Linear Algebra, Theory And Applications, 2012a

466 FIELDS AND FIELD EXTENSIONS
Definition F.5.5 AgroupG is a nonempty set with an operation, denoted here as · such
that the following axioms hold.
1. For α, β, γ ∈ G, (α · β) · γ = α · (β · γ) . We usually don’t bother to write the ·.
2. There exists ι ∈ G such that αι = ια = α
3. For every α ∈ G, there exists α −1 ∈ G such that αα −1 = α −1 α = ι.
Then why is G ≡ G (K, F) , where K is a finite extension of F, a group? If you simply
look at the automorphisms of K then it is obvious that this is a group with the operation
being composition. Also, from Theorem F.4.5 |G (K, F)| is finite. Clearly ι ∈ G. It is
just the automorphism which takes everything to itself. The operation in this case is just
composition. Thus the associative law is obvious. What about the existence of the inverse?
Clearly, you can define the inverse of α, but does it fix F? Ifα = ι, then the inverse is clearly
ι. Otherwise, consider α, α 2 , ··· . Since |G (K, F)| is finite, eventually there is a repeat. Thus
α m = α n , n>m.Simply multiply on the left by ( α −1) m
to get ι = αα n−m . Hence α −1 is
a suitable power of α and so α −1 obviously leaves F fixed. Thus G (K, F) which has been
called a group all along, really is a group.
Then the following corollary is the reason why separable polynomials are so important.
Also, one can show that if F contains a field which is isomorphic to Q then every polynomial
with coefficients in F is separable. This will be done later after presenting the big results.
This is equivalent to saying that the field has characteristic zero. In addition, the property
of being separable holds in other situations which are described later.
Corollary F.5.6 Let K be a splitting field of p (x) over the field F. Assume p (x) is
separable. Then
|G (K, F)| =[K : F]
Proof: Just note that K is also the splitting field of p 1 (x), the product of the distinct
irreducible factors and that from Lemma F.5.3, p 1 (x) has distinct roots. Thus the conclusion
follows from Theorem F.4.5.
What if L is an intermediate field between F and K? Thenp 1 (x) still has coefficients in
L and distinct roots in K and so it also follows that
|G (K, L)| =[K : L]
Definition F.5.7 Let G be a group of automorphisms of a field K. Then denote by K G the
fixed field of G. Thus
K G ≡{x ∈ K : σ (x) =x for all σ ∈ G}
Thus there are two new things, the fixed field of a group of automorphisms H denoted
by K H and the Gallois group G (K, L). How are these related? First here is a simple lemma
which comes from the definitions.
Lemma F.5.8 Let K be an algebraic extension of L (each element of L is a root of some
polynomial in L) forL, K fields. Then
G (K, L) =G ( K, K G(K,L)
)
Proof: It is clear that L ⊆ K G(K,L) because if r ∈ L then by definition, everything in
G (K, L) fixesr and so r is in K G(K,L) . Therefore,
G (K, L) ⊇ G ( K, K G(K,L)
)
.