefore there must exist c], C2, ... , Cn+l in F, not all zero, such that cIa + ... + cn+lan+1 = O. So a is a root of cIX + ... + Cn+lXn+1• 0 COROLLARY 22.9. If F ~ E and a E E is algebraic over F then F(a) is an algebraic extension of F. PROOF. F(a) is finite over F, hence algebraic over F. 0 This result is not a triviality. For it asserts that every element of F(a)-not just a-is algebraic over F. Example Here's an example of an algebraic extension that is not a fmite extension. For each integer n 2 I let an be the positive real2nth root of2. Let Qo = Q and for n 2 I let Qn = Qn-l(an). Then Since Qn = Q(an) (why?) and an is a root of X 2" - 2, Qn is algebraic over Q by Corollary 22.9. Since every element of E is in some Qn this implies that E is algebraic over Q. But E is not fmite over Q. For since X 2n - 2 is irreducible over Q by Eisenstein's Criterion, [Q (an) : Q] = 2n. Thus for every n 2 I there is a subset of E that has 2n elements and is linearly independent over Q. Therefore by Theorem 22.6 E cannot be finite over Q. Before we can establish the most useful fact about finite extensions, we need another result about linear independence. THEOREM 22.10 Suppose F ~ E and there exists a positive integer n such that every subset of E that is linearly independent over F has at most n elements. Then any linearly independent m-element subset {aJ, ... , am} of E can be extended to a basis for E over F, i.e. there is a basis for E over F that contains {aI, ... , am}. PROOF. If {aI, ... , am} spans E over Fthen it is already a basis for E over F. If {aJ, ... , am} does not span E over F there is some b l E E that is not a linear combination of a" ... , am with coefficients from F. It follows that {aI, ... , am, bd is independent over F, for if not we can write 234 Section 22. Extensions of Fields with CI, ... , em and din F, not all O. Since {aI, .. ,' a"J is independent we must have d *" 0, and thus we can solve for b l and contradict the fact that b l is not a linear combination of aI, , .. , am. If {a" ... , am, btl spans E over F then it is a basis for E over F. If {a" ... , am, b l } does not span E over F then we can repeat the above argument to show that there is b2 E E such that {aI, ... , am, b" b2} is independent over F. If we keep repeating this reasoning then, since no subset of E linearly independent over F can have more than n elements, we must arrive at some bl, b2" "', bk such that {aI, ... , am" b l , "', bd is a basis for E over F. 0 The most frequently used fact about finite extensions is the fact that "degrees multiply", THEOREM 22.11 Suppose F ~ E ~ K. Then K is finite over F iff E is finite over F and K is finite over E. Furthermore, if [E : F] = m and [K: E] = n then [K: F] = mn, and if {a" ... , am} is a basis for E over F and {bl, ... , bn } is a basis for Kover E thenS= {a;bj [I ::::i::::m, I ::::j::::n} isabasisforKoverF. PROOF Suppose first that K is finite over F and B is a basis for Kover F having, say, r elements. Then B spans Kover F and therefore B spans Kover E. Applying Theorem 22.6 twice, we see that no subset of E that is independent over F can have more than r elements, and that no subset of K that is independent over E can have more than r elements. Applying Theorem 22.10 twice, we see that the independent subset {I} of E can be extended to a basis for E over F with at most r elements, and that the independent subset {I} of K can be extended to a basis for Kover E with at most r elements. So E is finite over F and K is finite over E. Now assume that E is finite over F and K is finite over E. Let {aI, ... , am}, {b" .. " bn } and S be as given in the statement of the theorem. All our claims will be established if we can show that S is a basis for Kover F. For any v E Kthere exist d" .. " dn E E such that Each di can be written as a linear combination of aJ, ... , am with coefficients in F, and therefore v can be written as a linear combination of elements of S with coefficients in F. Thus S spans Kover F. To see that S is linearly independent over F, suppose we have for some cij's in F, i.e. Section 22. Extensions of Fields 235 For every bJ the coefficient c),a)+ ... +c""am on b, is in E, so since {b), .'" bn } is independent over E and bb ... , bn are distinct we conclude that c),a)+ ... + Cn"am = o for each}, 1 'S} 'S n. By the independence of {a), ... , am} over F we then conclude that all cij = 0.0 COROLLARY 22.12 If F c;;:; E and [E : Fl = n then, for every a E E, deg(alF) divides n. PROOF. We have F ~ F(a) ~ E, so by Theorem 22.11 [F(a) : F] divides n, i.e. deg(a/F) divides n.D Notation. If F c;;:; E and a), ... , an E E then F(a), ... , an) denotes the intersection of all subfields of E containing F u {a), ... , an}. Note that for elements a) and az, F(a), az) can also be described as F(a)(az), the extension obtained by adjoining az to F(a). Equally well, F(ab az) = F(az)(a). Examples 1. Since irr( Ji IrQ) = XZ - 2, deg( Ji IrQ) = 2 and every element of Q( Ji) can be written uniquely in the form a + b Ji , with a, b E Q. Thus .[3 'l Q( Ji), because if (a + b Ji )2 = 3 then If ab "# 0 we can solve for J2 and conclude that J2 E Q, contradicting deg( J2 IQ) = 2. So a = 0 or b = 0, and thus 2b2 = 3 or a2 = 3. But then J3i2 E Q or .J3 E Q, again a contradiction Since .[3 'l rQ( Ji) the polynomial XZ- 3 in Q( Ji )[X] is irreducible over Q( Ji) (it has degree 2 and has no root in Q( Ji », so deg( .[3 IQ( Ji» = 2 and {1,.[3 } is a basis for E = Q( Ji , .[3 ) over Q( J2). By Theorem 22.11 [E: rQ] = [Q( Ji) : Q] . [E : rQ( Ji )] and this equals deg( Ji IQ) . deg( .[3 IQ( Ji» = 2 . 2 = 4. Again by Theorem 22.11, {I' I, I . .[3, Ji . 1, Ji . .[3} = {I, .J3, Ji,.J6} is a basis for E over Q. It is interesting to note that E can be obtained by adjoining one element to Q. For note that since E is a field containing the number a = Ji + .[3 , Q(a) c;;:; E. On the other hand, E c;;:; Q(a) because lQ!(a) is a field containing both Ji and .J3 . To see this, note that 236 Section 22. Extensions of Fields (-Ii + fj/=2+3 +216 =5+216, and thus 16 E 3(X) = X2 +X +1. The other root of 1>3(X) in C is -I-;..j; ,
so since E ~ JR, 1>3(X) has no root in E. Since 1>3(X) has degree 2, 1>3(X) is
therefore irreducible over E, so since
[K :1(£] = [E : QJ][K : E],
[K :12′)
22.10 a) Prove the trigonometric identity cos(3B) = 4cos3 B-3 cos B.
b) Use part (a) to show that deg(cos(7V9)/Q) = 3.
22.11 Let p E Z+ be prime and let , 2 = cos(2111p2) + i sin(2111p2). Then , 2 is a root of the
p p
cyclotomic polynomial
Prove that 4> 2 (X) is irreducible over Q and that deg(‘ 2 IQ) = p(P-I).
p p
22.12 Suppose £ is an extension field of Zs and £ has exactly 78125 elements.
Find deg(alZs) for every a E £ – Zs.
22.13 Suppose £ is an extension field of Z7 and dEE – ~. Find deg(diZ7) if
a) d3 = 2
b) If = 2
22.14 Prove that ‘5 ~ Q( ‘7)’
22.15 Suppose £ is a finite extension of F and [£: F1 is prime. Prove that F(a) = £ for every
aE£-F.

Section 22. Extensions of Fields 239
22.16 Suppose E is an extension field ofQ and E has elements a and b such that a8 + 4a7
+2a l – 6a + 2 = 0 and b9 + 3b6 + 12b2 – 6b + 15 = O. Prove that Q(a) n Q(b) = Q.
22.17 Suppose F c;;;; E, a E E is algebraic over F and deg(a/F) is odd. Prove that F(a) =
F(a2).
22.18 Suppose F c;;;; E and let K = {a EEl a is algebraic over F}.
a) Prove that K is a subfield of E. (K is called the algebraic closure of F in E.)
b) Prove that if bEE and b is algebraic over K then b E K.
22.19 Suppose F c;;;; E and c and d are elements of E that are algebraic over F, such that
deg(clF) and deg(d/F) are relatively prime.
a) Prove that [F(c, d) : F] = deg(clF) . deg(d/F).
b) Prove that irr(d/F) is irreducible over F(c).
22.20 Suppose F c;;;; E and a and b are elements of E that are algebraic over F. Prove that
irr(a/F) is irreducible over F(b) iff irr(b/F) is irreducible over F(a).
22.21 a) If Pn is the nth positive prime, show that for every n
b) Prove that the infinite set {2112, 2113, 21.6, 2117, 21111, … } is linearly independent
over Q.
22.22 Suppose E is an algebraic extension of F and D is a subring of E containing F. Prove
that D is a field.
22.23 Prove that if F c;;;; E and a and b are elements of E that are not both algebraic over F
then ab and a + b cannot both be algebraic over F.
22.24 Suppose F c;;;; E and assume there are ai, … , an in E such that E = F(al, … , a,J Prove
that the following three statements are all equivalent to each other:
i ) E is finite over F.
ii) E is algebraic over F.
iii) Each of ai, … , all is algebraic over F.
22.25 Let E be an algebraic extension of F and let cp : E -> E be a one-to-one ring
homomorphism such that cp(c) = c for all c E F. Prove that cp is onto.
22.26 Suppose F c;;;; E and {Slo .. ” SII} is a finite subset of E that spans E over F. Prove that
there exists a subset of {SI’ … , SII} that is a basis for E over F.

SECTION 23
CONSTRUCTIONS WITH
STRAIGHTEDGE AND
COMPASS
The geometers of ancient Greece considered the problem of performing
geometric constructions using only a straightedge and compass. (A straightedge is
a ruler with no markings on it. Thus the straightedge can be used to draw the line
through two previously constructed points, but not to measure or mark off
distances. The compass can be used to draw the circle whose center is a previously
constructed point and whose radius is the distance between two previously
constructed points.) The Greeks succeeded, for example, in constructing regular 3-
gons (equilateral triangles), 4-gons (squares), pentagons and hexagons, and they
were familiar with the method for bisecting a given angle.
When it came to certain other problems, however, the Greeks were baffled.
They were unable to solve any of the following. (We use the word “construct” to
mean “construct with straightedge and compass”.)
Problem I. Given the side of a cube, construct the side of a larger cube with twice
the volume.
Problem II. Give a general method for trisecting angles. That is, show how,
starting with a given angle, to construct an angle one-third the size.
Problem Ill. Construct a regular heptagon (7-gon).
The Greeks worked on these problems over 2000 years ago, and in the
intervening centuries many people have sought in vain for solutions. Even today
there are people who still try, unaware that the task is utterly hopeless. For it was
proved in the nineteenth century that the constructions required by these problems
cannot be carried out using only straightedge and compass. Our purpose in this
section is to use what we know about field extensions to prove that the
constructions are impossible.
240

Section 23. Constructions with Straightedge and Compass 241
We shall be working with a certain subfield C, of C, called the field of
constructible numbers. To describe this subfield, we identify C with the set of
points in the xy-plane. The field IR of real numbers is identified with the set of
points on the x-axis.
We will define the set C, of constructible (or, more precisely, constructible
from 0 and 1) numbers to consist of all elements of C that can be obtained by
starting with 0 and 1 and using the following methods (A), (8), (C) any finite
number of times:
(A) Let LI be the line determined by two points that we already know are in C,
and let L2 be another such line that is not parallel to L I • Take the point of
intersection of LI and L2.
(8) Let L be the line determined by two points in Cn and let C be a circle whose
center is in C, and whose radius is the distance between some two points of C,.
Take the points of intersection of Land C.
(C) Let CI and C2 be two (distinct) circles satisfying the requirements on the circle
C in (8). Take the points of intersection of C I and C2•
DEFINITION Let z E Co Then z E C,. if and only if z can be obtained by starting
with 0 and 1 and applying methods (A), (B), (C) some finite number oftimes.
To get an idea of how this definition generates points, observe that the points
marked in each of the following diagrams are in C,o :
2
The points in C, correspond in a clear sense to the points we want to call
“constructible”, because we get them by starting with 0 and I and using the
straightedge (for lines) and compass (for circles).
THEOREM 23.1 C, is a subfield ofe.
PROOF. We first show that C, n IR is a subfield ofe. To show this it will suffice,
since 0 and I are in C,o , to establish (i) and (ii):

242 Section 23. Constructions with Straightedge and Compass
(i) If a, b E ICc n JR and b *- ° then a – b E IC, n lR.
(ii) If a, bE ICc n JR and b *- ° then alb E ICc n JR.
To establish (i) we use method (B), with the x-axis as the line L (the x-axis is
detennined by 0, I E ICc , so we can use it) and the circle with. center at a and
radius Ibl as C. (This circle is legitimate since its radius is the distance between 0,
b E ICc .) One of the two points of intersection of Land C is a-b.
To establish (ii) it suffices by (i) to handle the case where both a, b > O.
Notice that there are points of ICc on the y-axis, other than the origin. For example,
-1, 2 E ICc by (i), so if we draw the circles of radius 2 centered at ±1, their points
of intersection (± J3i ) are both in ICc . Thus we can use the y-axis as the line in
method (B), and in particular if we intersect the y-axis first with the circle of radius
a centered at the origin, and then with the circle of radius b centered at ai, we see
that the points ai and (a + b)i are both in ICc . By a similar argument we see that the
point 1 + ai is in ICc . If we draw the line L through (a + b)i and 1 + ai, then the
point 1 + c where L intersects the x-axis is in ICc:
Hence, by (i), C E ICc. But, using similar triangles,
a+b a
–=-
I +c c
and thus c = alb, so alb E ICc nlR.
We now prove that ICc itself is a subfield of C. We will use the fact that if z =
a + bi E IC then z E ICc iff both a and b are in ICc . (See Exercise 23.1.) Using this
fact we see that if z = a + bi arid w = c + di are in ICc then a, b, c, d E ICc , so a – c
and’b – d are in ICc since ICc n JR is a field, and thus z – w = (a – c) + (b – d)i is
in ICc . Finally, if w -:f- 0 then

Section 23. Constructions with Straightedge and Compass 243
z/w= a+bi = (a+bi)(c-di) =(aC+bd)+(bC-ad)i.
c+di (c+di)(c-di) C2 +d2 C2 +d2
Since C, n lR is a field, Exercise 23.1 implies that Z/W E CC, . This concludes the
proof. 0
It follows from Theorem 23.1 that Ql, the prime subfield of CC, is contained in
CC, . The key to the applications of CC, to the construction problems is the following
characterization of the elements of CC” which enables us to apply our results on
field extensions.
THEOREM 23.2 Let z E cc. Then z E CC, iff there exist Z], •.. , Zn E CC such that
ZT E Ql, zJ E Ql(z], … , Zj_]) for 2 ~j ~ n, and Z E Ql(z], … , zn).
PROOF. Assume first that there exist such numbers Z], … , Zn. We want to show
that Z E CC,. Ifwe knew the truth of the implication
then we could argue as follows: First, Z] E CC, because ZT E Ql; hence Ql(z]) ~ CCc
since CC, is a field. Then Z2 E CC,. since z~ E Ql(z]) ~ CCc, so Ql(z]’ Z2) ~ CC, since
CCc is a field. Continuing in this way, we could conclude by induction that Ql(zj, … ,
Zn) ~ CCc and hence z E CC, .
To verify the above implication we first deal with the case where w2 is a
positive real number and thus w E lR. We can assume w > 0 since CC, is a field.
The following diagram indicates how to construct w from w2:
The length of the vertical segment is w, as we see by using the fact that the two
small triangles are similar.
For the general case we use the fact that one square root of c + di is given by

244 Section 23. Constructions with Straightedge and Compass
~ +c + ).Jc2 +d2-c
-I
2 2
[23.1 )
(we take the “+” if d > 0 and the “-” if d < 0). It follows from the special case we have just dealt with and Exercise 23.1 that if c + di E C, then both square roots of c + di are in C,. This concludes the first halfofthe proof. For the second half it will be convenient to call a sequence z" ... , Zn of complex numbers a square root sequence if zI E Q and z} E Q(ZI, ... , Zj-l) for 2 :s j :s n. We let S be the set of complex numbers Z for which there exists a square root sequence z" "', Zn such that Z E Q(ZI, ... , Zn). We wantto show that C(" ~ S. We will frequently use the easily verified fact that if z" ... , Zn and w" ... , Wm are square root sequences then z" ... , Zm W" ... , Wm is a square root sequence. To prove that C" <;:;; S it will suffice to establish if Z = a + bi E C, then a E Sand b E S, [23.2) for then if a E Q (Z], .. " zn) and b E Q (WI, ... , wm ) for square root sequences ZI, ... , Zn and WI, ... , Wm we have a + bi E Q (zJ, .. " Z'" WI, .. " W m, i), Notice that if a line L is determined by two points that have all their coordinates in Q(z" .. " zn) then L has an equation of the form ax + by + c = 0 with a, b, cE Q(ZI, ... , zn). Likewise, if a circle C has center WI and radius the distance between W2 and W3, and all the coordinates of WI, Wz, W3 are in Q(z" .. " zn), then C has an equation of the form x2 + l + ax + by + c = 0 with a, b, c E Q(z" ... , zm). We now prove [23.2] by induction on the number of applications of methods (A), (8), and (C) used in constructing z. Ifno applications are used then Z = 0 or 1, and 0, I E Q <;:;; S. Assume now that Z is obtained by applying method (A) to two lines determined by points which, by inductive hypothesis, have all their coordinates in Q(z" .. " zn) for some square root sequence ZI, ... , Zn. Then these lines have equations ax + by + c = 0 and a x + by + c' = 0 with all the coefficients in Q(ZI, ... , Zn). The coordinates of Z are obtained by solving these two equations simultaneously and hence are in Q(z" ... , zn). If Z is obtained by applying method (8) then the coordinates of z are found by solving simultaneously two equations ax + by + c = 0 and x2 + l + a x + by + c '= 0 with all coefficients in Q(ZI, ... , Zn) for some square root sequence Zl, ... , Zn' Finding the coordinates of Z will require us to use a square root Zn+l of an element of Q(ZI, "', zn). Then Zl, "', Zn, Zn+l is a square root sequence and the coordinates of Z are in Q(ZI, ... , Zm Zn+l)' Section 23. Constructions with Straightedge and Compass 245 Finally, if z is obtained by using method (C) then the coordinates of z are found by solving simultaneously two equations x 2 + / + ax + by + c = 0 and x2 + / + a x + by + c' = 0 with all coefficients in Q (zl' ... , zn) for some square root sequence ZI' ... , Zw Solving these equations is equivalent to solving (a - a')x + (b - b')y + (c - c') = 0 and x2 + y + a'x + b'y + c' = 0, so we can conclude the argument as we did for method (B). 0 COROLLARY 23.3 If Z E C, then z is algebraic over Q and deg(zIQ) = 2' for some nonnegative integer r. PROOF. If z E Ce then by Theorem 23.2 we have z E Q(z\, ... , zn) for some square root sequence z\, ... , Zn. Then each extension in the chain is finite, of degree 1 or 2. Hence [Q(ZI, ... , Zn) : Q] = 2m for some m :S 11. By Corollary 22.12, deg(zlQ) = 2' for some r :S m. 0 It is worth mentioning that the converse of Corollary 23.2 is false. There exist z's in C such that deg(zIQ) = 4 but z (I' Ce. (See Exercise 24.28.) Applications 1. (Doubling the Cube) If we could construct from the side of a cube the side of a larger cube with twice the volume, then from 1 we could construct iff.. Since 1 E C, this would imply iff. E C e. But deg( iff. IQ) = 3 and therefore by Corollary 23.3 iff. ~ Ce. Thus there is no construction for doubling cubes. The problem of doubling the cube is sometimes called the Delian problem, because of a legend concerning its origin. As the story goes, there was a terrible plague in Athens about 2500 years ago, and the Athenians consulted the oracle at Delos about what steps they could take to end it. The oracle told them to double the cubical altar to Apollo. They proceeded to double its side, which of course multiplied the volume by 8. The plague continued, and it was surmised that perhaps the oracle had meant that they should double the volume. At this point it became rather a pressing matter to construct iff.. 2. (Trisecting Angles) In order to demonstrate the nonexistence of a general construction for trisecting angles with straightedge and compass, it suffices to 246 Section 23. Constructions with Straightedge and Compass exhibit one specific angle that cannot be so trisected. We use an angle of n!3 radians. It is easy to see that the point (6 = cos(n!3) + isin(n/3) is in C e. If it were possible to trisect an angle of n-l3 radians with straightedge and compass then the point (18 = cos(n/9)+isin(n/9) would be in Cn and so would cos(n/9). But by Exercise 22.10 deg(cos(n/9)/Q) = 3, so cos(n/9) ~ C e. 3. (Constructing a Regular Heptagon) We will prove the general result that if P is prime and P ~ 3 then the constructibility of a regular p-gon implies that p = 2'+ I for some r ~ I. From this it is immediate that it is impossible to construct a 7-gon. The constructibility of a regular p-gon implies that (p == cos(2n/p) + i sin(2n/p) is in C r (Exercise 23.5). But irr«(/Q) == … , qlv and a regular
n-gon is constructible. Then for any ri ~ 2, a regular q/ -gon is constructible.
(Why?) Thus
is in Ce so deg( Sq,’ /Q) = 2′ for some r ~ O. But by Exercise 22.11, this says that
q;(q; – 1) = 2′, and therefore q; = 2. Thus for all qi ~ 3 we must have rj = 1 and Sq,
E !CC> so qi must be a Fermat prime, as above. We have proved
THEOREM 23.4 If n ~ 3 and a regular n-gon can be constructed with straightedge
and compass then n = 2mPI . Pk where m ~ 0, k ~ 0 and PI, … , Pk are distinct
Fermat primes.
The conditions on n given in Theorem 23.4 are also sufficient for the
constructibility of a regular n-gon. (We will prove this in Section 25, using Galois
theory.) The sufficiency was first proved by Gauss towards the end of his now
classic book Disquisitiones Arithmeticae (1801). In this work Gauss also claimed

Section 23. Constructions with Straightedge and Compass 247
that he had a proof of the result in Theorem 23.4, but he never published a proof.
The first published proof was given by Pierre Wantzel in a paper that appeared in
1837. In this paper Wantzel also gave the first proofs of the impossibility of
doubling the cube or trisecting angles with straightedge and compass.
The ancient Greeks also considered constructions with tools other than
straightedge and compass. For example, they considered the possibility of allowing
the distance between any two constructed points to be marked off on the
straightedge. With this added ability it is possible to double the cube, trisect angles,
and construct a heptagon (see Exercises 23.12-23.14).
EXERCISES
23.1 Let z = a + bi E IC. Show that z E IC,. iff both a and b are in ICc,
23.2 For each of the following values of n determine whether a regular n-gon is
constructible with straightedge and compass: n = II; 13; 15; 18; 19; 112; 340.
23.3 Give an example of a nonzero angle that can be trisected using straightedge and
compass.
23.4 Show how to construct a regular hexagon using straightedge and compass.
23.5 Show that if a regular p-gon is constructible anywhere in the xy-plane then (p E ICc.
23.6 Prove that a regular decagon (IO-gon) is constructible, as follows. Consider the triangle
with vertices at 0, 1, and (10 = cos(n-!S) + isin(n-!S). Find the length of the side
connecting 1 and (10 by bisecting the angle of the triangle at (10 and using similar
triangles.
23.7 The result of Exercise 23.6 implies that a regular pentagon is constructible. Prove this
in another way by starting from the fact that if B = 2n15 then cos(2B) = cos(3B).
23.8 (Squaring the Circle) Lindemann proved in 1882 that l( is transcendental over Q.
Assuming this result, show that there is no construction with straightedge and compass
that will produce from the radius of a given circle the side of a square that has the same
area as that circle.
23.9 Prove that if r E Z+ and p = 2′ + I is prime then r = 211 for some nonnegative integer n.
23.10 Express cos 5B in terms of cos B and use your result to show that there is no general
method for cutting angles into tive equal pieces with straightedge and compass.
23.11 We have seen that an angle of 20° is not constructible with straightedge and compass.
a) Prove that neither an angle of 1 ° nor an angle of 2° is constructible.
b) Prove that if n E Z+ then an angle of nO is constructible iff 31n.

248 Section 23. Constructions with Straightedge and Compass
23.12 The following construction for trisecting angles using compass and marked
straightedge is due to Archimedes.
a) Consider an acute angle AOB, where A is the point 1, 0 is the origin, and B is a
point in the first quadrant one unit from O. Construct the circle of radius 1 with
center at O. Mark two points one unit apart on the straightedge. Then place
the straightedge so that its edge passes through B and one of the two marked
points is on the circle while the other is on the negative x-axis at some point C
outside the circle. Prove that angle ACB is one-third of angle
AOB.
b) Explain why being able to trisect acute angles enables us to trisect all
angles.
23.13 The following construction for doubling the cube using compass and marked
straightedge is due to Isaac Newton.
Construct a line L with negative slope that passes through the origin and makes an
angle of 30° with the positive x-axis. Mark two points one unit apart on the
straightedge. Then place the straightedge so that its edge passes through the point i
and one of the two marked points is on L while the other is at a point A on the
positive x-axis. Prove that the distance from A to i is ifi.
23.14 The following construction of a regular heptagon using compass and marked
straightedge was published by Crockett Johnson in the Mathematical Gazette,
volume 59 (1975), pages 17-21.
Construct a square ABCD with sides of length I. Then construct the circle centered at
B and passing through D, and construct the perpendicular bisector of AB at M. Mark
two points E and F one unit apart on the straightedge, and place the straightedge so
that E lies on the perpendicular bisector, F lies on the circle, and the straightedge
passes through A, with F between A and E.
Let x denote the length of BE and let 8 denote angle AEM. Note that 2x sin 8 = I and
apply the law of cosines to triangle BEF to show that sin 38= cos 48. Conclude that 8
= n/14 and thus we can construct the angle 4(n/14) = 2m’7.

SECTION 24
NORMAL AND
SEPARABLE EXTENSIONS
We will now consider certain special kinds of field extensions that are useful
in studying the roots of polynomials.
If F ~ E andfiX) E F[X] can be written as C(X-al)(X-a2) … (X-an) for some
c and some a;s in E (not necessarily distinct) then we say thatfiX) splits over E
and that E splits fiX). Note that then c E F, because c is the leading coefficient of
f{X) E F[X]. It is apparent that if K is any extension of E then f{X) also splits over
K, so there are many different extensions of F over which f{X) splits. We wish to
single out a standard extension that splits f{X), so that we can discuss f{X) in terms
of this one definite extension. The thing to do is to opt for economy and use a
minimal extension that splitsf{X).
DEFINITION Let F <;;; E, and letf{X) E F[X] be nonconstant. We say that E is a splitting field for f(X) over F if f{X) splits over E but does not split over any proper subfield of E containing F. THEOREM 24.1 Supposef{X) E F[X] has degree n ~ 1 and E is an extension of F such that in E[X] we have Then E is a splitting field for f{X) over F if and only if E = F(al, ... , an). PROOF. If E is a splitting field for f{X) over F then it follows from the fact that f{X) splits over F(aJ, ... , an) that F(a1' ... , an) cannot be a proper subfield of E. Hence E = F(aJ, ... , an), Conversely, if E = F(a1' .. " an) then to show that E is a splitting field for f{X) over F we must show that if K is a subfield of E containing F andj(X) splits over K, say asf{X) = c(X - bl) ... (X -- bn), then every a, is in K (for then K = E because E = F(al' ... , an». But since 0 = fiai) = c(ai - bl) ... (ai - bn), some ai - bj = 0 and thus ai = bj E K. 0 249 250 Section 24. Normal and Separable Extensions It follows from Corollary 20.4 and Theorem 24.1 that for any j(X) E F[X] there exists a splitting field for j(X) over F, and every splitting field for j(X) over F is a finite extension of F. But we can say a little more. THEOREM 24.2 Suppose j(X) E F[X] has degree n :::: l. Then there exists a splitting field for j(X) over F whose degree over F divides n!. PROOF. We argue by induction on n. If n = I then F itself is a splitting field for j(X) over F, of degree lover F. Assuming the result for polynomials of degree less than n, we consider two cases. First, if j(X) is irreducible over F of degree n then we consider an extension F(aj) of F, where aj is a root ofj(X). In F(aj)[X] we can writej(X) = (X - aj)g(X) with deg(g(X» = n - l. Then by the inductive hypothesis there is a splitting field E for g(X) over F(aj) such that [E: F(aj)] divides (n - I)L By Theorem 24.1, E = F(aj)(a2' ... , an) where a2, ... , an are roots of g(X). Sincej(X) splits over E and E = F(aj, ... , an), E is a splitting field for j(X) over F. Since [F(aj) : F] = n becausej(X) is irreducible over F, [E : F] = n . [E : F(aj)], and this divides n! because [E : F(aj)] divides (n - I)L Second, if j(X) factors in F[X] as j(X) = g(X)h(X), where deg(g(X» = m < n and deg(h(X» = k < n, then by inductive hypothesis there is a splitting field K = F(aj, ... , am) for g(X) over F such that [K : F] divides mL Again by inductive hypothesis there is a splitting field E = K(b j, ••• , bk) for heX) over K such that [E : K] divides kL Then E = F(aj, ... , am, bj, ... , bk) is a splitting field for j(X) over F, and [E : F] = [K : F][E : K], which divides m!k!, which divides nL (Note that n!lm!k! = n!lm!(n - m)! is an integer since it is the number of ways of choosing m elements from an n-element set.) 0 Examples 1. C is a splitting field for X2 + lover IR, since X2 + I = (X -i)(X + i) and C = JR.(i). We have [C : JR.] = 2. 2. Ifj(X) = X- 2 then the roots ofj(X) in Care ifi. (3 ifi, and (~ ifi. so E = 'O( ifi. (3 ifi, (~ ifi) = 'O( ifi. (3) is a splitting field for X - 2 over 'O. We saw in the third example following Corollary 22.12 that [E : '0] = 6. 3. Continuing the preceding example, we note that since ifi E R R«(3) is a splitting field for X3 -2 over R. Since (3 is a root of X2 +X +1, which is irreducible over R we have [JR.«(3) : R] = 2. R«(3) is also a splitting field for X2 + X + lover R 4. Ifj(X) = ~ - 5X2 + 6 = (X2 - 2)(X2- 3) then E = '0 (ji. J3) is a splitting field for j(X) over 'O. We saw in the first example following Corollary 22.12 that Section 24. Normal and Separable E:l$tensions 251 [E : Q] = 4. We also saw that E = Q( ..fi + -J3) and that irr( ( ..fi + -J3 )/ Q) = X' -10X2 + 1. The other roots of this polynomial in E are -(..fi + -J3) and ±( ..fi - -J3 ), and E is a splitting field for X'-1 OX2+ lover Q. 5. If n E Z+ and (n = cos(2n!n) + isin(2n!n) then the roots of X' - I in C are ('" s;,s~, ... ,S:-l, I so Q «(n) is a splitting field for X' - I over Q. Ifn is a prime p then, as we have seen, the splitting field Q«(p) has degree p - lover Q. We have indicated that we want a splitting field for j{X) over F to serve as a "standard" extension of Fthat splitsj{X). What we want to show, then, is that if E and K are both splitting fields for j{X) over F then E and K are essentiaIIy the "same" extension of F, in the sense that there is an isomorphism ep : E -> K such
that ep(e) = e for every e E F.
DEFINITION Let E,K be extension fields of F. E and K are isomorphic over F if
there exists an isomorphism ep from E onto K such that ep(e) = e for every e E F.
Such an isomorphism is caIIed an isomorphism of E and Kover F.
THEOREM 24.3 Let F be a field, j{X) E F[X]. Let E and K be splitting fields for
j{X) over F. Then E and K are isomorphic over F.
The idea of the proof is to proceed by induction on deg(f{X). We take an
irreducible factor g(X) of j{X) in F[X] and let a E E, b E K be roots of g(X). We
show that F(a) and F(b) are isomorphic over F via an isomorphism that sends a to
b, and then we want to apply the inductive hypothesis toj{X)/(X -a) E F(a)[X] and
the corresponding polynomialj{X)/(X -b) E F(b) [X]. Unfortunately we now have
two polynomials, with coefficients in two different fields. But the fields are
isomorphic, and under this isomorphism the two polynomials match up. To prove
Theorem 24.3, therefore, we first state it in a more general form.
THEOREM 24.4 Let F, F* be fields and let ep : F -> F* be an isomorphism. Let
j{X) E F[X] be nonconstant and let f”(X) E F* [X] be the polynomial obtained by
replacing each coefficient e; ofj{X) by ep(e,). Let E be a splitting field for j{X) over
F and let E* be a splitting field for f*(X) over F*. Then there exists an
isomorphism ep+ : E -> E* such that ep+(e) = ep(e) for every e E F.
We express the result of this theorem by saying that ep can be extended to an
isomorphism ep+ : E -> E*. Note that Theorem 24.3 follows immediately from
Theorem 24.4, by starting with the identity isomorphism from F onto F.
We aim to prove Theorem 24.4 by induction on deg(f{X). We need the
foIIowing result to carry out the induction step.
LEMMA 24.5 Let F, F*, ep, f{X),and f*(X) be as in Theorem 24.4 and assume that
fiX) is irreducible in F[X]. Then if a and b are roots off{X) andf*(X) in extensions

252 Section 24. Normal and Separable Extensions
of F and F*, respectively, there is an isomorphism CPl : F(a) ~ F*(b) such that
cpl(a) = band cpl(e) = cp(e) for every e E F.
PROOF. We use a generalized version of the proof of Corollary 22.5.
First, cP extends in an obvious way to an isomorphism from F[X] onto F*[X],
obtained by mapping every g(X) E F[X] to g*(X). If we let I be the principal ideal
(j(X) in F[X] and let 1* = (j*(X) in F*[X] then I is mapped onto 1* so we have an
isomorphism
If! : F[X]II ~ F*[X]II*
such that for every g(X) in F[X] the coset g(X) of g(X) in F[X]II is mapped to the
coset of g*(X) in F*[X]II*, which we will denote by [g*(X)].
Since.f(X) is irreducible over F, r(X) is irreducible over F* . . Note that I =
(irr(a/F) and 1*= (irr(b/F*». If we let CfJa : F[X]/I ~ F(a) and CfJb : F*[X]/I* ~
F*(b) be the isomorphisms provided by Theorem 22.4, and let CfJl = CfJb 0 lI’ 0 CfJa-1,
we see that CfJl : F(a) ~ F*(b) is an an isomorphism. We have
and for every e E F
D
Another version of this proof, using Theorem 22.3 instead of Theorem 22.4, is
requested in Exercise 24.14.
PROOF OF THEOREM 24.4. We proceed by induction on the degree offix). If
deg(j(X) = 1 there is nothing to prove, since E = F and E* = F*. So suppose
deg(j(X) = n and the theorem is true for all fields F, F* and all g(X) E F[X] of
degree less than n. Let heX) be an irreducible factor offiX) in F[X] and let h*(X) be
the corresponding irreducible factor of r(X) in F*[X]. Since fiX) splits over E
there is a root a of heX) in E (see Exercise 24.1). Likewise, there is a root b of
h*(X) in E*. By Lemma 24.5 there is an isomorphiam CPI : F(a) ~ F*(b) that
extends cP and maps a to b.
We now conclude the proof by applying the inductive hypothesis to the
isomorphism CPl. In order to do so, we write fiX) = (X-a)g(X) in F(a) [X]. This
yields r(X) = (X – b)g*(X), where g*(X) is obtained by applying CPI to all the
coefficients of g(X). If we can show that E is a splitting field for g(X) over F(a) and
E* is a splitting field for g*(X) over F*(b) then the inductive hypothesis will imply
that CPI can be extended to an isomorphism cP+ : E ~ E*, and cP+ will extend cP
because CPI does.

Section 24. Normal and Separable Extensions 253
Now g(X) splits over E, because f{X) does (Exercise 24.1). Furthermore, if
g(X) were to split over some proper subfield of E containing F(a) then so would
f{X), and this would contradict the fact that E is a splitting field for f{X) over F.
Thus E is a splitting field for g(X) over F(a), and similarly E* is a splitting field for
g*(X) over F*(b). This completes the proof. D
With this theorem behind us, we can speak of the splitting field of a
polynomial over a given field.
It is worthwhile to record the following consequences of Theorem 24.4 while
the ideas are still fresh.
COROLLARY 24.6 Let F be a field, f{X) E F[X], and let E be the splitting field
for f{X) over F. Let K,K* be subfields of E containing F, and let qJ : K -+ K* be an
isomorphism over F. Then qJ extends to an automorphism of E over F.
PROOF. E is the splitting field for f{X) over K, and E is also the splitting field for
f{X) over K*. Since qJ fixes every element of F,f{X) is the polynomial obtained by
applying qJ to all the coefficients of fiX). Thus, by Theorem 24.4, qJ extends to an
isomorphism from E onto E. D
DE}’INITION If F ~ E then two elements a and b of E are said to be conjugate
oveJ’ F if a and b are algebraic over F and irr(a/F) =irr(b/F).
COROLLARY 24.7 Let F, f{X), and E be as in the previous corollary and let a, b
E E. Then a and b are conjugate over F iff there exists an automorphism qJ of E
over F such that qJ(a) = b.
PROOF. Since E is finite over F, a and b are algebraic over F. If there exists an
automorphism qJ of E over F such that qJ(a) = b, then denoting irr(alF) by g(X) and
applying qJ to both sides of the equation g(a) = 0 yields g(b) = O. Thus g(X) is
irr(b/F).
Conversely, if a and b are conjugate over F then, applying Lemma 24.5 to the
irreducible polynomial of a and b over F, we see that the identity isomorphism
from F onto itself extends to an isomorphism from F(a) onto F(b) that maps a to b.
By Corollary 24.6 this isomorphism extends to an automorphism of E over F. D
Example The roots of X’ – 2 in C are ±ifi and ±i ifi, so the splitting field for X’
– 2 over Ql is E = Ql( ifi, i), and [E : Ql] = 4 . 2 = 8 since irr(iIQ( ifi» = X2 + 1.
If b is any of the four roots of X – 2 we know by Lemma 24.5 that there is an
isomorphism qJl : Ql( ifi) -+ Ql(b) over Ql mapping ifi to b. Since irr(±i/Ql( ifi»
= X2 + I and applying qJl to the coefficients of X2 + 1 leaves the polynomial
unchanged, Lemma 24.5 tells us that qJl can be extended to an automorphism of E
over Ql mapping i to either of the roots ±i of X2 + 1. So we have obtained eight
automorphisms of E over Ql. Since any automorphism of E over Ql is completely

254 Section 24. Normal and Separable Extensions
determined by its effects on ifi and i. and has the same effects on ifi and i as one
of the eight automorphisms we have indicated, we have found all the
automorphisms of E over IQ.
As another illustration of Lemma 24.5, consi,der (ii)=> (iii)=> (i).
(i)=>(ii): If at • …. ak are the distinct roots ofj(X) in E then E = F(at. …. ak)’ If
E ~ K and

(iii): Suppose g(X) is irreducible in F[X] and a E E is a root of g(X). To
show that g(X) splits over E we will find an extension K of E such that g(X) splits
over K and K is the splitting field over F for some polynomial heX) E F[X]. We can
then argue as follows: For any root b of g(X) in K, a and b are conjugate over F, so
by Corollary 24.7 there is an automorphism

(i): Let {bl, … , br } be a basis for E over F and let g,(X) = irr(b/F) for
each i. Each g,(X) has a root in E, so by (iii) each g,(X) splits over E. Thus j{X) =
gl(X)g2(X) … g,(X) splits over E, and E is in fact the splitting field for j{X) over F
since if F (;: L (;: E andflX) splits over L then in particular h[, … , hr are all in L so L
= E. 0
DEFINITION A finite extension E of F is called a normal extension of F if it
satisfies the conditions of Theorem 24.8.
The proof that (ii)=>(iii) in Theorem 24.8 includes a proof of the following
fact, which we record for future use.
LEMMA 24.9 If F(bJ, … , br ) is a finite extension of Fthen the splitting field over
F(b[, … , br) of irr(b/F) ….. irr(b/F) is a finite normal extension of F.
It turns out that for algebraic extensions that are not tinite, (ii) and (iii) in
Theorem 24.8 are still equivalent, and (i) gets replaced by the statement that E is
the splitting field over F for a set of polynomials.
Normal extensions will be tied in with normal subgroups when we come to the
main theorem of Galois theory.
Examples
1. Extensions of degree 2 are normal. For let [E : F] = 2 and let g(X) be an
irreducible polynomial over F that has a root a E E. Then [F(a) : fl divides 2, i.e.
deg(g(X)) is I or 2. Thus g(X) splits over E, so E is normal over Fusing
characterization (iii) of normality.
2. A normal extension of a normal extension of F need not be a normal
extension of F. For consider
We have [1Qi( ..fi) : IQiJ = 2 = [1Qi( ifi) : 1Qi( ..fi)], so both extensions are normal.
But iQ( ifi) is not normal over iQ, since X’ -2 is irreducible over IQi and has a root
in iQ( ifi) but does not split over iQ( ifi ).
3. We have seen that there are automorphisms of K = iQ( ifi, i) over iQ that
send ifi to i ifi and thus do not map iQ( ifi) onto itself. Thus property (ii) of
Theorem 24.8 fails for the extension iQ( ifi) of iQ, showing again that iQ( ifi) is
not normal over iQ.

256 Section 24. Normal and Separable Extensions
4. If a is an element of an extension of Z7 such that a3 = 2, then the elements a,
2a, and 4a of Z7(a) are roots of X’ – 2. So X’ – 2 splits over Z7(a) and Z7(a) is the
splitting field for X’- 2 over Z7. Thus Z7(a) is normal over Z7.
The polynomial g(X) = X’ + X + 1 is irreducible over Z7 (since it has degree 3
and has no roots in Z7), but the element a2 + a of Z7(a) is a root of g(X). It follows
from property (iii) of Theorem 24.8 that g(.x) must split over Z7(a). (It turns out
that the other two roots of g(X) in Z7(a) are 4a2 + 2a and 2a2 + 4a.)
If E is the splitting field over F for a nonconstant polynomialj(X) E F[X] and
we write j{X) = c(X – at) … (X – an) in E[X], then at, … , an need not all be
distinct. This is a significant issue when we try to use E to study j{X).
DEFINITION Ifj(X) E F[X] and a is a root ofj{X) in an extension K of F, we say
that a is a root of multiplicity m if in K[X] we can write j{X) = (X – a)mg(X) with
g(a)”* O.
The meaning of the definition is unchanged if we replace K[X] by L[X], for
any extension L of K, because ifj{X) = (X – a)mg(X) in L[X] then g(X) E K[X]. In
particular, if L is the splitting field for j{X) over K andj{X) = c(X – at) … (X – an)
in L[X] then a is a root of multiplicity m if and only if there are exactly m factors X
– ai such that ai = a.
DEFINITIONS Roots of multiplicity 1 are called simple roots, and roots of
multiplicity at least 2 are called multiple roots. A polynomial is said to have
distinct roots if all of its roots are simple.
There is an easy way to determine whether a root a of a polynomial fiX) is a
multiple root, by using the formal derivative fiX). If fiX) = Co +clX +C2X2 +. . .
+cnX”, then the formal derivative off(X) is defined to be
f'(X) = CI + 2C2 X + … + nc,x-I.
This is called the “formal” derivative because it is not defined by a limit, as
derivatives are defined in calculus. (We have no notion of “limit” available, since
we are working over an arbitrary field F.) Nevertheless, formal derivatives obey
the usual sum, product, and power rules for derivatives in calculus. (See Exercise
19.17.)
THEOREM 24.10 Letj{X) E F[X] be nonconstant.
i) If F ~ K and a is a root of fiX) in K then a is a mUltiple root if and only if
f(a) = O.

Section 24. Normal and Separable Extensions 257
ii) Suppose fiX) is irreducible over F. Then there exists an extension K of F
containing a multiple root of fiX) if and only if f(X) is the zero polynomial, and in
this case every root of jeX) in every exten.5ion K of F is a multiple root.
PROOF. i) Exercise 19.18(b).
ii) First supposef(X) has a multiple root a in some extension K of F. We want
to show that f(X) is the zero polynomial. By part (i), f(a) = 0, so by Theorem
22.2 fiX) divides f(X) in F[X]. But this is impossible if f(X) is not the zero
polynomial, because then f(X) has a degree that is less than that of fiX). So f(X) is
the zero polynomial.
Now suppose f(X) is the zero polynomial. Then f(a) = 0 for every root a of
fiX) in any extension K of F, so by part (i) every root off(X) is a multiple root. D
COROLLARY 24.11 If F is of characteristic 0 andj(X) E F[X] is irreducible over
F, thenj(X) has distinct roots.
PROOF. If the leading coefficient of fiX) is Cn * 0 then the leading coefficient of
f(X) is nCn * 0 (since F is of characteristic 0), so f(X) is not the zero polynomial.
Thusf(X) has distinct roots. D
The assumption thatj(X) is irreducible is indispensable in Corollary 24.11. For
example, the polynomial (X – 1)2 in Q![X] clearly does not have distinct roots.
In characteristic p, even an irreducible polynomial can have multiple roots.
Example Consider Z2(X), the quotient field of Z2[~\1. The polynomial Y 2 – X in
Z2(X) [ Y] is irreducible over Z2(X) because it has degree 2 and has no roots in
Z2(X)’ (If g(X), heX) E Z2 [X] and (g(X)/h(X)2 = X, then, mUltiplying both sides by
h(X)2 we get an equation in which the left side has even degree and the right side
has odd degree.) Since the formal derivative of y2 – X is 2Y – 0 = 0, every root of
Y 2 – X is a mUltiple root. Indeed, if we adjoin a root a then y2 X = (Y – ai since
we are in characteristic 2.
DEFINITIONS Suppose F <;;; E. If a E E then a is said to be separable over F if a is algebraic over F and irr(a/F) has distinct roots. E is called a separable extension of F if every element of E is separable over F. It follows from Corollary 24.11 that if F has characteristic 0 then every algebraic extension of F is separable. In characteristic p, some algebraic extensions are separable and some are not. (The extension Z2(X)(a) of Z2(X) in the preceding example is not separable.) DEFINITIONS If F <;;; E then E is called a simple extension of F if there exists a E E such that E = F(a). The element a is then called a primitive element for E over F. 258 Section 24. Normal and Separable Extensions One nice thing about finite separable extensions is that they are always simple extensions. To see this we need a preliminary result that is interesting in its own right. THEOREM 24.12 If F is a field then every finite subgroup of the multiplicative group F - {OJ is cyclic. PROOF Let H be a finite subgroupof F - {OJ under multiplication. For every positive integer n, the polynomial X" - 1 has at most n roots in F by Corollary 19.4, so hn = 1 has at most n solutions h E H. Thus H is cyclic by Exercise 5.27. (Exercise 5.27 can be proved directly, or by applying Theorem 14.2.)0 And now for our result about finite separable extensions. THEOREM 24.13 (Primitive Element Theorem) If F is a field then every finite separable extension E of F is simple. PROOF. If F is a finite field, then since E is finite over F, E has only finitely many elements. Therefore the multiplicative group E -{OJ is cyclic by Theorem 24.12. If a is a generator for E - {OJ then E = F(a), so E is a simple extension of F. For the remainder of the proof we can assume that F has infinitely many elements. Since E is a finite extension of F, there are elements aI, ... , a r in E such that E = F(at. ... , ar). If we can show that E is a simple extension of F in the case where r = 2 then the result for all r's will follow by an easy induction. So suppose E = F(a, b). Let [F(a) : F] = m and [E : F(a)] = n, so that [E : F] = mn. We will show that there is some c E F such that E = F(a + cb). The idea is that we can find c E F such that [F(a + cb) : F] ? mn, and therefore F(a + cb) = E. Let K be the splitting field for irr(a/F) . irr(b/F) over F. We claim that we can find embeddings Tt. ... , Tmn of E into Kover F such that for some c E F the elements Ti(a + cb), 1 ~ i ~ mn, are all distinct. It will then follow that irr«a + cb)/F) has at least mn distinct roots in K, and therefore deg«a + cb)/F) ? mn, as desired. To find Tt. ... , T""" note that irr(a/F) splits over K and has distinct roots aI, ... , am in K since a is separable over F. For any ai there is by Lemma 24.5 an isomorphism qJi : F(a) - F(ai) over F such that qJi(a) = ai. IffiX) = irr(b/F(a» then there is some g(X) E F(a)[X] such that irr(b/F) = fiX)g(X), so if we denote by fi(X) and glx) the polynomials obtained by applying qJi to the coefficients of fiX) and g(X) then irr(b/F) = fi(X)glx), with deg(fi(X) = n. Since irr(b/F) splits over K and has distinct roots, it follows thatfi(X) has n distinct roots bi!, ... , bin in K. For any bij there is by Lemma 24.5 an extension of qJi to an embedding qJij of E into K such that qJij(b) = bij . The embeddings qJij, 1 ~ i ~ m, 1 ~ j ~ n are mn distinct embeddings of E into Kover F. Call these embeddings Tt. ... , Tmn. Section 24. Normal and Separable Extensions 259 If i '* j then either r;(a) '* ria) or ri(b) '* r)Cb). We want to show that there is some e E F such that riCa + cb) '* ria + cb) whenever i '* j, i.e. rtCa) - rtCa) '* eCr/b) - r,Cb» if either riCa) '* rj{a) or riCb) '* rtCb). Ifri(b) = rib) this is clearly true for every c, for then riCa) '* ria). So all we need is and since F is infinite it is clear that such a c exists. 0 The Primitive Element Theorem will be useful to us in our discussion of Galois theory in the next section. The argument that produced rl, ... , rmn has the following consequence, which will also be significant. If j(X) E F[X] we say j(X) is separable over F if every irreducible factor of j(X) in F[X] has distinct roots. THEOREM 24.14 If j(X) is separable over F and E is the splitting field for j(X) over F, then there are exactly [E : F] automorphisms of E over F. PROOF. We have E = F(ah ... , an), where the a;'s are the roots ofj(X) in E. For each a;, irr(alF) divides.f{X) in F[X] and thus has distinct roots. If [F(aJ) : F] = mJ and [F(ah ... , aj) : F(aJ, ... , aj-D] = mj for 2 s.j S. n then as in the proof of the Primitive Element Theorem we see that there are mJ embeddings of F(aJ) into E over F; that each of these has m2 extensions to an embedding of F(aJ, a2) into E over F; that each of these extensions has m3 extensions to an embedding of F(ah aZ, a3) into E over F; and so on. This gives us mrm2 ..... mn = [E : F] embeddings of E into E over F. Since.f{X) splits over the image of each of these embeddings (because.f{X) splits over E), the image cannot be a proper subfield of the splitting field E. So each of these embeddings is onto and is therefore an automorphism of E over F. By considering aJ, ... , an in that order we see that every automorphism of E over F must have the same effect on aJ, ... , an as one of our [E : F] embeddings, hence must be one of these embeddings. 0 We conclude this section by taking a closer look at finite fields. By Theorem 22.13, there exists a finite field with exactly k elements if and only if k is a power of a prime. If F is any field with pn elements then the multiplicative group F - {O} has pn - 1 elements, so by Theorem 10.4 every nonzero element of F is a root of X p"-J - 1. Therefore every element of F, including 0, is a root of X p" - X, and F is a splitting field for X p" _·X over the prime subfield. It follows, by Theorem 24.4, that if two fields each have pn elements then the obvious isomorphism of their prime subfields extends to an isomorphism between the fields themselves. We have proved 260 Section 24. Normal and Separable Extensions THEOREM 24.15 If F is a field with exactly pn elements then every element of F is a root of X I,n_ X , and F is a splitting field for X p"- X over the prime subfield. If two finite fields have the same number of elements then they are isomorphic. Notice that if F has pn elements then F has an extension of degree r, for any integer r ~ 1, obtained as a splitting field E for X pnr - X over F. For since pn - 1 divides pnr - 1, apnr -I = 1 for every nonzero a E F, so every element of F is a root of X pnr - X . Therefore E is also a splitting field for X pnr - X over the prime subfield of F, so lEI = pnr . Since iFl = pn , a basis for E over F must have exactly r elements, so [E : F] = r. In addition, any extension of F of degree r must have exactly pnr elements, and must therefore be a splitting field for X pnr_ X over the prime subfield, hence over F. Thus any two such extensions are isomorphic over F, and we have THEOREM 24.16 Let F be a field with exactly pn elements. Then for any integer r ~ 1, F has an extension of degree r, which is obtained as a splitting field for X pnr_ X over F. Any extension of F of degree r has exactly pnr elements, and any two extensions of F of degree r are isomorphic over F. Furthermore, any such extension is normal and separable over F. The statement about separability follows from the fact that for every element a of the extension, irr(a/F) divides X pllr_ X in F[X], hence has distinct roots. EXERCISES 24.1 Suppose K is a field, fiX) E K[X] and g(X) is a nonconstant factor of fiX) in K[X]. Prove that iff(X) splits over K then g(X) splits over K. 24.2 For each polynomial, find the splitting field over Q and its degree over Q. a)~-l b)X'-l c)XS-1 d)X'-8 e)~+1 .f}X'+1 g)~+2 h)~+4 i) X' - 2, where p is prime j) (X2 - 3)~ - 2) k) 24~ - 26X2 + 9X - I l);t + 5X2 + 6 m);t-5X2+36 n)X'+~+l 24.3 Find all multiple roots of XS + X-I in its splitting field over 1Z3. 24.4 Find all multiple roots of 3X' + 2X + 3 in its splitting field over 1Z19· Section 24. Normal and Separable Extensions 261 24.5 How many automorphisms are there of Q( ifi ) over Q? How many are there of Q ( ifi, (3) over Q? 24.6 Give an example of a finite extension that is not normal but has the property of normal extensions indicated in Corollary 24.6. 24.7 Show again that Q( ifi ) is not normal over Q by exhibiting two elements a, b of Q ( ifi) that are conjugate over Q for which there is no automorphism ofQ( ifi) over Q mapping a to b. 24.8 Prove that Q( ~2 + ji ) is normal over Q. 24.9 Suppose F ~ E ~ K with K finite and normal over F. Must E be normal over F? Must K be normal over E? 24.10 Let E be a finite normal extension of F. Must condition (iii) of Theorem 24.8 continue to hold if we delete the word "irreducible"? 24.11 Give an example of a separable extension that is not normal, and give an example of a normal extension that is not separable. 24.12 Suppose F ~ E ~ K and K is separable over F. Must E be separable over F? Must K be separable over E? 24.13 Suppose that E is a separable extension of F and there exists a positive integer n such that deg(a/F) ~ n for all a E E. Prove that E is a finite extension of F and [E : FJ ~ n. 24.14 Give another proof of Lemma 24.S, by using Theorem 22.3 instead of Theorem 22.4. 24.15 A field F is called perfect if every finite extension of F is separable. Show that every field of characteristic 0 is perfect and that every finite field is perfect. 24.16 Let Fbe a field of prime characteristic p and letj(X) E F[XJ be irreducible over F. a) Show that there is a polynomial g(X) E F[XJ such thatj(X) = g(X P') for some integer r ~ 0 and g(X) has distinct roots. b) Show that there is an integer r ~ 0 such that all the roots ofj(X) have multiplicity pro c) Suppose F ~ E and a E E is algebraic over F. Show that there is an integer r ~ 0 such that apr is separable over F. 24.17 Let F be a field of characteristic p, where p is prime. Prove that F is perfect if and only if every element of F has a pth root in F. 24.18 Let F be the quotient field of Zz[XJ[Y], and consider the extension F(a, b), where aZ = Xand bZ = Y. 262 Section 24. Normal and Separable Extensions a) Show that [F(a, b) : Fl = 4 and F(a, b) is not a simple extension of F. b) Show that F(a, b) is a normal extension of F, but F(a, b) is not the splitting field over F of any irreducible polynomial. 24.19 Let F be a finite field and letj{X), g(X) be two irreducible polynomials in F[X] of the same degree. Prove that the splitting field forj{X) over F is also the splitting field for g(X) over F. 24.20 Let F be a finite field such that IFI = pn and let k E Z+. Prove that F has a subfield with exactly pk elements if and only if kin, and that if such a subfield exists it is unique. 24.21 Let p be prime and let g(X) be irreducible over the field Z[J" Prove that if n E Z+ then g(X) divides Xpn - X in Zp[X] if and only if deg(g(X)) divides n. 24.22 (Automorphisms of a finite field) Let F be the finite field such that IFI = pn and let G be the group of automorphisms of F over its prime subfield, under composition. a) The Frobenius automorphism of F is the mapping (J : F --+ F given by u(a) = d'. Verify that (J E G. b) Show that (J has order n in G. c) Show that G =< (J> •
d) Let E be an extension of F of degree r. Show that the group of automorphisms of E over
F is cyclic of order r, generated by the automorphism that maps every a E E to apn .
24.23 Let F be a finite field. Prove that every element of F can be written as the sum of two
squares, i.e. for every a E F there exist b, c E F such that a = b2 + c2.
24.24 Suppose E is a finite extension of a field F and K\ and K2 are subfields of E that
contain F and are normal over F. Prove that K\ n K2 is normal over F.
24.25 Suppose F s;;; K\ s;;; E, E is finite over F and K\ is normal over F. Prove that if F s;;; K2
s;;; E and no proper subfield of E contains both K\ and K2 then E is normal over K2.
24.26 Prove that if E is a finite extension of F then E is normal over F if and only if for
every polynomialj{X) E F[X] that is irreducible over F, all the irreducible factors of
j{X) in E[X] have the same degree.
24.27 Let E be a finite extension of F. Prove that E is a simple extension of F if and only if
there are only finitely many distinct subfields of E containing F. [Hints: Show that if
E = F(a) then any subfield K of E containing F is obtained by adjoining to F the
coefficients of irr(a/K), and irr(a/K) divides irr(a/F) in E[X]. For the converse, it
suffices to show that for any two elements b, dEE, F(b, d) is a simple extension of F.

Section 24. Normal and Separable Extensions 263
coefficients of irr(alK), and irr(alK) divides irr(a/f) in E[X]. For the converse, it
suffices to show that for any two elements b, dEE, F(b, d) is a simple extension of F.
Consider the subfields F(b + cd), for c E F. If F is infinite there must exist Cj *- C2 in F
such that F(b + Cjd) = F(b + czd).]
24.28 Suppose the complex number z is in Ce, so z E Q(aj, … , an) for some square root
sequence aj, … , an of complex numbers. By Lemma 24.9 let K be an extension of
Q(a\, … , an) that is a finite normal extension ofQ.
a) Prove that for every root z’ of irr(zIQ) in K, there is a sequence d], … , dn of
elements of K such that z’ E Q(d\, … , dn), d l2 E Q, and d: E Q(dj, … , dj-\)
for all 2 :5;j :5; n.
b) Prove that if E is the splitting field for irr(zIQ) over Q then [E: Q] = 2m for
some integer m.
c) Prove that ifw E C and deg(wIQ) = 4 and the splitting field for irr(w;Q)
over Q has degree 12 or 24 over Q then w e Ce. This indicates how the
converse of Corollary 23.3 can fail. (For an example where this happens see
Exercise 26.13.)
The remaining sequence of exercises develops further aspects of the idea of
separability.
24.29 Let E be a finite extension of F. By Lemma 24.9 let K be an extension of E that is a
finite normal extension of F. Define the separable degree [E: Fls of E over F to be
the number of embeddings of E into Kover F.
a) Show that [E: Fls:5; [E : Fl·
b) Show that [E: Fls does not depend on our choice of K.
24.30 Prove that if F s:; E s:; K and K is finite over F then
[K: Fl, = [E : F)., .. [K : E).”,
24.31 Prove that if E is a finite extension of F then [E : F)., divides [E: F). [Suggestion:
Consider the case E = F(a) first, and use Exercise 24.16 in characteristic p.]
24.32 Prove that if E is a finite extension of F then E is a separable extension of F if and
only if [E: F), = [E : F).
24.33 Prove that if F s:; E s:; K are finite extensions such that E is separable over F and K is
separable over E then K is separable over F.
24.34 Show that F(aj, … , all) is a separable extension of F if and only if each a, is separable
over F.

264 Section 24. Normal and Separable Extensions
24.35 Let F ~ E. Define the separable closure SE(F) of Fin E to consist of all elements of
E that are separable over F.
a) Prove that SE(F) is a subfield of E containing F.
b) Prove that if E is finite over F then [SE(F) : fl = [E: fls’

SECTION 25
GALOIS THEORY
Galois theory brings together ideas from group theory (subgroups, normal
subgroups, quotient groups, indices) and ideas from field theory (subfields, normal
and separable extensions, degrees, automorphisms). It is often referred to as one of
the most beautiful parts of mathematics, both because it reveals a close and precise
interplay between these two sets of ideas and because this interplay has fascinating
consequences. We will see in this section that it enables us to complete the
determination of which regular polygons are constructible with straightedge and
compass by turning a question about a field extension into an answerable question
about a group. In the next section we will see that the same kind of translation
allows us to answer questions about formulas for finding the roots of polynomials.
Our first step is to introduce the main ingredients of the theory. If F s E then
the set of automorphisms of E over F is a group under composition. We call this
group the Galois group of E over F and denote it by r(E/F). On the other hand, if
S is a set of automorphisms of a field E, then by the fixed field of S we mean the
subset {a EEl cp(a) = a for every cp E S}. It is easy to verify that this subset is a
sub field of E. We denote it by (S).
It is obviously true that
F s (f(E/F» [25.1 J
since every element of F is fixed (not moved) by every automorphism of E that
fixes all of F. The finite extensions that provide the setting for Galois theory are
those for which equality holds in [25.1]. These extensions can be characterized in
several equivalent ways.
THEOREM 25.1 Let E be a finite extension of F. The following are equivalent:
i) F = (f(E/F).
ii) E is normal and separable over F.
265