218 CHAPTER 9. SOME ITEMS WHICH RESEMBLE LINEAR ALGEBRA

When the roots of p̄(x) = p(x) are distinct, equality holds in the above. If the roots arelisted according to multiplicity, the automorphisms are determined by the permutations ofthe roots. When the roots are distinct, |G(K,F)| = n!. Also, G(K,F) is a group for theoperation being composition.

Proof: So how large is |G(K,F)| in case p(x) is a polynomial of degree n which has ndistinct roots? Let p(x) be a monic polynomial with roots in K, {r1, · · · ,rn} and supposethat none of the ri is in F. Thus

p(x) = xn +a1xn−1 +a2xn−2 + · · ·+an =n

∏k=1

(x− rk) , ai ∈ F

ThusK= F [r1, · · · ,rn]. Let σ be a mapping from {r1, · · · ,rn} to {r1, · · · ,rn} , say r j→ ri j .In other words σ produces a permutation of these roots. Consider the following way ofobtaining something in G(K,F) from σ . If you have a typical thing in K, you can obtainanother thing inK by replacing each r j with ri j in an element of F [r1, · · · ,rn], a polynomialwhich has coefficients in F. Furthermore, if you do this, then the resulting map fromK toKis an automorphism, preserving the operations of multiplication and addition. Does it keepF fixed? Of course it does because you don’t change the coefficients of the polynomialswhich are always in F. Thus every permutation of the roots determines an automorphismof K.

Now suppose σ is an automorphism of K and the roots of p(x) are distinct. Does σ

determine a permutation of the roots? If ri is a root, what of σ (ri)? Is it also a root simplydue to σ being an automorphism? Note that σ (0) = 0 and so σ (0) = 0 = σ (p(ri)) =p(σ (ri)), the last from the assumption that σ is an automorphism. Thus σ maps roots toroots. Since it is one to one and the roots are distinct, it must be a permutation. It followsthat |G(K,F)| equals the number of permutations of {r1, · · · ,rn} which is n! and that thereis a one to one correspondence between the permutations of the roots and G(K,F) . It isalways the case that an automorphism takes roots to roots, but if the roots are repeated, thenthere may be fewer than n! of these automorphisms.

Now consider the claim about G(K,F) being a group.The associative law (α ·β ) · γ = α · (β · γ) is obvious. This is just the way composition

acts.The identity ι is just the identity map, clearly an automorphism which fixes F.Each automorphism is, by definition one to one and onto. Therefore, the inverse must

also be an automorphism. Indeed, if σ (x) ,σ (y) are two generic things in K, then

σ−1 (σ (x)σ (y)) = σ

−1 (σ (xy)) = xy = σ−1 (σ (x))σ

−1 (σ (y))

That σ−1 is an automorphism with respect to addition goes the same way. The estimate onthe size is from 9.22. Does the inverse fix F? Consider α,α2, · · · . Because of the estimateon the size of G(K,F) , you must have αm = αn for some m < n. Hence multiply on theleft by

(α−1

)m to get ι = αn−m. Thus α−1 = α(n−1)−m which is α raised to a nonnegativepower. The right leaves F fixed and so the left does also. ■

In the above, there is a field which is a finite extension of a smaller field and the groupof automorphisms which leave the given smaller field fixed was discussed. Next is a moregeneral notion in which there is given a group of automorphisms. This group will determinea smaller field called a fixed field.