Milne J. - Fileds and Galois Theory [jnl article].pdf - Algebra - Mathematics2014

FIELDS AND GALOIS THEORY

J.S. MILNE

Abstract. These are the notes for the second part of Math 594, University of Michigan,

Winter 1994, exactly as they were handed out during the course except for some minor

corrections.

Please send comments and corrections to me at jmilne@umich.edu using “Math594” as

the subject.

v2.01 (August 21, 1996). First version on the web.

v2.02 (May 27, 1998). About 40 minor corrections (thanks to Henry Kim).

Contents

1. Extensions of Fields

1.1. Deﬁnitions

1.2. The characteristic of a ﬁeld

1.3. The polynomial ring F [ X ]

1.4. Factoring polynomials

1.5. Extension ﬁelds;degrees

1.6. Construction of some extensions

1.7. Generators of extension ﬁelds

1.8. Algebraic and transcendental elements

1.9. Transcendental numbers

1.10. Constructions with straight-edge and compass.

2. Splitting Fields;Algebraic Closures

2.1. Maps from simple extensions.

2.2. Splitting ﬁelds

2.3. Algebraic closures

3. The Fundamental Theorem of Galois Theory

3.1. Multiple roots

3.2. Groups of automorphisms of ﬁelds

3.3. Separable, normal, and Galois extensions

3.4. The fundamental theorem of Galois theory

3.5. Constructible numbers revisited

3.6. Galois group of a polynomial

3.7. Solvability of equations

J.S. MILNE

4. Computing Galois Groups.

4.1. When is G f ⊂

A n ?

4.2. When is G f transitive?

4.3. Polynomials of degree

≤

4.4. Quartic polynomials

4.5. Examples of polynomials with S p as Galois group over

4.6. Finite ﬁelds

4.7. Computing Galois groups over

5. Applications of Galois Theory

5.1. Primitive element theorem.

5.2. Fundamental Theorem of Algebra

5.3. Cyclotomic extensions

5.4. Independence of characters

5.5. Hilbert’s Theorem 90.

5.6. Cyclic extensions.

5.7. Proof of Galois’s solvability theorem

5.8. The general polynomial of degree n

Symmetric polynomials

The general polynomial

Abriefhistory

5.9. Norms and traces

5.10. Inﬁnite Galois extensions (sketch)

6. Transcendental Extensions

FIELDS AND GALOIS THEORY

1. Extensions of Fields

1.1. Deﬁnitions. A ﬁeld is a set F with two composition laws + and

such that

(a) ( F, +) is an abelian group;

(b) let F × = F −{ 0 } ;then ( F × ,· ) is an abelian group;

Equivalently, a ﬁeld is a nonzero commutative ring (meaning with 1) such that every nonzero

element has an inverse. A ﬁeld contains at least two distinct elements, 0 and 1. The smallest,

and one of the most important, ﬁelds is

2 =

/ 2

{

0 , 1

}

Lemma 1.1.

A commutative ring R is a ﬁeld if and only if it has no ideals other than (0)

and R.

Proof. Suppose R is a ﬁeld, and let I beanonzeroidealin R .If a isanonzeroelement

of I ,then1= a − 1 a

∈

=0,then( a )= R , which means that there is a b in F such that

ab =1.

Example 1.2.

The following are ﬁelds:

F p =

A homomorphism of ﬁelds α : F →F is simply a homomorphism of rings, i.e., it is a map

with the properties

α ( a + b )= α ( a )+ α ( b ) , α ( ab )= α ( a ) α ( b ) , α (1) = 1 , all a,b

∈

Such a homomorphism is always injective, because the kernel is a proper ideal (it doesn’t

contain 1), which must therefore be zero.

1.2. The characteristic of a ﬁeld. The map

Z →

F, n

→

1 F +1 F +

···

+1 F ( n times) ,

is a homomorphism of rings.

Case 1: Kernel = (0);then n

1 F =0 =

⇒

n =0(in

). The map

Z →

F extends to a

homomorphism

→

F , n →

( m

1 F )( n

1 F ) − 1 .Thus F contains a copy of

.Inthiscase,

we say that F has characteristic zero .

Case 2: Kernel

= 1. The smallest such n will be a

prime p (else F will have nonzero zero-divisors), and p generates the kernel. In this case,

{m· 1 F |m∈ Z }≈ F p ,and F contains a copy of F p .Wesaythat F has characteristic p.

The ﬁelds F p , p prime, and

= (0), i.e., n

1 F =0some n

are called the prime ﬁelds. Every ﬁeld contains a copy of

one of them.

Remark 1.3. The binomial theorem

( a + b ) m = a m + m

a m− 1 b +

···

+ m

a m−r b r +

···

+ b m

holds in any ring. If p is prime, then p| r for all r ,1

1. Therefore, when F has

characteristic p ,( a + b ) p = a p + b p . Hence a → a p is a homomorphism F → F , called the

Frobenius endomorphism of F .When F is ﬁnite, it is an isomorphism, called the Frobenius

automorphism.

≤ r ≤ p−

I ,andso I = R . Conversely, suppose R is a commutative ring with

no nontrivial ideals;if a

J.S. MILNE

1.3. The polynomial ring F [ X ] . I shall assume everyone knows the following (see Jacob-

son Chapter II, or Math 593).

(a) Let I beanonzeroidealin F [ X ]. If f ( X ) is a nonzero polynomial of least degree in I ,

then I =( f ( X )). When we choose f to be monic, i.e., to have leading coeEcient one, it is

uniquely determined by I . There is a one-to-one correspondence between the nonzero ideals

of F [ X ] and the monic polynomials in F [ X ]. The prime ideals correspond to the irreducible

monic polynomials.

(b) Division algorithm :given f ( X )and g ( X )

∈

F [ X ]with g

= 0, we can ﬁnd q ( X )and

F [ X ]withdeg( r ) < deg( g ) such that f = gq + r ;moreover, q ( X )and r ( X )are

uniquely determined. Thus the ring F [ X ] is a Euclidean domain.

∈

F [ X ]havegcd d ( X );the algorithm gives polynomials

a ( X )and b ( X ) such that

a ( X ) ·f ( X )+ b ( X ) ·g ( X )= d ( X ) , deg( a ) ≤ deg( g ) , deg( b ) ≤ deg( f ) .

Recall how it goes. Using the division algorithm, we construct a sequence of quotients and

remainders:

f = q 0 g + r 0

g = q 1 r 0 + r 1

r 0 = q 2 r 1 + r 2

···

r n− 2 = q n r n− 1 + r n

r n− 1 = q n +1 r n .

= af + bg.

Maple knows Euclid’s algorithm—to learn its syntax, type “?gcdex;”.

(d) Since F [ X ] is an integral domain, we can form its ﬁeld of fractions F ( X ). It consists

of quotients f ( X ) /g ( X ), f and g polynomials, g

−

q n r n− 1 = r n− 2

−

q n ( r n− 3

−

q n− 1 r n− 2 )=

···

=0 .

1.4. Factoring polynomials. It will frequently be important for us to know whether a

polynomial is irreducible and, if it isn’t, what its factors are. The following results help.

Proposition 1.4.

Suppose r = d , c, d∈ Z , gcd( c, d )=1 , is a root of a polynomial

a m X m + a m− 1 X m− 1 +

···

+ a 0 , i

∈ Z

Then c|a 0 and d|a m .

Proof.

It is clear from the equation

a m c m + a m− 1 c m− 1 d +

···

+ a 0 d m =0

that d

a m c m , and therefore, d

a m . The proof that c

a 0 is similar.

Example 1.5.

The polynomial X 3

−

3 X

−

1 is irreducible in

[ X ] because its only possible

roots are

1 (and they aren’t).

[ X ] be such that its coe % cients have greatest common

divisor 1 .Iff ( X ) factors nontrivially in

Proposition 1.6.

Let f ( X )

∈ Z

[ X ] , then it factors nontrivially in

[ X ] ; moreover,

if f ( X )

∈ Z

[ X ] is monic, then any monic factor of f ( X ) in

[ X ] lies in

[ X ] .

r ( X )

Then r n =gcd( f,g ), and

r n = r n− 2

FIELDS AND GALOIS THEORY

Proof.

Use Gauss’s lemma (see Jacobson, 2.16, or Math 593).

Proposition 1.7.

(Eisenstein criterion) Let

f = a m X m + a m− 1 X m− 1 +

···

+ a 0 , i ∈ Z

;

suppose that there is a prime p such that:

p does not divide a m ,

p divides a m− 1 ,...,a 0 ,

p 2 does not divide a 0 .

Then f is irreducible in

[ X ] .

Proof. We may remove any common factor from the coeEcients f , and hence assume

that they have gcd = 1. Therefore, if f ( X ) factors in

[ X ], it factors in

[ X ]:

+ c 0 ) , i ,c i ∈ Z , n, r<m.

Since p , but not p 2 , divides a 0 = b 0 c 0 , p must divide exactly one of b 0 , c 0 ,say p divides b 0 .

Now from the equation

···

+ a 0 =( b n X n +

···

+ b 0 )( c r X r +

···

a 1 = b 0 c 1 + b 1 c 0 ,

we see that p

b 1 . Now from the equation

a 2 = b 0 c 2 + b 1 c 1 + b 2 c 0 ,

b 2 . By continuing in this way, we ﬁnd that p divides b 0 ,b 1 ,...,b n ,which

contradicts the fact that p does not divide a m .

The above three propositions hold with

replaced by any unique factorization domain.

Proposition 1.8.

There is an algorithm for factoring a polynomial in

[ X ] .

[ X ]. Multiply f ( X ) by an integer, so that it is monic, and

then replace it by D deg( f ) f ( D ), D = a common denominator for the coeEcients of f ,toobtain

a monic polynomial with integer coeEcients. Thus we need consider only polynomials

f ( X )= X m + a 1 X m− 1 +

Proof.

Consider f ( X )

∈ Q

From the fundamental theorem of algebra (see later), we know that f splits completely in

···

+ a m , i ∈ Z

[ X ]:

f ( X )=

( X

−

α i ) , α i

∈ C

i =1

From the equation f ( α i ) = 0, it follows that |α i | is less than some bound M depending on

a 1 ,...,a m .Nowif g ( X )isamonicfactorof f ( X ), then its roots in C are certain of the α i ,

and its coeEcients are symmetric polynomials in its roots. Therefore the absolute values of

the coeEcients of g ( X ) are bounded. Since they are also integers (by 1.6), we see that there

are only ﬁnitely many possibilities for g ( X ). Thus, to ﬁnd the factors of f ( X ) we (better

Maple) only have to do a ﬁnite amount of checking.

One other observation is sometimes useful: Suppose that the leading coeEcient of f ( X ) ∈

[ X ] is not divisible by the prime p ;if f ( X ) is irreducible in

F p [ X ], then it is irreducible

[ X ]. Unfortunately, this test is not always eﬀective: for example, X 4

−

10 X 2

+1 is

reducible 1

modulo every prime, but it is irreducible in

[ X ].

1 I don’t know an elementary proof of this. One proof uses that its Galois group is

≈

Z /

) 2 .

a m X m + a m− 1 X m− 1 +

we see that p

(

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