## Galois Theory: Field Automorphisms

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 $f(a + b) = f(a) + f(b), f(ab) = f(a)f(b)$.

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. $\textrm{Aut}(L/K) = \{f \in \textrm{Aut}(L): \forall a \in K, f(a) = a\}$.

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 $a_{1}f_{1} + a_{2}f_{2} + ... + a_{n}f_{n} = 0 \Leftrightarrow a_{1} = ... = a_{n} = 0$ with $a_{i} \in L$ and $f_{i} = f_{j} \Leftrightarrow i = j$.

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 $f_{1}(b) \neq f_{n}(b)$. For every a in L, $a_{1}f_{1}(ab) + a_{2}f_{2}(ab) + ... + a_{n}f_{n}(ab) = 0$ and $f_{n}(b)(a_{1}f_{1}(a) + a_{2}f_{2}(a) + ... + a_{n}f_{n}(a)) = 0$; subtracting the two equations, we get $(a_{1}(f_{1}(b) - f_{n}(b))f_{1}(a) + ... + a_{n-1}(f_{n-1}(b) - f_{n}(b))f_{n-1}(a) = 0$, contradicting the minimality of n. Note that $f_{1}(b) - f_{n}(b)$ 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 $L^{S}$, 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  $f_{1}(b_{i})x_{1} + f_{2}(b_{i})x_{2} + ... + f_{n+1}(b_{i})x_{n+1} = 0$ 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, $x_{1}f_{1}(k_{1}b_{1} + ... + k_{n}b_{n}) + ... + x_{n+1}f_{n+1}(k_{1}b_{1} + ... + k_{n}b_{n}) = 0$ for all k(i) in L^S; since every element of L can be written as $k_{1}b_{1} + ... + k_{n}b_{n}$, 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 $f_{j}(b_{1})x_{1} + ... + f_{j}(b_{n+1})x_{n+1} = 0$ 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 $b_{1}x_{1} + ... + b_{m}x_{m} = 0$ 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 $f_{2}(f_{j}(b_{1})x_{1} + ... + f_{j}(b_{m})x_{m}) = 0$; since S is a group, we can relabel the index gotten from composing f2 and f(j) as j, to get $f_{j}(b_{1})f_{2}(x_{1}) + ... + f_{j}(b_{m})f_{2}(x_{m}) = 0$. 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.

### 5 Responses to Galois Theory: Field Automorphisms

1. Pseudonym says:

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?

2. Alon Levy says:

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).

3. Pseudonym says:

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.

4. Alon Levy says:

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.

5. […] Levy has a bunch of cool posts up, but here are my pics: a couple of cool posts on Galois Theory, and an interesting expression of pessimism toward the left […]