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 *x*1 = *x*2 = … = *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 *x*1 is not in L^S and *f*2(*x*1) != *x*1 = *f*1(*x*1). For every *j*, we have ; since S is a group, we can relabel the index gotten from composing *f*2 and *f*(*j*) as *j*, to get . But *x*(*m*) is in L^S while *x*1 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.

Does a field homomorphism also have to satisfy f(a/b) = f(a)/f(b)? If so, is this automatic, or must it be proven?

It’s automatic, subject to the caveat that the homomorphism is nonzero. In that case 1 has to map to 1, because we have 1*f(a) = f(a) = f(1*a) = f(1)*f(a) so either f(a) = 0 for all a, or f(1) = 1. But f(a) = 0 iff the homomorphism is zero, so by assumption, f(1) = 1.

Then, 1 = f(1) = f(a*(1/a)) = f(a)f(1/a), so f(1/a) = 1/f(a), and f(a/b) = f(a)f(1/b) = f(a)/f(b).

Gotcha, thanks. I figured this was the case, but I managed to get you to do the work for me. 🙂

Incidentally, this stuff isn’t necessarily obvious. A homomorphism between two geometric algebras, for example, is required to preserve outer product but not necessarily inner product.

Well, it depends on the algebraic structure you deal with, obviously. But it’s true for homomorphisms of groups (f(gh) = f(g)f(h) implies f(1) = 1 and f(g^(-1)) = f(g)^(-1)), so whenever you have a group structure, as you do on fields with respect to both addition and multiplication, homomorphisms preserve inverse operations.

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