Page 271 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes Definition: Let F be an extension field of K. The set
{   Aut(F) | (a) = a for all a  K }

is called the Galois group of F over K, denoted by Gal(F/K).
Definition: Let K be a field, let f(x)  K[x], and let F be a splitting field for f(x) over K. Then
Gal(F/K) is called the Galois group of f(x) over K, or the Galois group of the equation f(x) = 0
over K.
Proposition states that if F is an extension field of K, and f(x)  K[x], then any element of
Gal(F/K) defines a permutation of the roots of f(x) that lie in F. The next theorem is extremely
important.
Theorem 1: Let K be a field, let f(x)  K[x] have positive degree, and let F be a splitting field for
f(x) over K. If no irreducible factor of f(x) has repeated roots, then j Gal(F=K)j = [F : K].
This result can be used to compute the Galois group of any finite extension of any finite field, but
first we need to review the structure of finite fields. If F is a finite field of characteristic p, then it
is a vector space over its prime subfield Z , and so it has p elements, where [F : Z ] = n. The
n
p
p
structure of F is determined by the following theorem.
Theorem 2: If F is a finite field with p elements, then F is the splitting field of the polynomial
n
x p n – x over the prime subfield of F.
The description of the splitting field of x p n – x over Z shows that for each prime p and each
p
positive integer n, there exists a field with p elements. The uniqueness of splitting fields shows
n
that two finite fields are isomorphic iff they have the same number of elements. The field with
p elements is called the Galois field of order pn, denoted by GF(p ). Every finite field is a simple
n
n
extension of its prime subfield, since the multiplicative group of nonzero elements is cyclic, and
this implies that for each positive integer n there exists an irreducible polynomial of degree n in
Z [x].
p
p
n
If F is a field of characteristic p, and n  Z+, then {a  F | a = a} is a subfield of F, and this
observation actually produces all subfields. In fact, Proposition 6.5.5 has the following statement:
Let F be a field with p elements. Each subfield of F has p elements for some divisor m of n.
n
m
Conversely, for each positive divisor m of n there exists a unique subfield of F with pm elements.
If F is a field of characteristic p, consider the function  : F  F defined by (x) = xp. Since F has
characteristic p, we have (a + b) = (a + b) = a + b = (a) + (b), because in the binomial expansion
p
p
p
of (a + b) each coefficient except those of ap and bp is zero. (The coefficient (p!)/(k!(p – k)!)
p
contains p in the numerator but not the denominator since p is prime, and so it must be equal to
zero in a field of characteristic p.) It is clear that  preserves products, and so  is a ring
homomorphism. Furthermore, since it is not the zero mapping, it must be one-to-one. If F is
finite, then  must also be onto, and so in this case  is called the Frobenius automorphism of F.
n
n
Note that  (x) = x (Inductively,  (x) = ( (x)) = (x p n-1 ) p = x .) Using an appropriate power
p
p
p
n
n-1
n
of the Frobenius automorphism, we can prove that the Galois group of any finite field must be
cyclic.
Theorem 3: Let K be a finite field and let F be an extension of K with [F : K] = m. Then
Gal(F/K) is a cyclic group of order m.
Outline of the proof: We start with the observation that F has pn elements, for some positive
integer n. Then K has pr elements, for r = n/m, and F is the splitting field of x p n  x over its
prime subfield, and hence over K. Since f(x) has no repeated roots, to conclude that |Gal(F/K)|
= m. Now define  : F  F to be the rth power of the Frobenius automorphism. That is, define

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