An automorphism of any mathematical object is a bijective homomorphism from the object to itself. It’s easy to see that the automorphisms of any object form a group: the composition of homomorphisms is itself a homomorphism, function composition is associative, there exists an identity function, and the inverse of a bijective homomorphism is a homomorphism. Recall that a homomorphism of fields is one that satisfies .
The automorphism group of a field L is denoted Aut(L). If L/K is a field extension, then the automorphism group of the extension, Aut(L/K), is the subgroup of Aut(L) consisting of all automorphisms of L that fix every element of K, i.e. .
First, elements of Aut(L) are linearly independent over L. Denoting elements by a, b, c, etc., and automorphisms by f, g, etc., we get that with and .
To see why, suppose otherwise and let n be the minimal integer for which the equation holds for some nonzero a(i)’s. Clearly, this implies that all a(i)’s are nonzero. As the automorphisms are distinct, let . For every a in L, and ; subtracting the two equations, we get , contradicting the minimality of n. Note that is a nonzero constant in L rather than an automorphism.
Every subset of Aut(L) has a fixed field, consisting of all elements of L fixed by every element in the subset. That the fixed field is indeed a field follows trivially from the definition of a homomorphism. The fixed field of the subset S of Aut(L) is denoted , but I’ll sometimes ASCIIfy it to L^S. The pair of fields – L and L^S – form an extension, so we can try relating the extension to the subset.
First, if S is finite, then [L:L^S] >= |S|. To see why, let [L:L^S] = n and suppose on the contrary that |S| > n. Let b(i) from i = 1 to n be a basis for L over L^S, and let f(j) from j = 1 to n + 1 be distinct elements of S. The system of equations from i = 1 to n in the variables x(j) in L has n + 1 unknowns and n equations, so it has infinitely many solutions. In particular, it has a solution other than x1 = x2 = … = x(n + 1) = 0.
For that solution, for all k(i) in L^S; since every element of L can be written as , this implies that every x(j) = 0 by the linear independence of automorphism. That’s clearly a contradiction, so [L:L^S] >= |S|.
Second, if S is a subgroup rather than just a subset, then in fact [L:L^S] = |S|. To see why, suppose on the contrary that |S| < [L:L^S]. Let |S| = n; then there exist n + 1 elements of L that are linearly independent over L^S, say b(i). The system of equations has n equations and n + 1 unknowns, so it has a nonzero solution. Choose a solution with a minimal number of nonzero values of x(i), say m, and relabel the b(i)’s to make these the first m variables.
Now, S is a group, so it contains the identity, say f1. The linear independence of the b(i)’s and the resulting equality imply that not all x(i)’s are in L^S. So divide throughout to get x(m) = 1, which is in L^S, and suppose x1 is not in L^S and f2(x1) != x1 = f1(x1). For every j, we have ; since S is a group, we can relabel the index gotten from composing f2 and f(j) as j, to get . But x(m) is in L^S while x1 isn’t, so by subtraction we get a solution with at least 1 but at most m – 1 nonzero values of x(i), contradicting the minimality of m. Thus [L:L^S] = |S|.
The above results help us relate subgroups of a finite subgroup G of Aut(L) to intermediate fields of L/L^G. To see why, if H is a subgroup of G, then [L:L^H] = |H|, and by the tower law, [L^H:L^G] = [G:H]. Further, Aut(L/L^H) = H, because every element of H clearly fixes L^H, and if Aut(L/L^H) has any additional elements, then [L:L^S] = |S| implies that it has a smaller fixed field than L^H, a contradiction.
Conversely, if E is an intermediate field of L/L^G, then L^Aut(L/E) = E, because by construction L^Aut(L/E) contains E, and if it’s any bigger then we’ll get Aut(L/E) has fewer than [L:E] elements, another contradiction.