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in the process, in other words, the rationals contain a copy of the integers, namely, the
rationals of the form a/1, a в€€ Z. So a natural question is whether S в€’1 R contains a copy
of R.

2.8.3 Proposition
Deп¬Ѓne f : R в†’ S в€’1 R by f (a) = a/1. Then f is a ring homomorphism. If S has no zero
divisors then f is a monomorphism, and we say that R can be embedded in S в€’1 R. In
particular:
(i) A commutative ring R can be embedded in its complete (or full ) ring of fractions
(S в€’1 R, where S consists of all non-divisors of zero in R).
(ii) An integral domain can be embedded in its quotient п¬Ѓeld.
Proof. We have f (a + b) = a+b = a + 1 = f (a) + f (b), f (ab) = ab = a 1 = f (a)f (b),
b b
1 1 1 1
and f (1) = 1 , proving that f is a ring homomorphism. If S has no zero divisors and
1
f (a) = a/1 = 0/1, then for some s в€€ S we have sa = 0, and since s cannot be a zero
divisor, we have a = 0. Thus f is a monomorphism. в™Ј

2.8.4 Corollary
The quotient п¬Ѓeld F of an integral domain R is the smallest п¬Ѓeld containing R.
Proof. By (2.8.3), we may regard R as a subset of F , so that F is a п¬Ѓeld containing R.
But if L is any п¬Ѓeld containing R, then all fractions a/b, a, b в€€ R, must belong to L. Thus
F вЉ† L. в™Ј
30 CHAPTER 2. RING FUNDAMENTALS

Problems For Section 2.8
1. If the integral domain D is in fact a п¬Ѓeld, what is the quotient п¬Ѓeld of D?
2. If D is the set F [X] of all polynomials over a п¬Ѓeld F , what is the quotient п¬Ѓeld of D?
3. Give a detailed proof that addition in a ring of fractions is associative.
4. Give a detailed proof that the distributive laws hold in a ring of fractions.
5. Let R be an integral domain with quotient п¬Ѓeld F , and let h be a ring monomorphism
from R to a п¬Ѓeld L. Show that h has a unique extension to a monomorphism from F
to L.
6. Let h be the ring homomorphism from Z to Zp , p prime, given by h(x) = x mod p. Why
canвЂ™t the analysis of Problem 5 be used to show that h extends to a monomorphism of
the rationals to Zp ? (This canвЂ™t possibly work since Zp is п¬Ѓnite, but what goes wrong?)
7. Let S be a multiplicative subset of the commutative ring R, with f : R в†’ S в€’1 R deп¬Ѓned
by f (a) = a/1. If g is a ring homomorphism from R to a commutative ring R and g(s)
is a unit in R for each s в€€ S, we wish to п¬Ѓnd a ring homomorphism g : S в€’1 R в†’ R such
that g(f (a)) = g(a) for every a в€€ R, i.e., such that the diagram below is commutative.
Show that there is only one conceivable way to deп¬Ѓne g.

GR
g
R
x`
xx
xx
x
f
 xx g
S в€’1 R

8. Show that the mapping you have deп¬Ѓned in Problem 7 is a well-deп¬Ѓned ring homomor-
phism.

2.9 Irreducible Polynomials
In (2.6.1) we deп¬Ѓned an irreducible element of a ring; it is a nonzero nonunit which cannot
be represented as a product of nonunits. If R is an integral domain, we will refer to an
irreducible element of R[X] as an irreducible polynomial. Now in F [X], where F is a п¬Ѓeld,
the units are simply the nonzero elements of F (Section 2.1, Problem 2). Thus in this
case, an irreducible element is a polynomial of degree at least 1 that cannot be factored
into two polynomials of lower degree. A polynomial that is not irreducible is said to be
reducible or factorable. For example, X 2 + 1, regarded as an element of R[X], where R
is the п¬Ѓeld of real numbers, is irreducible, but if we replace R by the larger п¬Ѓeld C of
в€љ
complex numbers, X 2 + 1 is factorable as (X в€’ i)(X + i), i = в€’1. We say that X 2 + 1
is irreducible over R but not irreducible over C.
Now consider D[X], where D is a unique factorization domain but not necessarily
a п¬Ѓeld, for example, D = Z. The polynomial 12X + 18 is not an irreducible element
of Z[X] because it can be factored as the product of the two nonunits 6 and 2X + 3.
2.9. IRREDUCIBLE POLYNOMIALS 31

It is convenient to factor out the greatest common divisor of the coeп¬ѓcients (6 in this
case). The result is a primitive polynomial, one whose content (gcd of coeп¬ѓcients) is 1.
A primitive polynomial will be irreducible if and only if it cannot be factored into two
polynomials of lower degree.
In this section, we will compare irreducibility over a unique factorization domain D
and irreducibility over the quotient п¬Ѓeld F of D. Here is the key result.

2.9.2 Proposition
Let D be a unique factorization domain with quotient п¬Ѓeld F . Suppose that f is a nonzero
polynomial in D[X] and that f can be factored as gh, where g and h belong to F [X].
Then there is a nonzero element О» в€€ F such that О»g в€€ D[X] and О»в€’1 h в€€ D[X]. Thus if f
is factorable over F , then it is factorable over D. Equivalently, if f is irreducible over D,
then f is irreducible over F .

Proof. The coeп¬ѓcients of g and h are quotients of elements of D. If a is the least common
denominator for g (technically, the least common multiple of the denominators of the
coeп¬ѓcients of g), let g в€— = ag в€€ D[X]. Similarly, let hв€— = bh в€€ D[X]. Thus abf = g в€— hв€—
with g в€— , hв€— в€€ D[X] and c = ab в€€ D.
Now if p is a prime factor of c, we will show that either p divides all coeп¬ѓcients of g в€—
or p divides all coeп¬ѓcients of hв€— . We do this for all prime factors of c to get f = g0 h0
with g0 , h0 в€€ D[X]. Since going from g to g0 involves only multiplication or division by
nonzero constants in D, we have g0 = О»g for some nonzero О» в€€ F . But then h0 = О»в€’1 h,
as desired.
Now let

g в€— (X) = g0 + g1 X + В· В· В· + gs X s , hв€— (X) = h0 + h1 X + В· В· В· + ht X t .

Since p is a prime factor of c = ab and abf = g в€— hв€— , p must divide all coeп¬ѓcients of g в€— hв€— .
If p does not divide every gi and p does not divide every hi , let gu and hv be the coeп¬ѓcients
of minimum index not divisible by p. Then the coeп¬ѓcient of X u+v in g в€— hв€— is

g0 hu+v + g1 hu+vв€’1 + В· В· В· + gu hv + В· В· В· + gu+vв€’1 h1 + gu+v h0 .

But by choice of u and v, p divides every term of this expression except gu hv , so that p
cannot divide the entire expression. So there is a coeп¬ѓcient of g в€— hв€— not divisible by p, a

The technique of the above proof yields the following result.

2.9.3 GaussвЂ™ Lemma
Let f and g be nonconstant polynomials in D[X], where D is a unique factorization
domain. If c denotes content, then c(f g) = c(f )c(g), up to associates. In particular, the
product of two primitive polynomials is primitive.
32 CHAPTER 2. RING FUNDAMENTALS

Proof. By deп¬Ѓnition of content we may write f = c(f )f в€— and g = c(g)g в€— where f в€— and g в€—
are primitive. Thus f g = c(f )c(g)f в€— g в€— . It follows that c(f )c(g) divides every coeп¬ѓcient
of f g, so c(f )c(g) divides c(f g). Now let p be any prime factor of c(f g); then p divides
c(f )c(g)f в€— g в€— , and the proof of (2.9.2) shows that either p divides every coeп¬ѓcient of
c(f )f в€— or p divides every coeп¬ѓcient of c(g)g в€— . If, say, p divides every coeп¬ѓcient of c(f )f в€— ,
then (since p is prime) either p divides c(f ) or p divides every coeп¬ѓcient of f в€— . But f в€— is
primitive, so that p divides c(f ), hence p divides c(f )c(g). We conclude that c(f g) divides
c(f )c(g), and the result follows. в™Ј

2.9.4 Corollary of the Proof of (2.9.3)
If h is a nonconstant polynomial in D[X] and h = ahв€— where hв€— is primitive and a в€€ D,
then a must be the content of h.

Proof. Since a divides every coeп¬ѓcient of h, a must divide c(h). If p is any prime factor
of c(h), then p divides every coeп¬ѓcient of ahв€— , and as in (2.9.3), either p divides a or
p divides every coeп¬ѓcient of hв€— , which is impossible by primitivity of hв€— . Thus c(h)
divides a, and the result follows. в™Ј

Proposition 2.9.2 yields a precise statement comparing irreducibility over D with ir-
reducibility over F .

2.9.5 Proposition
Let D be a unique factorization domain with quotient п¬Ѓeld F . If f is a nonconstant
polynomial in D[X], then f is irreducible over D if and only if f is primitive and irreducible
over F .

Proof. If f is irreducible over D, then f is irreducible over F by (2.9.2). If f is not
primitive, then f = c(f )f в€— where f в€— is primitive and c(f ) is not a unit. This contradicts
the irreducibility of f over D. Conversely, if f = gh is a factorization of the primitive
polynomial f over D, then g and h must be of degree at least 1. Thus neither g nor h is
a unit in F [X], so f = gh is a factorization of f over F . в™Ј

Here is another basic application of (2.9.2).

2.9.6 Theorem
If R is a unique factorization domain, so is R[X].

Proof. If f в€€ R[X], f = 0, then f can be factored over the quotient п¬Ѓeld F as f =
f1 f2 . . . fk , where the fi are irreducible polynomials in F [X]. (Recall that F [X] is a
Euclidean domain, hence a unique factorization domain.) By (2.9.2), for some nonzero
О»1 в€€ F we may write f = (О»1 f1 )(О»в€’1 f2 . . . fk ) with О»1 f1 and О»в€’1 f2 . . . fk in R[X]. Again
1 1
by (2.9.2), we have

О»в€’1 f2 . . . fk = f2 О»в€’1 f3 . . . fk = (О»2 f2 )(О»в€’1 О»в€’1 f3 . . . fk )
1 1 2 1
2.9. IRREDUCIBLE POLYNOMIALS 33

with О»2 f2 and О»в€’1 О»в€’1 f3 . . . fk в€€ R[X]. Continuing inductively, we express f as i=1 О»i fi
k
2 1
where the О»i fi are in R[X] and are irreducible over F . But О»i fi is the product of its
content and a primitive polynomial (which is irreducible over F , hence over R by (2.9.5)).
Furthermore, the content is either a unit or a product of irreducible elements of the UFD
R, and these elements are irreducible in R[X] as well. This establishes the existence of a
factorization into irreducibles.
Now suppose that f = g1 В· В· В· gr = h1 В· В· В· hs , where the gi and hi are nonconstant
irreducible polynomials in R[X]. (Constant polynomials cause no diп¬ѓculty because R is
a UFD.) By (2.9.5), the gi and hi are irreducible over F , and since F [X] is a UFD, we
have r = s and, after reordering if necessary, gi and hi are associates (in F [X]) for each
i. Now gi = ci hi for some constant ci в€€ F , and we have ci = ai /bi with ai , bi в€€ R. Thus
bi gi = ai hi , with gi and hi primitive by (2.9.5). By (2.9.4), bi gi has content bi and ai hi
has content ai . Therefore ai and bi are associates, which makes ci a unit in R, which in
turn makes gi and hi associates in R[X], proving uniqueness of factorization. в™Ј

The following result is often used to establish irreducibility of a polynomial.

2.9.7 EisensteinвЂ™s Irreducibility Criterion
Let R be a UFD with quotient п¬Ѓeld F , and let f (X) = an X n + В· В· В· + a1 X + a0 be a
polynomial in R[X], with n в‰Ґ 1 and an = 0. If p is prime in R and p divides ai for
0 в‰¤ i < n, but p does not divide an and p2 does not divide a0 , then f is irreducible over
F . Thus by (2.9.5), if f is primitive then f is irreducible over R.

Proof. If we divide f by its content to produce a primitive polynomial f в€— , the hypothesis
still holds for f в€— . (Since p does not divide an , it is not a prime factor of c(f ), so it must
divide the ith coeп¬ѓcient of f в€— for 0 в‰¤ i < n.) If we can prove that f в€— is irreducible
over R, then by (2.9.5), f в€— is irreducible over F , and therefore so is f . Thus we may
assume without loss of generality that f is primitive, and prove that f is irreducible
over R.
Assume that f = gh, with g(X) = g0 +В· В· В·+gr X r and h(X) = h0 +В· В· В·+hs X s . If r = 0
then g0 divides all coeп¬ѓcients ai of f , so g0 divides c(f ), hence g(= g0 ) is a unit. Thus we
may assume that r в‰Ґ 1, and similarly s в‰Ґ 1. By hypothesis, p divides a0 = g0 h0 but p2
does not divide a0 , so p cannot divide both g0 and h0 . Assume that p fails to divide h0 ,
so that p divides g0 ; the argument is symmetrical in the other case. Now gr hs = an , and
by hypothesis, p does not divide an , so that p does not divide gr . Let i be the smallest
integer such that p does not divide gi ; then 1 в‰¤ i в‰¤ r < n (since r + s = n and s в‰Ґ 1).
Now

ai = g0 hi + g1 hiв€’1 + В· В· В· + gi h0

and by choice of i, p divides g0 , . . . , giв€’1 . But p divides the entire sum ai , so p must divide
the last term gi h0 . Consequently, either p divides gi , which contradicts the choice of i, or
p divides h0 , which contradicts our earlier assumption. Thus there can be no factorization
of f as a product of polynomials of lower degree; in other words, f is irreducible over R. в™Ј
34 CHAPTER 2. RING FUNDAMENTALS

Problems For Section 2.9
1. (The rational root test, which can be useful in factoring a polynomial over Q.)
Let f (X) = an X n + В· В· В· + a1 X + a0 в€€ Z[X]. If f has a rational root u/v where u and v
are relatively prime integers and v = 0, show that v divides an and u divides a0 .
2. Show that for every positive integer n, there is at least one irreducible polynomial of
degree n over the integers.
3. If f (X) в€€ Z[X] and p is prime, we can reduce all coeп¬ѓcients of f modulo p to obtain
a new polynomial fp (X) в€€ Zp [X]. If f is factorable over Z, then fp is factorable over
Zp . Therefore if fp is irreducible over Zp , then f is irreducible over Z. Use this idea
to show that the polynomial X 3 + 27X 2 + 5X + 97 is irreducible over Z. (Note that
Eisenstein does not apply.)
4. If we make a change of variable X = Y + c in the polynomial f (X), the result is a new
polynomial g(Y ) = f (Y + c). If g is factorable over Z, so is f since f (X) = g(X в€’ c).
Thus if f is irreducible over Z, so is g. Use this idea to show that X 4 +4X 3 +6X 2 +4X+4
is irreducible over Z.
5. Show that in Z[X], the ideal n, X , n в‰Ґ 2, is not principal, and therefore Z[X] is a
UFD that is not a PID.
aij X i Y j , aij в€€ F ,
6. Show that if F is a п¬Ѓeld, then F [X, Y ], the set of all polynomials
is not a PID since the ideal X, Y is not principal.
7. Let f (X, Y ) = X 2 + Y 2 + 1 в€€ C[X, Y ], where C is the п¬Ѓeld of complex numbers.
Write f as Y 2 + (X 2 + 1) and use EisensteinвЂ™s criterion to show that f is irreducible
over C.
8. Show that f (X, Y ) = X 3 + Y 3 + 1 is irreducible over C.
Chapter 3

Field Fundamentals

3.1 Field Extensions
If F is a п¬Ѓeld and F [X] is the set of all polynomials over F, that is, polynomials with
coeп¬ѓcients in F , we know that F [X] is a Euclidean domain, and therefore a principal ideal
domain and a unique factorization domain (see Sections 2.6 and 2.7). Thus any nonzero
polynomial f in F [X] can be factored uniquely as a product of irreducible polynomials.
Any root of f must be a root of one of the irreducible factors, but at this point we have
no concrete information about the existence of roots and how they might be found. For
example, X 2 + 1 has no real roots, but if we consider the larger п¬Ѓeld of complex numbers,
we get two roots, +i and в€’i. It appears that the process of passing to a larger п¬Ѓeld may
help produce roots, and this turns out to be correct.

3.1.1 Deп¬Ѓnitions
If F and E are п¬Ѓelds and F вЉ† E, we say that E is an extension of F , and we write F в‰¤ E,
or sometimes E/F .
If E is an extension of F , then in particular E is an abelian group under addition, and
we may multiply the вЂњvectorвЂќ x в€€ E by the вЂњscalarвЂќ О» в€€ F , and the axioms of a vector
space are satisп¬Ѓed. Thus if F в‰¤ E, then E is a vector space over F . The dimension of this
vector space is called the degree of the extension, written [E : F ]. If [E : F ] = n < в€ћ, we
say that E is a п¬Ѓnite extension of F , or that the extension E/F is п¬Ѓnite, or that E is of
degree n over F .
If f is a nonconstant polynomial over the п¬Ѓeld F , and f has no roots in F , we can
always produce a root of f in an extension п¬Ѓeld of F . We do this after a preliminary
result.

3.1.2 Lemma
Let f : F в†’ E be a homomorphism of п¬Ѓelds, i.e., f (a + b) = f (a) + f (b), f (ab) = f (a)f (b)
(all a, b в€€ F ), and f (1F ) = 1E . Then f is a monomorphism.

1
2 CHAPTER 3. FIELD FUNDAMENTALS

Proof. First note that a п¬Ѓeld F has no ideals except {0} and F . For if a is a nonzero
member of the ideal I, then ab = 1 for some b в€€ F , hence 1 в€€ I, and therefore I = F .
Taking I to be the kernel of f , we see that I cannot be all of F because f (1) = 0. Thus
I must be {0}, so that f is injective. в™Ј

3.1.3 Theorem
Let f be a nonconstant polynomial over the п¬Ѓeld F . Then there is an extension E/F and
an element О± в€€ E such that f (О±) = 0.
Proof. Since f can be factored into irreducibles, we may assume without loss of generality
that f itself is irreducible. The ideal I = f (X) in F [X] is prime (see (2.6.1)), in fact
maximal (see (2.6.9)). Thus E = F [X]/I is a п¬Ѓeld by (2.4.3). We have a problem at this
point because F need not be a subset of E, but we can place an isomorphic copy of F
inside E via the homomorphism h : a в†’ a + I; by (3.1.2), h is a monomorphism, so we
may identify F with a subп¬Ѓeld of E. Now let О± = X + I; if f (X) = a0 + a1 X + В· В· В· + an X n ,
then

f (О±) = (a0 + I) + a1 (X + I) + В· В· В· + an (X + I)n
= (a0 + a1 X + В· В· В· + an X n ) + I
= f (X) + I

which is zero in E. в™Ј
The extension E is sometimes said to be obtained from F by adjoining a root О± of f .
Here is a further connection between roots and extensions.

3.1.4 Proposition
Let f and g be polynomials over the п¬Ѓeld F . Then f and g are relatively prime if and
only if f and g have no common root in any extension of F .
Proof. If f and g are relatively prime, their greatest common divisor is 1, so there are
polynomials a(X) and b(X) over F such that a(X)f (X) + b(X)g(X) = 1. If О± is a
common root of f and g, then the substitution of О± for X yields 0 = 1, a contradiction.
Conversely, if the greatest common divisor d(X) of f (X) and g(X) is nonconstant, let E
be an extension of F in which d(X) has a root О± (E exists by (3.1.3)). Since d(X) divides
both f (X) and g(X), О± is a common root of f and g in E. в™Ј

3.1.5 Corollary
If f and g are distinct monic irreducible polynomials over F , then f and g have no common
roots in any extension of F .
Proof. If h is a nonconstant divisor of the irreducible polynomials f and g, then up
to multiplication by constants, h coincides with both f and g, so that f is a constant
multiple of g. This is impossible because f and g are monic and distinct. Thus f and g
are relatively prime, and the result follows from (3.1.4). в™Ј
3.1. FIELD EXTENSIONS 3

If E is an extension of F and О± в€€ E is a root of a polynomial f в€€ F [X], it is often
of interest to examine the п¬Ѓeld F (О±) generated by F and О±, in other words the smallest
subп¬Ѓeld of E containing F and О± (more precisely, containing all elements of F along
with О±). The п¬Ѓeld F (О±) can be described abstractly as the intersection of all subп¬Ѓelds of
E containing F and О±, and more concretely as the collection of all rational functions
a0 + a1 О± + В· В· В· + am О±m
b0 + b1 О± + В· В· В· + b n О± n
with ai , bj в€€ F, m, n = 0, 1, . . . , and b0 + b1 О± + В· В· В· + bn О±n = 0. In fact there is a much
less complicated description of F (О±), as we will see shortly.

If E is an extension of F , the element О± в€€ E is said to be algebraic over F is there is a
nonconstant polynomial f в€€ F [X] such that f (О±) = 0; if О± is not algebraic over F , it is
said to be transcendental over F . If every element of E is algebraic over F , then E is said
to be an algebraic extension of F .
Suppose that О± в€€ E is algebraic over F , and let I be the set of all polynomials g
over F such that g(О±) = 0. If g1 and g2 belong to I, so does g1 В± g2 , and if g в€€ I and
c в€€ F [X], then cg в€€ I. Thus I is an ideal of F [X], and since F [X] is a PID, I consists of
all multiples of some m(X) в€€ F [X]. Any two such generators must be multiples of each
other, so if we require that m(X) be monic, then m(X) is unique. The polynomial m(X)
has the following properties:

(1) If g в€€ F [X], then g(О±) = 0 if and only if m(X) divides g(X).
(2) m(X) is the monic polynomial of least degree such that m(О±) = 0.
(3) m(X) is the unique monic irreducible polynomial such that m(О±) = 0.

Property (1) follows because g(О±) = 0 iп¬Ђ g(X) в€€ I, and I = m(X) , the ideal generated
by m(X). Property (2) follows from (1). To prove (3), note that if m(X) = h(X)k(X)
with deg h and deg k less than deg m, then either h(О±) = 0 or k(О±) = 0, so that by (1),
either h(X) or k(X) is a multiple of m(X), which is impossible. Thus m(X) is irreducible,
and uniqueness of m(X) follows from (3.1.5).
The polynomial m(X) is called the minimal polynomial of О± over F , sometimes written
as min(О±, F ).

3.1.7 Theorem
If О± в€€ E is algebraic over F and the minimal polynomial m(X) of О± over F has degree
n, then F (О±) = F [О±], the set of polynomials in О± with coeп¬ѓcients in F . In fact, F [О±]
is the set Fnв€’1 [О±] of all polynomials of degree at most n в€’ 1 with coeп¬ѓcients in F , and
1, О±, . . . , О±nв€’1 form a basis for the vector space F [О±] over the п¬Ѓeld F . Consequently,
[F (О±) : F ] = n.

Proof. Let f (X) be any nonzero polynomial over F of degree n в€’ 1 or less. Then since
m(X) is irreducible and deg f < deg m, f (X) and m(X) are relatively prime, and there
4 CHAPTER 3. FIELD FUNDAMENTALS

are polynomials a(X) and b(X) over F such that a(X)f (X) + b(X)m(X) = 1. But
then a(О±)f (О±) = 1, so that any nonzero element of Fnв€’1 [О±] has a multiplicative inverse.
It follows that Fnв€’1 [О±] is a п¬Ѓeld. (This may not be obvious, since the product of two
polynomials of degree n в€’ 1 or less can have degree greater than n в€’ 1, but if deg g > n в€’ 1,
then divide g by m to get g(X) = q(X)m(X) + r(X) where deg r(X) < deg m(X) = n.
Replace X by О± to get g(О±) = r(О±) в€€ Fnв€’1 [О±]. Less abstractly, if m(О±) = О±3 + О± + 1 = 0,
then О±3 = в€’О± в€’ 1, О±4 = в€’О±2 в€’ О±, and so on.)
Now any п¬Ѓeld containing F and О± must contain all polynomials in О±, in particular
all polynomials of degree at most n в€’ 1. Therefore Fnв€’1 [О±] вЉ† F [О±] вЉ† F (О±). But F (О±)
is the smallest п¬Ѓeld containing F and О±, so F (О±) вЉ† Fnв€’1 [О±], and we conclude that
F (О±) = F [О±] = Fnв€’1 [О±]. Finally, the elements 1, О±, . . . , О±nв€’1 certainly span Fnв€’1 [О±], and
they are linearly independent because if a nontrivial linear combination of these elements
were zero, we would have a nonzero polynomial of degree less than that of m(X) with О±
as a root, contradicting (2) of (3.1.6). в™Ј

We now prove a basic multiplicativity result for extensions, after a preliminary dis-
cussion.

3.1.8 Lemma
Suppose that F в‰¤ K в‰¤ E, the elements О±i , i в€€ I, form a basis for E over K, and the
elements ОІj , j в€€ J, form a basis for K over F . (I and J need not be п¬Ѓnite.) Then the
products О±i ОІj , i в€€ I, j в€€ J, form a basis for E over F .

Proof. If Оі в€€ E, then Оі is a linear combination of the О±i with coeп¬ѓcients ai в€€ K, and
each ai is a linear combination of the ОІj with coeп¬ѓcients bij в€€ F . It follows that the О±i ОІj
span E over F . Now if i,j О»ij О±i ОІj = 0, then i О»ij О±i = 0 for all j, and consequently
О»ij = 0 for all i, j, and the О±i ОІj are linearly independent. в™Ј

3.1.9 The Degree is Multiplicative
If F в‰¤ K в‰¤ E, then [E : F ] = [E : K][K : F ]. In particular, [E : F ] is п¬Ѓnite if and only if
[E : K] and [K : F ] are both п¬Ѓnite.

Proof. In (3.1.8), we have [E : K] = |I|, [K : F ] = |J|, and [E : F ] = |I||J|. в™Ј

We close this section by proving that every п¬Ѓnite extension is algebraic.

3.1.10 Theorem
If E is a п¬Ѓnite extension of F , then E is an algebraic extension of F .

Proof. Let О± в€€ E, and let n = [E : F ]. Then 1, О±, О±2 , . . . , О±n are n + 1 vectors in an
n-dimensional vector space, so they must be linearly dependent. Thus О± is a root of a
nonzero polynomial with coeп¬ѓcients in F , which means that О± is algebraic over F . в™Ј
3.2. SPLITTING FIELDS 5

Problems For Section 3.1
1. Let E be an extension of F , and let S be a subset of E. If F (S) is the subп¬Ѓeld of E
generated by S over F , in other words, the smallest subп¬Ѓeld of E containing F and S,
describe F (S) explicitly, and justify your characterization.
2. If for each i в€€ I, Ki is a subп¬Ѓeld of the п¬Ѓeld E, the composite of the Ki (notation
i Ki ) is the smallest subп¬Ѓeld of E containing every Ki . As in Problem 1, describe
the composite explicitly.
3. Assume that О± is algebraic over F , with [F [О±] : F ] = n. If ОІ в€€ F [О±], show that
[F [ОІ] : F ] в‰¤ n, in fact [F [ОІ] : F ] divides n.
в€љ
4. The minimal polynomial of 2 over the rationals Q is X 2 в€’ 2, by (3) of (3.1.6). Thus
в€љ в€љ
Q[ 2] consists of all numbers of the form a0 + a1 2, where a0 and a1 are rational.
в€љ
By Problem 3, we know that в€’1 + 2 has a minimal polynomial over Q of degree at
most 2. Find this minimal polynomial.
5. If О± is algebraic over F and ОІ belongs to F [О±], describe a systematic procedure for
п¬Ѓnding the minimal polynomial of ОІ over F .
6. If E/F and the element О± в€€ E is transcendental over F , show that F (О±) is isomorphic
to F (X), the п¬Ѓeld of rational functions with coeп¬ѓcients in F .
7. Theorem 3.1.3 gives one method of adjoining a root of a polynomial, and in fact there
is essentially only one way to do this. If E is an extension of F and О± в€€ E is algebraic
over F with minimal polynomial m(X), let I be the ideal m(X) вЉ† F [X]. Show that
F (О±) is isomorphic to F [X]/I. [Deп¬Ѓne П• : F [X] в†’ E by П•(f (X)) = f (О±), and use
the п¬Ѓrst isomorphism theorem for rings.]
8. In the proof of (3.1.3), we showed that if f is irreducible in F [X], then I = f is a
maximal ideal. Show that conversely, if I is a maximal ideal, then f is irreducible.
9. Suppose that F в‰¤ E в‰¤ L, with О± в€€ L. What is the relation between the minimal
polynomial of О± over F and the minimal polynomial of О± over E?
10. If О±1 , . . . , О±n are algebraic over F , we can successively adjoin the О±i to F to obtain
the п¬Ѓeld F [О±1 , . . . , О±n ] consisting of all polynomials over F in the О±i . Show that
n
[F [О±1 , . . . , О±n ] : F ] в‰¤ [F (О±i ) : F ] < в€ћ
i=1

3.2 Splitting Fields
If f is a polynomial over the п¬Ѓeld F , then by (3.1.3) we can п¬Ѓnd an extension E1 of F
containing a root О±1 of f . If not all roots of f lie in E1 , we can п¬Ѓnd an extension E2
of E1 containing another root О±2 of f . If we continue the process, eventually we reach a
complete factorization of f . In this section we examine this idea in detail.
If E is an extension of F and О±1 , . . . , О±k в€€ E, we will use the notation F (О±1 , . . . , О±k )
for the subп¬Ѓeld of E generated by F and the О±i . Thus F (О±1 , . . . , О±k ) is the smallest
subп¬Ѓeld of E containing all elements of F along with the О±i . ( вЂњSmallestвЂќ means that
F (О±1 , . . . , О±k ) is the intersection of all such subп¬Ѓelds.) Explicitly, F (О±1 , . . . , О±k ) is the
collection of all rational functions in the О±i with nonzero denominators.
6 CHAPTER 3. FIELD FUNDAMENTALS

If E is an extension of F and f в€€ F [X], we say that f splits over E if f can be written
as О»(X в€’ О±1 ) В· В· В· (X в€’ О±k ) for some О±1 , . . . , О±k в€€ E and О» в€€ F .
(There is a subtle point that should be mentioned. We would like to refer to the О±i
as вЂњtheвЂќ roots of f , but in doing so we are implicitly assuming that if ОІ is an element of
some extension E of E and f (ОІ) = 0, then ОІ must be one of the О±i . This follows upon
substituting ОІ into the equation f (X) = О»(X в€’ О±1 ) В· В· В· (X в€’ О±k ) = 0.)
If K is an extension of F and f в€€ F [X], we say that K is a splitting п¬Ѓeld for f over
F if f splits over K but not over any proper subп¬Ѓeld of K containing F .
Equivalently, K is a splitting п¬Ѓeld for f over F if f splits over K and K is generated
over F by the roots О±1 , . . . , О±k of f , in other words, F (О±1 , . . . , О±k ) = K. For if K is a
splitting п¬Ѓeld for f , then since f splits over K we have all О±j в€€ K, so F (О±1 , . . . , О±k ) вЉ† K.
But f splits over F (О±1 , . . . , О±k ), and it follows that F (О±1 , . . . , О±k ) cannot be a proper
subп¬Ѓeld; it must coincide with K. Conversely, if f splits over K and F (О±1 , . . . , О±k ) = K,
let L be a subп¬Ѓeld of K containing F . If f splits over L then all О±i belong to L, so
K = F (О±1 , . . . , О±k ) вЉ† L вЉ† K, so L = K.
If f в€€ F [X] and f splits over the extension E of F , then E contains a unique splitting
п¬Ѓeld for f , namely F (О±1 , . . . , О±k ).

3.2.2 Proposition
If f в€€ F [X] and deg f = n, then f has a splitting п¬Ѓeld K over F with [K : F ] в‰¤ n!.
Proof. We may assume that n в‰Ґ 1. (If f is constant, take K = F .) By (3.1.3), F has an
extension E1 containing a root О±1 of f , and the extension F (О±1 )/F has degree at most
n. (Since f (О±1 ) = 0, the minimal polynomial of О±1 divides f ; see (3.1.6) and (3.1.7).) We
may then write f (X) = (X в€’ О±1 )r1 g(X), where О±1 is not a root of g and deg g в‰¤ n в€’ 1.
If g is nonconstant, we can п¬Ѓnd an extension of F (О±1 ) containing a root О±2 of g, and the
extension F (О±1 , О±2 ) will have degree at most n в€’ 1 over F (О±1 ). Continue inductively and
use (3.1.9) to reach an extension of degree at most n! containing all the roots of f . в™Ј
If f в€€ F [X] and f splits over E, then we may pick any root О± of f and adjoin it to F
to obtain the extension F (О±). Roots of the same irreducible factor of f yield essentially
the same extension, as the next result shows.

3.2.3 Theorem
If О± and ОІ are roots of the irreducible polynomial f в€€ F [X] in an extension E of F , then
F (О±) is isomorphic to F (ОІ) via an isomorphism that carries О± into ОІ and is the identity
on F .
Proof. Without loss of generality we may assume f monic (if not, divide f by its leading
coeп¬ѓcient). By (3.1.6), part (3), f is the minimal polynomial of both О± and ОІ. By (3.1.7),
the elements of F (О±) can be expressed uniquely as a0 + a1 О± + В· В· В· + anв€’1 О±nв€’1 , where the
ai belong to F and n is the degree of f . The desired isomorphism is given by
a0 + a1 О± + В· В· В· + anв€’1 О±nв€’1 в†’ a0 + a1 ОІ + В· В· В· + anв€’1 ОІ nв€’1 . в™Ј
3.2. SPLITTING FIELDS 7

If f is a polynomial in F [X] and F is isomorphic to the п¬Ѓeld F via the isomorphism i,
we may regard f as a polynomial over F . We simply use i to transfer f . Thus if
f = a0 + a1 X + В· В· В· an X n , then f = i(f ) = i(a0 ) + i(a1 )X + В· В· В· + i(an )X n . There is only
a notational diп¬Ђerence between f and f , and we expect that splitting п¬Ѓelds for f and f
should also be essentially the same. We prove this after the following deп¬Ѓnition.

3.2.4 Deп¬Ѓnition
If E and E are extensions of F and i is an isomorphism of E and E , we say that i is
an F -isomorphism if i п¬Ѓxes F , that is, if i(a) = a for every a в€€ F . F -homomorphisms,
F -monomorphisms, etc., are deп¬Ѓned similarly.

3.2.5 Isomorphism Extension Theorem
Suppose that F and F are isomorphic, and the isomorphism i carries the polynomial
f в€€ F [X] to f в€€ F [X]. If K is a splitting п¬Ѓeld for f over F and K is a splitting п¬Ѓeld
for f over F , then i can be extended to an isomorphism of K and K . In particular, if
F = F and i is the identity function, we conclude that any two splitting п¬Ѓelds of f are
F -isomorphic.

Proof. Carry out the construction of a splitting п¬Ѓeld for f over F as in (3.2.2), and perform
exactly the same steps to construct a splitting п¬Ѓeld for f over F . At every stage, there is
only a notational diп¬Ђerence between the п¬Ѓelds obtained. Furthermore, we can do the п¬Ѓrst
construction inside K and the second inside K . But the comment at the end of (3.2.1)
shows that the splitting п¬Ѓelds that we have constructed coincide with K and K . в™Ј

3.2.6 Example
We will п¬Ѓnd a splitting п¬Ѓeld for f (X) = X 3 в€’ 2 over the rationals Q. в€љ
If О± is the positive cube root of 2, then the roots of f are О±, О±(в€’ 1 + i 1 3) and
в€љ 2 2
О±(в€’ 2 в€’ i 2 3). The polynomial f is irreducible, either by EisensteinвЂ™s criterion or by the
1 1

observation that if f were factorable, it would have a linear factor, and there is no rational
number whose cube is 2. Thus f is the minimal polynomial of О±, so [Q(О±) : Q] в€љ 3. Now
=
в€љ
since О± and i 3 generate all the roots of f , the splitting п¬Ѓeld is K = Q(О±, i 3). (We
в€љ
complex numbers C.) Since i 3 в€љ Q(О±)
в€€
regard all п¬Ѓelds in this example as subп¬Ѓelds of the в€љ /
(because Q(О±) is a subп¬Ѓeld of the reals), [Q(О±, iв€љ 3) : Q(О±)] is at least 2. But i 3 is a
root of X 2 + 3 в€€ Q(О±)[X], so the degree of Q(О±, i 3) over Q(О±) is a most 2, and therefore
is exactly 2. Thus
в€љ в€љ
[K : Q] = [Q(О±, i 3) : Q] = [Q(О±, i 3) : Q(О±)][Q(О±) : Q] = 2 Г— 3 = 6.

Problems For Section 3.2
1. Find a splitting п¬Ѓeld for f (X) = X 2 в€’ 4X + 4 over Q.
2. Find a splitting п¬Ѓeld K for f (X) = X 2 в€’ 2X + 4 over Q, and determine the degree of
K over Q.
8 CHAPTER 3. FIELD FUNDAMENTALS

3. Find a splitting п¬Ѓeld K for f (X) = X 4 в€’ 2 over Q, and determine [K : Q].
4. Let C be a family of polynomials over F , and let K be an extension of F . Show that
the following two conditions are equivalent:
(a) Each f в€€ C splits over K, but if F в‰¤ K < K, then it is not true that each f в€€ C
splits over K .
(b) Each f в€€ C splits over K, and K is generated over F by the roots of all the
polynomials in C.
If one, and hence both, of these conditions are satisп¬Ѓed, we say that K is a splitting
п¬Ѓeld for C over F .
5. Suppose that K is a splitting п¬Ѓeld for the п¬Ѓnite set of polynomials {f1 , . . . , fr } over F .
Express K as a splitting п¬Ѓeld for a single polynomial f over F .
6. If m and n are в€љ в€љ square-free positive integers greater than 1, show that the
distinct
splitting п¬Ѓeld Q( m, n) of (X 2 в€’ m)(X 2 в€’ n) has degree 4 over Q.

3.3 Algebraic Closures
If f is a polynomial of degree n over the rationals or the reals, or more generally over
the complex numbers, then f need not have any rational roots, or even real roots, but we
know that f always has n complex roots, counting multiplicity. This favorable situation
can be duplicated for any п¬Ѓeld F , that is, we can construct an algebraic extension C of F
with the property that any polynomial in C[X] splits over C. There are many ways to
express this idea.

3.3.1 Proposition
If C is a п¬Ѓeld, the following conditions are equivalent:

(1) Every nonconstant polynomial f в€€ C[X] has at least one root in C.
(2) Every nonconstant polynomial f в€€ C[X] splits over C.
(3) Every irreducible polynomial f в€€ C[X] is linear.
(4) C has no proper algebraic extensions.

If any (and hence all) of these conditions are satisп¬Ѓed, we say that C is algebraically closed.

Proof. (1) implies (2): By (1) we may write f = (X в€’ О±1 )g. Proceed inductively to show
that any nonconstant polynomial is a product of linear factors.
(2) implies (3): If f is an irreducible polynomial in C[X], then by (2.9.1), f is non-
constant. By (2), f is a product of linear factors. But f is irreducible, so there can be
only one such factor.
(3) implies (4): Let E be an algebraic extension of C. If О± в€€ E, let f be the minimal
polynomial of О± over C. Then f is irreducible and by (3), f is of the form X в€’ О±. But
then О± в€€ C, so E = C.
3.3. ALGEBRAIC CLOSURES 9

(4) implies (1): Let f be a nonconstant polynomial in C[X], and adjoin a root О± of f
to obtain C(О±), as in (3.1.3). But then C(О±) is an algebraic extension of C, so by (4),
О± в€€ C. в™Ј
It will be useful to embed an arbitrary п¬Ѓeld F in an algebraically closed п¬Ѓeld.

An extension C of F is an algebraic closure of F if C is algebraic over F and C is
algebraically closed.
Note that C is minimal among algebraically closed extensions of F . For if F в‰¤ K в‰¤ C
and О± в€€ C, О± в€€ K, then since О± is algebraic over F it is algebraic over K. But since
/
О± в€€ K, the minimal polynomial of О± over K is a nonlinear irreducible polynomial in
/
K[X]. By (3) of (3.3.1), K cannot be algebraically closed.
If C is an algebraic extension of F , then in order for C to be an algebraic closure of F
it is suп¬ѓcient that every polynomial in F [X] (rather than C[X]) splits over C. To prove
this, we will need the following result.

3.3.3 Proposition
If E is generated over F by п¬Ѓnitely many elements О±1 , . . . , О±n algebraic over F (so that
E = F (О±1 , . . . , О±n )), then E is a п¬Ѓnite extension of F .
Proof. Set E0 = F and Ek = F (О±1 , . . . , О±k ), 1 в‰¤ k в‰¤ n (so En = E). Then Ek =
Ekв€’1 (О±k ), where О±k is algebraic over F and hence over Ekв€’1 . But by (3.1.7), [Ek : Ekв€’1 ]
is the degree of the minimal polynomial of О±k over Ekв€’1 , which is п¬Ѓnite. By (3.1.9),
n
[E : F ] = k=1 [Ek : Ekв€’1 ] < в€ћ. в™Ј

3.3.4 Corollary
If E is an extension of F and A is the set of all elements in E that are algebraic over F
(the algebraic closure of F in E), then A is a subп¬Ѓeld of E.
Proof. If О±, ОІ в€€ A, then the sum, diп¬Ђerence, product and quotient (if ОІ = 0) of О± and ОІ
belong to F (О±, ОІ), which is a п¬Ѓnite extension of F by (3.3.3), and therefore an algebraic
extension of F by (3.1.10). But then О± + ОІ, О± в€’ ОІ, О±ОІ and О±/ОІ belong to A, proving that
A is a п¬Ѓeld. в™Ј

3.3.5 Corollary (Transitivity of Algebraic Extensions)
If E is algebraic over K (in other words, every element of E is algebraic over K), and K
is algebraic over F , then E is algebraic over F .
Proof. Let О± в€€ E, and let m(X) = b0 + b1 X + В· В· В· + bnв€’1 X nв€’1 + X n be the minimal
polynomial of О± over K. The bi belong to K and are therefore algebraic over F . If
L = F (b0 , b1 , . . . , bnв€’1 ), then by (3.3.3), L is a п¬Ѓnite extension of F . Since the coeп¬ѓcients
of m(X) belong to L, О± is algebraic over L, so by (3.1.7), L(О±) is a п¬Ѓnite extension of L.
By (3.1.9), L(О±) is a п¬Ѓnite extension of F . By (3.1.10), О± is algebraic over F . в™Ј
10 CHAPTER 3. FIELD FUNDAMENTALS

Now we can add another condition to (3.3.1).

3.3.6 Proposition
Let C be an algebraic extension of F . Then C is an algebraic closure of F if and only if
every nonconstant polynomial in F [X] splits over C.

Proof. The вЂњonly ifвЂќ part follows from (2) of (3.3.1), since F вЉ† C. Thus assume that
every nonconstant polynomial in F [X] splits over C. If f is a nonconstant polynomial
in C[X], we will show that f has at least one root in C, and it will follow from (1) of
(3.3.1) that C is algebraically closed. Adjoin a root О± of f to obtain the extension C(О±).
Then C(О±) is algebraic over C by (3.1.7), and C is algebraic over F by hypothesis. By
(3.3.5), C(О±) is algebraic over F , so О± is algebraic over F . But then О± is a root of some
polynomial g в€€ F [X], and by hypothesis, g splits over C. By deп¬Ѓnition of вЂњsplitsвЂќ (see
(3.2.1)), all roots of g lie in C, in particular О± в€€ C. Thus f has at least one root in C. в™Ј

To avoid a lengthy excursion into formal set theory, we argue intuitively to establish
the following three results. (For complete proofs, see the appendix to Chapter 3.)

3.3.7 Theorem
Every п¬Ѓeld F has an algebraic closure.

Informal argument. Well-order F [X] and use transп¬Ѓnite induction, beginning with the
п¬Ѓeld F0 = F . At stage f we adjoin all roots of the polynomial f by constructing a
splitting п¬Ѓeld for f over the п¬Ѓeld F<f that has been generated so far by the recursive
procedure. When we reach the end of the process, we will have a п¬Ѓeld C such that every
polynomial f in F [X] splits over C. By (3.3.6), C is an algebraic closure of F . в™Ј

3.3.8 Theorem
Any two algebraic closures C and C of F are F -isomorphic.

Informal argument. Carry out the recursive procedure described in (3.3.7) in both C
and C . At each stage we may use the fact that any two splitting п¬Ѓelds of the same
polynomial are F -isomorphic; see (3.2.5). When we п¬Ѓnish, we have F -isomorphic algebraic
closures of F , say D вЉ† C and D вЉ† C . But an algebraic closure is a minimal algebraically
closed extension by (3.3.2), and therefore D = C and D = C . в™Ј

3.3.9 Theorem
If E is an algebraic extension of F , C is an algebraic closure of F , and i is an embedding
(that is, a monomorphism) of F into C, then i can be extended to an embedding of E
into C.
3.4. SEPARABILITY 11

Informal argument. Each О± в€€ E is a root of some polynomial in F [X], so if we allow О± to
range over all of E, we get a collection S of polynomials in F [X]. Within C, carry out the
recursive procedure of (3.3.7) on the polynomials in S. The resulting п¬Ѓeld lies inside C
and contains an F -isomorphic copy of E. в™Ј

Problems For Section 3.3
1. Show that the converse of (3.3.3) holds, that is, if E is a п¬Ѓnite extension of F , then E
is generated over F by п¬Ѓnitely many elements that are algebraic over F .
2. An algebraic number is a complex number that is algebraic over the rational п¬Ѓeld Q.
A transcendental number is a complex number that is not algebraic over Q. Show that
there only countably many algebraic numbers, and consequently there are uncountably
many transcendental numbers.
3. Give an example of an extension C/F such that C is algebraically closed but C is not
an algebraic extension of F .
4. Give an example of an extension E/F such that E is an algebraic but not a п¬Ѓnite
extension of F .
5. In the proof of (3.3.7), why is C algebraic over F ?
6. Show that the set A of algebraic numbers is an algebraic closure of Q.
7. If E is an algebraic extension of the inп¬Ѓnite п¬Ѓeld F , show that |E| = |F |.
8. Show that any set S of nonconstant polynomials in F [X] has a splitting п¬Ѓeld over F .
9. Show that an algebraically closed п¬Ѓeld must be inп¬Ѓnite.

3.4 Separability
If f is a polynomial in F [X], we can construct a splitting п¬Ѓeld K for f over F , and all
roots of f must lie in K. In this section we investigate the multiplicity of the roots.

An irreducible polynomial f в€€ F [X] is separable if f has no repeated roots in a splitting
п¬Ѓeld; otherwise f is inseparable. If f is an arbitrary polynomial, not necessarily irreducible,
then we call f separable if each of its irreducible factors is separable.
Thus if f (X) = (X в€’ 1)2 (X в€’ 3) over Q, then f is separable, because the irreducible
factors (X в€’ 1) and (X в€’ 3) do not have repeated roots. We will see shortly that over a
п¬Ѓeld of characteristic 0 (for example, the rationals), every polynomial is separable. Here
is a method for testing for multiple roots.

3.4.2 Proposition
If

f (X) = a0 + a1 X + В· В· В· + an X n в€€ F [X],
12 CHAPTER 3. FIELD FUNDAMENTALS

let f be the derivative of f , deп¬Ѓned by

f (X) = a1 + 2a2 X + В· В· В· + nan X nв€’1 .

[Note that the derivative is a purely formal expression; we completely ignore questions
about existence of limits. One can check by brute force that the usual rules for diп¬Ђeren-
tiating a sum and product apply].
If g is the greatest common divisor of f and f , then f has a repeated root in a splitting
п¬Ѓeld if and only if the degree of g is at least 1.

Proof. If f has a repeated root, we can write f (X) = (X в€’О±)r h(X) where r в‰Ґ 2. Applying
the product rule for derivatives, we see that (X в€’ О±) is a factor of both f and f , and
consequently deg g в‰Ґ 1. Conversely, if deg g в‰Ґ 1, let О± be a root of g in some splitting
п¬Ѓeld. Then (X в€’ О±) is a factor of both f and f . We will show that О± is a repeated root
of f . If not, we may write f (X) = (X в€’ О±)h(X) where h(О±) = 0. Diп¬Ђerentiate to obtain
f (X) = (X в€’ О±)h (X) + h(X), hence f (О±) = h(О±) = 0. This contradicts the fact that
(X в€’ О±) is a factor of f . в™Ј

3.4.3 Corollary
(1) Over a п¬Ѓeld of characteristic zero, every polynomial is separable.
(2) Over a п¬Ѓeld F of prime characteristic p, the irreducible polynomial f is inseparable
if and only if f is the zero polynomial. Equivalently, f is a polynomial in X p ; we
abbreviate this as f в€€ F [X p ].

Proof. (1) Without loss of generality, we can assume that we are testing an irreducible
polynomial f . The derivative of X n is nX nв€’1 , and in a п¬Ѓeld of characteristic 0, n cannot
be 0. Thus f is a nonzero polynomial whose degree is less than that of f . Since f is
irreducible, the gcd of f and f is either 1 or f , and the latter is excluded because f
cannot possibly divide f . By (3.4.2), f is separable.
(2) If f = 0, the argument of (1) shows that f is separable. If f = 0, then
gcd(f, f ) = f , so by (3.4.2), f is inseparable. In characteristic p, an integer n is zero if
and only if n is a multiple of p, and it follows that f = 0 iп¬Ђ f в€€ F [X p ]. в™Ј

By (3.4.3), part (1), every polynomial over the rationals (or the reals or the complex
numbers) is separable. This pleasant property is shared by п¬Ѓnite п¬Ѓelds as well. First
note that a п¬Ѓnite п¬Ѓeld F cannot have characteristic 0, since a п¬Ѓeld of characteristic 0
must contain a copy of the integers (and the rationals as well), and we cannot squeeze
inп¬Ѓnitely many integers into a п¬Ѓnite set. Now recall the binomial expansion modulo p,
p
which is simply (a + b)p = ap + bp , since p divides ( k ) for 1 в‰¤ k в‰¤ p в€’ 1. [By induction,
n n n
(a + b)p = ap + bp for every positive integer n.] Here is the key step in the analysis.

3.4.4 The Frobenius Automorphism
Let F be a п¬Ѓnite п¬Ѓeld of characteristic p, and deп¬Ѓne f : F в†’ F by f (О±) = О±p . Then f is
an automorphism. In particular, if О± в€€ F then О± = ОІ p for some ОІ в€€ F .
3.4. SEPARABILITY 13

Proof. We have f (1) = 1 and

f (О± + ОІ) = (О± + ОІ)p = О±p + ОІ p = f (О±) + f (ОІ),
f (О±ОІ) = (О±ОІ)p = О±p ОІ p = f (О±)f (ОІ)

so f is a monomorphism. But an injective function from a п¬Ѓnite set to itself is automati-
cally surjective, and the result follows. в™Ј

3.4.5 Proposition
Over a п¬Ѓnite п¬Ѓeld, every polynomial is separable.

Proof. Suppose that f is an irreducible polynomial over the п¬Ѓnite п¬Ѓeld F with repeated
roots in a splitting п¬Ѓeld. By (3.4.3), part (2), f (X) has the form a0 + a1 X p + В· В· В· + an X np
with the ai в€€ F . By (3.4.4), for each i there is an element bi в€€ F such that bp = ai . But
i
then

(b0 + b1 X + В· В· В· + bn X n )p = bp + bp X p + В· В· В· + bp X np = f (X)
n
0 1

which contradicts the irreducibility of f . в™Ј

Separability of an element can be deп¬Ѓned in terms of its minimal polynomial.

If E is an extension of F and О± в€€ E, then О± is separable over F if О± is algebraic over F
and min(О±, F ) is a separable polynomial. If every element of E is separable over F , we say
that E is a separable extension of F or the extension E/F is separable or E is separable
over F . By (3.4.3) and (3.4.5), every algebraic extension of a п¬Ѓeld of characteristic zero
or a п¬Ѓnite п¬Ѓeld is separable.

3.4.7 Lemma
If F в‰¤ K в‰¤ E and E is separable over F , then K is separable over F and E is separable
over K.

Proof. Since K is a subп¬Ѓeld of E, K/F is separable. If О± в€€ E, then since О± is a root of
min(О±, F ), it follows from (1) of (3.1.6) that min(О±, K) divides min(О±, F ). By hypothesis,
min(О±, F ) has no repeated roots in a splitting п¬Ѓeld, so neither does min(О±, K). Thus E/K
is separable. в™Ј

The converse of (3.4.7) is also true: If K/F and E/K are separable, then E/F is
separable. Thus we have transitivity of separable extensions. We will prove this (for п¬Ѓnite
extensions) in the exercises.
In view of (3.4.6), we can produce many examples of separable extensions. Inseparable
extensions are less common, but here is one way to construct them.
14 CHAPTER 3. FIELD FUNDAMENTALS

3.4.8 Example
Let F = Fp (t) be the set of rational functions (in the indeterminate t) with coeп¬ѓcients
in the п¬Ѓeld with p elements (the integers mod p). Thus an element of F looks like
a0 + a1 t + В· В· В· + am tm
.
b0 + b1 t + В· В· В· + b n t n
в€љ
with the ai and bj in Fp . Adjoin p t, that is, a root of X p в€’ t, to create the extension E.
Note that X p в€’ t is irreducible by Eisenstein, because t is irreducible in Fp [t]. (The
product of two nonconstant polynomials in t cannot possibly be t.) The extension E/F
is inseparable, since
в€љ в€љ
X p в€’ t = X p в€’ ( t)p = (X в€’ t)p ,
p p

which has multiple roots.

Problems For Section 3.4
1. Give an example of a separable polynomial f whose derivative is zero. (In view of
(3.4.3), f cannot be irreducible.)
2. Let О± в€€ E, where E is an algebraic extension of a п¬Ѓeld F of prime characteristic p.
Let m(X) be the minimal polynomial of О± over the п¬Ѓeld F (О±p ). Show that m(X)
splits over E, and in fact О± is the only root, so that m(X) is a power of (X в€’ О±).
3. Continuing Problem 2, if О± is separable over the п¬Ѓeld F (О±p ), show that О± в€€ F (О±p ).
4. A п¬Ѓeld F is said to be perfect if every polynomial over F is separable. Equivalently,
every algebraic extension of F is separable. Thus п¬Ѓelds of characteristic zero and
п¬Ѓnite п¬Ѓelds are perfect. Show that if F has prime characteristic p, then F is perfect
if and only if every element of F is the pth power of some element of F . For short we
write F = F p .
In Problems 5-8, we turn to transitivity of separable extensions.
5. Let E be a п¬Ѓnite extension of a п¬Ѓeld F of prime characteristic p, and let K = F (E p )
be the subп¬Ѓeld of E obtained from F by adjoining the pth powers of all elements of
E. Show that F (E p ) consists of all п¬Ѓnite linear combinations of elements in E p with
coeп¬ѓcients in F .
6. Let E be a п¬Ѓnite extension of the п¬Ѓeld F of prime characteristic p, and assume that
E = F (E p ). If the elements y1 , . . . , yr в€€ E are linearly independent over F , show
p p
that y1 , . . . , yr are linearly independent over F .
7. Let E be a п¬Ѓnite extension of the п¬Ѓeld F of prime characteristic p. Show that the
extension is separable if and only if E = F (E p ).
8. If F в‰¤ K в‰¤ E with [E : F ] < в€ћ, with E separable over K and K separable over F ,
show that E is separable over F .
9. Let f be an irreducible polynomial in F [X], where F has characteristic p > 0. Express
m
f (X) as g(X p ), where the nonnegative integer m is a large as possible. (This makes
0
sense because X p = X, so m = 0 always works, and f has п¬Ѓnite degree, so m is
bounded above.) Show that g is irreducible and separable.
3.5. NORMAL EXTENSIONS 15

m
10. Continuing Problem 9, if f has only one distinct root О±, show that О±p в€€ F .
11. If E/F , where char F = p > 0, and the element О± в€€ E is algebraic over F , show that
n
the minimal polynomial of О± over F has only one distinct root if and only if О±p в€€ F
for some nonnegative integer n. (In this case we say that О± is purely inseparable over
F .)

3.5 Normal Extensions
Let E/F be a п¬Ѓeld extension. In preparation for Galois theory, we are going to look at
monomorphisms deп¬Ѓned on E, especially those which п¬Ѓx F . First we examine what an
F -monomorphism does to the roots of a polynomial in F [X].

3.5.1 Lemma
Let Пѓ : E в†’ E be an F -monomorphism, and assume that the polynomial f в€€ F [X] splits
over E. If О± is a root of f in E, then so is Пѓ(О±). Thus Пѓ permutes the roots of f .

Proof. If b0 + b1 О± + В· В· В· + bn О±n = 0, with the bi в€€ F , apply Пѓ and note that since Пѓ is an
F -monomorphism, Пѓ(bi ) = bi and Пѓ(О±i ) = (Пѓ(О±))i . Thus

b0 + b1 Пѓ(О±) + В· В· В· + bn (Пѓ(О±))n = 0. в™Ј

Now let C be an algebraic closure of E. It is convenient to have C available because
it will contain all the roots of a polynomial f в€€ E[X], even if f does not split over E. We
are going to count the number of embeddings of E in C that п¬Ѓx F , that is, the number
of F -monomorphisms of E into C. Here is the key result.

3.5.2 Theorem
Let E/F be a п¬Ѓnite separable extension of degree n, and let Пѓ be an embedding of F
in C. Then Пѓ extends to exactly n embeddings of E in C; in other words, there are
exactly n embeddings П„ of E in C such that the restriction П„ |F of П„ to F coincides
with Пѓ. In particular, taking Пѓ to be the identity function on F , there are exactly n
F -monomorphisms of E into C.

Proof. An induction argument works well. If n = 1 then E = F and there is nothing to
prove, so assume n > 1 and choose an element О± that belongs to E but not to F . If f is
the minimal polynomial of О± over F , let g = Пѓ(f ). (This is a useful shorthand notation,
indicating that if ai is one of the coeп¬ѓcients of f , the corresponding coeп¬ѓcient of g is
Пѓ(ai ).) Any factorization of g can be translated via the inverse of Пѓ to a factorization
of f , so g is separable and irreducible over the п¬Ѓeld Пѓ(F ). If ОІ is any root of g, then there
is a unique isomorphism of F (О±) and (Пѓ(F ))(ОІ) that carries О± into ОІ and coincides with
Пѓ on F . Explicitly,

b0 + b1 О± + В· В· В· + br О±r в†’ Пѓ(b0 ) + Пѓ(b1 )ОІ + В· В· В· + Пѓ(br )ОІ r .
16 CHAPTER 3. FIELD FUNDAMENTALS

Now if deg g = r, then [F (О±) : F ] = deg f = deg g = r as well, so by (3.1.9), [E : F (О±)] =
n/r < n. By separability, g has exactly r distinct roots in C, so there are exactly r
possible choices of ОІ. In each case, by the induction hypothesis,the resulting embedding
of F (О±) in C has exactly n/r extensions to embeddings of E in C. This produces n
distinct embeddings of E in C extending Пѓ. But if П„ is any embedding of F in C that
extends Пѓ, then just as in (3.5.1), П„ must take О± to a root of g, i.e., to one of the ОІвЂ™s.
If there were more than n possible П„ вЂ™s, there would have to be more than n/r possible
extensions of at least one of the embeddings of F (О±) in C. This would contradict the
induction hypothesis. в™Ј

3.5.3 Example
в€љ
Q(
Adjoin the positive cube root of 2 to the rationals to get E = в€љ 3 2). The roots of the
в€љ в€љ
irreducible polynomial f (X) = X 3 в€’ 2 are 3 2, П‰ 3 2 and П‰ 2 3 2, where П‰ = ei2ПЂ/3 =
в€љ в€љ
в€’ 1 + 1 i 3 and П‰ 2 = ei4ПЂ/3 = в€’ 1 в€’ 1 i 3.
2 2 2 2
Notice that the polynomial f has a root in E but does not split in E (because the
other two roots are complex and E consists entirely of real numbers). We give a special
name to extensions that do not have this annoying drawback.

3.5.4 Deп¬Ѓnition
The algebraic extension E/F is normal (we also say that E is normal over F ) if every
irreducible polynomial over F that has at least one root in E splits over E. In other
words, if О± в€€ E, then all conjugates of О± over F (i.e., all roots of the minimal polynomial
of О± over F ) belong to E.
Here is an equivalent condition.

3.5.5 Theorem
The п¬Ѓnite extension E/F is normal if and only if every F -monomorphism of E into an
algebraic closure C is actually an F -automorphism of E. (The hypothesis that E/F is
п¬Ѓnite rather than simply algebraic can be removed, but we will not need the more general
result.)
Proof. If E/F is normal, then as in (3.5.1), an F -monomorphism П„ of E into C must map
each element of E to one of its conjugates. Thus by hypothesis, П„ (E) вЉ† E. But П„ (E) is
an isomorphic copy of E , so it must have the same degree as E over F . Since the degree
is assumed п¬Ѓnite, we have П„ (E) = E. (All we are saying here is that an m-dimensional
subspace of an m-dimensional vector space is the entire space.) Conversely, let О± в€€ E, and
let ОІ be any conjugate of О± over F . As in the proof of (3.5.2), there is an F -monomorphism
of E into C that carries О± to ОІ. If all such embeddings are F -automorphisms of E, we
must have ОІ в€€ E, and we conclude that E is normal over F . в™Ј

3.5.6 Remarks
In (3.5.2) and (3.5.5), the algebraic closure can be replaced by any п¬Ѓxed normal extension
of F containing E; the proof is the same. Also, the implication П„ (E) вЉ† E в‡’ П„ (E) = E
3.5. NORMAL EXTENSIONS 17

holds for any F -monomorphism П„ and any п¬Ѓnite extension E/F ; normality is not involved.
The next result yields many explicit examples of normal extensions.

3.5.7 Theorem
The п¬Ѓnite extension E/F is normal if and only if E is a splitting п¬Ѓeld for some polynomial
f в€€ F [X].

Proof. Assume that E is normal over F . Let О±1 , . . . , О±n be a basis for E over F , and
let fi be the minimal polynomial of О±i over F, i = 1, . . . , n. Since fi has a root О±i in E,
fi splits over E, hence so does f = f1 В· В· В· fn . If f splits over a п¬Ѓeld K with F вЉ† K вЉ† E,
then each О±i belongs to K, and therefore K must coincide with E. Thus E is a splitting
п¬Ѓeld for f over F . Conversely, let E be a splitting п¬Ѓeld for f over F , where the roots of f
are О±i , i = 1, . . . , n. Let П„ be an F -monomorphism of E into an algebraic closure. As in
(3.5.1), П„ takes each О±i into another root of f , and therefore П„ takes a polynomial in the
О±i to another polynomial in the О±i . But F (О±1 , . . . , О±n ) = E, so П„ (E) вЉ† E. By (3.5.6), П„
is an automorphism of E, so by (3.5.5), E/F is normal. в™Ј

3.5.8 Corollary
Let F в‰¤ K в‰¤ E, where E is a п¬Ѓnite extension of F . If E/F is normal, so is E/K.

Proof. By (3.5.7), E is a splitting п¬Ѓeld for some polynomial f в€€ F [X], so that E is
generated over F by the roots of f . But then f в€€ K[X] and E is generated over K by
the roots of f . Again by (3.5.7), E/K is normal. в™Ј

If E/F is normal and separable, it is said to be a Galois extension; we also say that E is
Galois over F . It follows from (3.5.2) and (3.5.5) that if E/F is a п¬Ѓnite Galois extension,
then there are exactly [E : F ] F -automorphisms of E. If E/F is п¬Ѓnite and separable but
not normal, then at least one F -embedding of E into an algebraic closure must fail to be
an automorphism of E. Thus in this case, the number of F -automorphisms of E is less
than the degree of the extension.
If E/F is an arbitrary extension, the Galois group of the extension, denoted by
Gal(E/F ), is the set of F -automorphisms of E. (The set is a group under composition of
functions.)

3.5.10 Example
в€љ
Let E = Q( 3 2), as in (3.5.3). The Galois group of the extension consists of the identity
в€љ
automorphism alone. For any Q-monomorphism Пѓ of E must take 3 в€љ into a root of
2
X в€’ 2. Since the other two roots are complex and do not belong to E, 3 2 must map to
3
в€љ
itself. But Пѓ is completely determined by its action on 3 2, and the result follows.
If E/F is not normal, we can always enlarge E to produce a normal extension of F .
If C is an algebraic closure of E, then C contains all the roots of every polynomial in
F [X], so C/F is normal. Let us try to look for a smaller normal extension.
18 CHAPTER 3. FIELD FUNDAMENTALS

3.5.11 The Normal Closure
Let E be a п¬Ѓnite extension of F , say E = F (О±1 , . . . , О±n ). If N вЉ‡ E is any normal extension
of F , then N must contain the О±i along with all conjugates of the О±i , that is, all roots of
min(О±i , F ), i = 1, . . . , n. Thus if f is the product of these minimal polynomials, then N
must contain the splitting п¬Ѓeld K for f over F . But K/F is normal by (3.5.7), so K must
be the smallest normal extension of F that contains E. It is called the normal closure of
E over F .
We close the section with an important result on the structure of п¬Ѓnite separable
extensions.

3.5.12 Theorem of the Primitive Element
If E/F is a п¬Ѓnite separable extension, then E = F (О±) for some О± в€€ E. We say that О± is
a primitive element of E over F .

Proof. We will argue by induction on n = [E : F ]. If n = 1 then E = F and we can take
О± to be any member of F . If n > 1, choose О± в€€ E \ F . By the induction hypothesis, there
is a primitive element ОІ for E over F (О±), so that E = F (О±, ОІ). We are going to show
that if c в€€ F is properly chosen, then E = F (О± + cОІ). Now by (3.5.2), there are exactly n
F -monomorphisms of E into an algebraic closure C, and each of these maps restricts to
an F -monomorphism of F (О± + cОІ) into C. If F (О± + cОІ) = E, then [F (О± + cОІ) : F ] < n,
and it follows from (3.5.2) that at least two embeddings of E, say Пѓ and П„ , must coincide
when restricted. Therefore

Пѓ(О±) + cПѓ(ОІ) = П„ (О±) + cП„ (ОІ),

hence
Пѓ(О±) в€’ П„ (О±)
c= . (1)
П„ (ОІ) в€’ Пѓ(ОІ)

(If П„ (ОІ) = Пѓ(ОІ) then by the previous equation, П„ (О±) = Пѓ(О±). But an F -embedding
of E is determined by what it does to О± and ОІ, hence Пѓ = П„ , a contradiction.) Now
an F -monomorphism must map О± to one of its conjugates over F , and similarly for ОІ.
Thus there are only п¬Ѓnitely many possible values for the ratio in (1). If we select c to
be diп¬Ђerent from each of these values, we reach a contradiction of our assumption that
F (О± + cОІ) = E. The proof is complete if F is an inп¬Ѓnite п¬Ѓeld. We must leave a gap here,
to be п¬Ѓlled later (see (6.4.4)). If F is п¬Ѓnite, then so is E (since E is a п¬Ѓnite-dimensional
vector space over F ). We will show that the multiplicative group of nonzero elements of
a п¬Ѓnite п¬Ѓeld E is cyclic, so if О± is a generator of this group, then E = F (О±). в™Ј

Problems For Section 3.5
1. Give an example of п¬Ѓelds F в‰¤ K в‰¤ E such that E/F is normal but K/F is not.
в€љ
2. Let E = Q( a), where a is an integer that is not a perfect square. Show that E/Q is
normal.
3.5. NORMAL EXTENSIONS 19

3. Give an example of п¬Ѓelds F в‰¤ K в‰¤ E such that E/K and K/F are normal, but E/F
is not. Thus transitivity fails for normal extensions.
4. Suppose that in (3.5.2), the hypothesis of separability is dropped. State and prove an
appropriate conclusion.
в€љв€љ
5. Show that E = Q( 2, 3) is a Galois extension of Q.
6. In Problem 5, п¬Ѓnd the Galois group of E/Q.
7. Let E be a п¬Ѓnite extension of F , and let K be a normal closure (= minimal normal
extension) of E over F , as in (3.5.11). Is K unique?
8. If E1 and E2 are normal extensions of F , show that E1 в€© E2 is normal over F .

Appendix To Chapter 3
In this appendix, we give a precise development of the results on algebraic closure treated
informally in the text.

A3.1 Lemma
Let E be an algebraic extension of F , and let Пѓ : E в†’ E be an F -monomorphism. Then
Пѓ is surjective, hence Пѓ is an automorphism of E.

Proof. Let О± в€€ E, and let f (X) be the minimal polynomial of О± over F . We consider the
subп¬Ѓeld L of E generated over F by the roots of f that lie in E. Then L is an extension
of F that is п¬Ѓnitely generated by algebraic elements, so by (3.3.3), L/F is п¬Ѓnite. As in
(3.5.1), Пѓ takes a root of f to a root of f , so Пѓ(L) вЉ† L. But [L : F ] = [Пѓ(L) : F ] < в€ћ
(Пѓ maps a basis to a basis), and consequently Пѓ(L) = L. But О± в€€ L, so О± в€€ Пѓ(L). в™Ј

The following result, due to Artin, is crucial.

A3.2 Theorem
If F is any п¬Ѓeld, there is an algebraically closed п¬Ѓeld E containing F .

Proof. For each nonconstant polynomial f in F [X], we create a variable X(f ). If T is the
collection of all such variables, we can form the ring F [T ] of all polynomials in all possible
п¬Ѓnite sets of variables in T , with coeп¬ѓcients in F . Let I be the ideal of F [T ] generated
by the polynomials f (X(f )), f в€€ F [X]. We claim that I is a proper ideal. If not,
then 1 в€€ I, so there are п¬Ѓnitely many polynomials f1 , . . . , fn in F [X] and polynomials
n
h1 , . . . , hn in F [T ] such that i=1 hi fi (X(fi )) = 1. Now only п¬Ѓnitely many variables
Xi = X(fi ), i = 1, . . . , m, can possibly appear in the hi , so we have an equation of the
form
n
hi (X1 , . . . , Xm )fi (Xi ) = 1 (1)
i=1
20 CHAPTER 3. FIELD FUNDAMENTALS

where m в‰Ґ n. Let L be the extension of F formed by successively adjoining the roots of
f1 , . . . , fn . Then each fi has a root О±i в€€ L. If we set О±i = 0 for n в‰¤ i < m and then set
Xi = О±i for each i in (1), we get 0 = 1, a contradiction.
Thus the ideal I is proper, and is therefore contained in a maximal ideal M. Let E1
be the п¬Ѓeld F [T ]/M. Then E1 contains an isomorphic copy of F , via the map taking
a в€€ F to a + M в€€ E1 . (Note that if a в€€ M, a = 0, then 1 = aв€’1 a в€€ M, a contradiction.)
Consequently, we can assume that F в‰¤ E1 . If f is any nonconstant polynomial in F [X],
then X(f ) + M в€€ E1 and f (X(f ) + M) = f (X(f )) + M = 0 because f (X(f )) в€€ I вЉ† M.
Iterating the above procedure, we construct a chain of п¬Ѓelds F в‰¤ E1 в‰¤ E2 в‰¤ В· В· В· such
that every polynomial of degree at least 1 in En [X] has a root in En+1 . The union E of
all the En is a п¬Ѓeld, and every nonconstant polynomial f in E[X] has all its coeп¬ѓcients
in some En . Therefore f has a root in En+1 вЉ† E. в™Ј

A3.3 Theorem
Every п¬Ѓeld F has an algebraic closure.

Proof. By (3.5.12), F has an algebraically closed extension L. If E is the algebraic closure
of F in L (see 3.3.4), then E/F is algebraic. Let f be a nonconstant polynomial in E[X].
Then f has a root О± in L (because L is algebraically closed). We now have О± algebraic
over E (because f в€€ E[X]), and E algebraic over F . As in (3.3.5), О± is algebraic over F ,
hence О± в€€ E. By (3.3.1), E is algebraically closed. в™Ј

A3.4 Problem
Suppose that Пѓ is a monomorphism of F into the algebraically closed п¬Ѓeld L. Let E be
an algebraic extension of F , and О± an element of E with minimal polynomial f over F .
We wish to extend Пѓ to a monomorphism from F (О±) to L. In how many ways can this
be done?
Let Пѓf be the polynomial in (ПѓF )[X] obtained from f by applying Пѓ to the coeп¬ѓcients
of f . Any extension of f is determined by what it does to О±, and as in (3.5.1), the image
of О± is a root of Пѓf . Now the number of distinct roots of f in an algebraic closure of F ,
call it t, is the same as the number of distinct roots of Пѓf in L; this follows from the
isomorphism extension theorem (3.2.5). Thus the number of extensions is at most t.
But if ОІ is any root of Пѓf , we can construct an extension of Пѓ by mapping the element
h(О±) в€€ F (О±) to (Пѓh)(ОІ); in particular, О± is mapped to ОІ. To show that the deп¬Ѓnition
makes sense, suppose that h1 (О±) = h2 (О±). Then (h1 в€’ h2 )(О±) = 0, so f divides h1 в€’ h2
in F [X]. Consequently, Пѓf divides Пѓh1 в€’ Пѓh2 in (ПѓF )[X], so (Пѓh1 )(ОІ) = (Пѓh2 )(ОІ).
We conclude that the number of extensions of Пѓ is the number of distinct roots of f
in an algebraic closure of F .
Rather than extend Пѓ one element at a time, we now attempt an extension to all of E.

A3.5 Theorem
Let Пѓ : F в†’ L be a monomorphism, with L algebraically closed. If E is an algebraic
extension of F , then Пѓ has an extension to a monomorphism П„ : E в†’ L.
3.5. NORMAL EXTENSIONS 21

Proof. Let G be the collection of all pairs (K, Вµ) where K is an intermediate п¬Ѓeld between
F and E and Вµ is an extension of Пѓ to a monomorphism from K to L. We partially
order G by (K1 , Вµ) в‰¤ (K2 , ПЃ) iп¬Ђ K1 вЉ† K2 and ПЃ restricted to K1 coincides with Вµ. Since
(F, Пѓ) в€€ G, we have G = в€…. If the pairs (Ki , Вµi ), i в€€ I, form a chain, there is an upper
bound (K, Вµ) for the chain, where K is the union of the Ki and Вµ coincides with Вµi on
each Ki . By ZornвЂ™s lemma, G has a maximal element (K0 , П„ ). If K0 вЉ‚ E, let О± в€€ E \ K0 .
By (3.5.12), П„ has an extension to K0 (О±), contradicting maximality of (K0 , П„ ). в™Ј

A3.6 Corollary
In (3.5.12), if E is algebraically closed and L is algebraic over Пѓ(F ), then П„ is an isomor-
phism.

Proof. Since E is algebraically closed, so is П„ (E). Since L is algebraic over Пѓ(F ), it is
algebraic over the larger п¬Ѓeld П„ (E). By (1) в‡ђв‡’ (4) in (3.3.1), L = П„ (E). в™Ј

A3.7 Theorem
Any two algebraic closures L and E of a п¬Ѓeld F are F -isomorphic.

Proof. We can assume that F is a subп¬Ѓeld of L and Пѓ : F в†’ L is the inclusion map.
By (3.5.12), Пѓ extends to an isomorphism П„ of E and L, and since П„ is an extension of Пѓ,
it is an F -monomorphism. в™Ј

A3.8 Theorem (=Theorem 3.3.9)
If E is an algebraic extension of F and C is an algebraic closure of F , then any embedding
of F into C can be extended to an embedding of E into C.

Proof. Repeat the proof of (3.5.12), with the mapping Вµ required to be an embedding. в™Ј

A3.9 Remark
The argument just given assumes that E is a subп¬Ѓeld of C. This can be assumed without
loss of generality, by (3.5.12), (3.5.12) and (3.5.12). In other words, we can assume that
an algebraic closure of F contains a speciп¬Ѓed algebraic extension of F .

A3.10 Theorem
Let E be an algebraic extension of F , and let L be the algebraic closure of F containing E
(see 3.5.12). If Пѓ is an F -monomorphism from E to L, then Пѓ can be extended to an
automorphism of L.

Proof. We have L algebraically closed and L/E algebraic, so by (3.5.12) with E replaced
by L and F by E, Пѓ extends to a monomorphism from L to L, an F -monomorphism by
hypothesis. The result follows from (3.5.12). в™Ј
Chapter 4

Module Fundamentals

4.1 Modules and Algebras
A vector space M over a п¬Ѓeld R is a set of objects called vectors, which can be added,
subtracted and multiplied by scalars (members of the underlying п¬Ѓeld). Thus M is an
abelian group under addition, and for each r в€€ R and x в€€ M we have an element rx в€€ M .
Scalar multiplication is distributive and associative, and the multiplicative identity of the
п¬Ѓeld acts as an identity on vectors. Formally,

r(x + y) = rx + ry; (r + s)x = rx + sx; r(sx) = (rs)x; 1x = x

for all x, y в€€ M and r, s в€€ R. A module is just a vector space over a ring. The formal
deп¬Ѓnition is exactly as above, but we relax the requirement that R be a п¬Ѓeld, and instead
allow an arbitrary ring. We have written the product rx with the scalar r on the left, and
technically we get a left R-module over the ring R. The axioms of a right R-module are

(x + y)r = xr + yr; x(r + s) = xr + xs; (xs)r = x(sr), x1 = x.

вЂњModuleвЂќ will always mean left module unless stated otherwise. Most of the time, there
is no reason to switch the scalars from one side to the other (especially if the underlying
ring is commutative). But there are cases where we must be very careful to distinguish
between left and right modules (see Example 6 of (4.1.3)).

4.1.2 Some Basic Properties of Modules
Let M be an R-module. The technique given for rings in (2.1.1) can be applied to establish
the following results, which hold for any x в€€ M and r в€€ R. We distinguish the zero vector
0M from the zero scalar 0R .
(1) r0M = 0M [r0M = r(0M + 0M ) = r0M + r0M ]
(2) 0R x = 0M [0R x = (0R + 0R )x = 0R x + 0R x]

1
2 CHAPTER 4. MODULE FUNDAMENTALS

(3) (в€’r)x = r(в€’x) = в€’(rx) [as in (2) of (2.1.1) with a replaced by r and b by x]

(4) If R is a п¬Ѓeld, or more generally a division ring, then rx = 0M implies that either
r = 0R or x = 0M . [If r = 0, multiply the equation rx = 0M by rв€’1 .]

4.1.3 Examples
1. If M is a vector space over the п¬Ѓeld R, then M is an R-module.

2. Any ring R is a module over itself. Rather than check all the formal requirements,
think intuitively: Elements of a ring can be added and subtracted, and we can
certainly multiply r в€€ R by x в€€ R, and the usual rules of arithmetic apply.

3. If R is any ring, then Rn , the set of all n-tuples with components in R, is an
R-module, with the usual deп¬Ѓnitions of addition and scalar multiplication (as in
Euclidean space, e.g., r(x1 , . . . , xn ) = (rx1 , . . . , rxn ), etc).

4. Let M = Mmn (R) be the set of all m Г— n matrices with entries in R. Then M is
an R-module, where addition is ordinary matrix addition, and multiplication of the
scalar c by the matrix A means multiplication of each entry of A by c.

5. Every abelian group A is a Z-module. Addition and subtraction is carried out
according to the group structure of A; the key point is that we can multiply x в€€ A
by the integer n. If n > 0, then nx = x + x + В· В· В· + x (n times); if n < 0, then
nx = в€’x в€’ x в€’ В· В· В· в€’ x (|n| times).

In all of these examples, we can switch from left to right modules by a simple notational
change. This is deп¬Ѓnitely not the case in the next example.

6. Let I be a left ideal of the ring R; then I is a left R-module. (If x в€€ I and r в€€ R
then rx (but not necessarily xr) belongs to I.) Similarly, a right ideal is a right
R-module, and a two-sided ideal is both a left and a right R-module.

An R-module M permits addition of vectors and scalar multiplication. If multiplica-
tion of vectors is allowed, we have an R-algebra.

Let R be a commutative ring. We say that M is an algebra over R, or that M is an
R-algebra, if M is an R-module that is also a ring (not necessarily commutative), and the
ring and module operations are compatible, i.e.,

r(xy) = (rx)y = x(ry) for all x, y в€€ M and r в€€ R.

4.1.5 Examples
1. Every commutative ring R is an algebra over itself (see Example 2 of (4.1.3)).

2. An arbitrary ring R is always a Z-algebra (see Example 5 of (4.1.3)).
4.1. MODULES AND ALGEBRAS 3

3. If R is a commutative ring, then Mn (R), the set of all n Г— n matrices with entries
in R, is an R-algebra (see Example 4 of (4.1.3)).
4. If R is a commutative ring, then the polynomial ring R[X] is an R-algebra, as
is the ring R[[X]] of formal power series; see Examples 5 and 6 of (2.1.3). The
compatibility condition is satisп¬Ѓed because an element of R can be regarded as a
polynomial of degree 0.
5. If E/F is a п¬Ѓeld extension, then E is an algebra over F . This continues to hold if E
is a division ring, and in this case we say that E is a division algebra over F .

To check that a subset S of a vector space is a subspace, we verify that S is closed
under addition of vectors and multiplication of a vector by a scalar. Exactly the same
idea applies to modules and algebras.

If N is a nonempty subset of the R-module M , we say that N is a submodule of M
(notation N в‰¤ M ) if for every x, y в€€ N and r, s в€€ R, we have rx + sy в€€ N . If M is an
R-algebra, we say that N is a subalgebra if N is a submodule that is also a subring.
For example, if A is an abelian group (= Z-module), the submodules of A are the
subsets closed under addition and multiplication by an integer (which amounts to addition
also). Thus the submodules of A are simply the subgroups. If R is a ring, hence a
module over itself, the submodules are those subsets closed under addition and also under
multiplication by any r в€€ R, in other words, the left ideals. (If we take R to be a right
R-module, then the submodules are the right ideals.)
We can produce many examples of subspaces of vector spaces by considering kernels
and images of linear transformations. A similar idea applies to modules.

Let M and N be R-modules. A module homomorphism (also called an R-homomorphism)
from M to N is a map f : M в†’ N such that

f (rx + sy) = rf (x) + sf (y) for all x, y в€€ M and r, s в€€ R.

Equivalently, f (x + y) = f (x) + f (y) and f (rx) = rf (x) for all x, y в€€ M and r в€€ R.
The kernel of a homomorphism f is ker f = {x в€€ M : f (x) = 0}, and the image of f
is {f (x) : x в€€ M }.
If follows from the deп¬Ѓnitions that the kernel of f is a submodule of M , and the image
of f is a submodule of N .
If M and N are R-algebras, an algebra homomorphism or homomorphism of algebras
from M to N is an R-module homomorphism that is also a ring homomorphism.

4.1.8 Another Way to Describe an Algebra
Assume that A is an algebra over the commutative ring R, and consider the map r в†’ r1
of R into A. The commutativity of R and the compatibility of the ring and module
4 CHAPTER 4. MODULE FUNDAMENTALS

operations imply that the map is a ring homomorphism. To see this, note that if r, s в€€ R
then

(rs)1 = (sr)1 = s(r1) = s[(r1)1] = (r1)(s1).

Furthermore, if y в€€ A then

(r1)y = r(1y) = r(y1) = y(r1)

so that r1 belongs to the center of A, i.e., the set of elements that commute with everything
in A.
Conversely, if f is a ring homomorphism from the commutative ring R to the center
of the ring A, we can make A into an R-module via rx = f (r)x. The compatibility
conditions are satisп¬Ѓed because

r(xy) = f (r)(xy) = (f (r)x)y = (rx)y

and

(f (r)x)y = (xf (r))y = x(f (r)y) = x(ry).

Because of this result, the deп¬Ѓnition of an R-algebra is sometimes given as follows. The
ring A is an algebra over the commutative ring R if there exists a ring homomorphism of
R into the center of A. For us at this stage, such a deп¬Ѓnition would be a severe overdose
of abstraction.
Notational Convention: We will often write the module {0} (and the ideal {0} in
a ring) simply as 0.

Problems For Section 4.1
1. If I is an ideal of the ring R, show how to make the quotient ring R/I into a left
R-module, and also show how to make R/I into a right R-module.
2. Let A be a commutative ring and F a п¬Ѓeld. Show that A is an algebra over F if and
only if A contains (an isomorphic copy of) F as a subring.

Problems 3, 4 and 5 illustrate that familiar properties of vector spaces need not hold
for modules.
3. Give an example of an R-module M with nonzero elements r в€€ R and x в€€ M such
that rx = 0.
4. Let M be the additive group of rational numbers. Show that any two elements of M
are linearly dependent (over the integers Z).
5. Continuing Problem 4, show that M cannot have a basis, that is, a linearly independent
spanning set over Z.
6. Prove the modular law for subgroups of a given group G: With the group operation
written multiplicatively,