220 CHAPTER 9. SOME ITEMS WHICH RESEMBLE LINEAR ALGEBRA

This is because σ (1) = 1 if σ is an automorphism. It is of the form Mσ r (x) = 0. Theoriginal system in 9.24 is of the form

σ1 (u1)x1 + · · ·+σ1 (uk)1+ · · ·+σ1 (ul)xl + · · ·+σ1 (um)xm = 0σ2 (u1)x1 + · · ·+σ2 (uk)1+ · · ·+σ2 (ul)xl + · · ·+σ2 (um)xm = 0

...σn (u1)x1 + · · ·+σn (uk)1+ · · ·+σn (ul)xl + · · ·+σn (um)xm = 0

which will be denoted as Mx= 0. Thus M (σ r (x)−x) = 0 where y ≡ σ r (x)−x ̸= 0.If any xk is 0, then σ r (xk) = 0. Thus all zero entries of x remain 0 in y and yk = 0 whereasxk ̸= 0 so y has fewer nonzero entries than x contradicting the choice of x as the one withfewest nonzero entries such that Mx= 0. ■

With the above estimate, here is another relation between the fixed fields and subgroupsof automorphisms.

Proposition 9.4.16 Let H be a finite group of automorphisms defined on a field K. Thenfor KH the fixed field,

G(K,KH) = H

Proof: If σ ∈H, then by definition of KH , σ ∈G(K,KH) so H ⊆G(K,KH) . Then byTheorem 9.4.15 and Theorem 9.4.12,

|H| ≥ [K :KH ]≥ |G(K,KH)| ≥ |H|

and so H = G(K,KH). ■For H a group of automorphisms of G(K,F) , let Hx be all hx for h ∈ H. Thus Hx = x

means hx = x for all h ∈ H. KH = {x ∈K : Hx = x}.Note how this proposition shows G(K,F) = G

(K,KG(K,F)

). Thus

|G(K,F)|=∣∣G(K,KG(K,F)

)∣∣= [K :KG(K,F)].

Is KG(K,F) = F? If x ∈ F,G(K,F)x = x so by definition, x ∈KG(K,F) and so F⊆KG(K,F).However, if x ∈KG(K,F) so G(K,F) fixes x, it is not at all clear that x ∈ F. Maybe G(K,F)fixes more things than F. Later a situation is given in which KG(K,F) = F.

Summary 9.4.17 The following are now available.

1. LetK be the splitting field of p(x). Then |G(K,F)| ≤ [K : F] . If the roots of p(x) areunique, then these are equal.

2. |G(K,F)| ≤ n! and when the roots of p(x) are distinct, |G(K,F)|= n!.

3. If H is a finite group of automorphisms on an arbitrary field K, then it follows thatG(K,KH) = H where KH is the fixed field of H.

4. F⊆KG(K,F) but it is not clear that these are equal.