Must an automorphism which fixes two fields necessarily fix their composite?

Question

Suppose $K$ and $F$ are subfields of some field $L$, and $\sigma$ is an automorphism of $L$ which fixes $K$ and $F$. Must $\sigma$ fix the composite $KF$?

Reading through chapter 14 of Dummit it seems the answer is yes, but I cannot see why. $\sigma$ clearly fixes $K\cap F$, but could not there be some element of $KF$ which is in neither $K$ nor $F$, which is not fixed by $\sigma$? When $KF/K\cap F$ is Galois, the answer is yes by the Fundamental Theorem of Galois Theory, but I cannot reason out why it must be true in general.

I have tried looking at things like $K=\mathbf{Q}(\sqrt 2)$, $F=\mathbf{Q}(\pi)$, $L = \mathbf{C}$, but I realized I am not sure what automorphisms which fix $F$ look like, or what $KF$ looks like.

Answer 1

$KF$ is the smallest subfield of $L$ containing $K$ and $F$. The field $L^{\sigma}$ contains $K$ and $F$ by hypothesis, so, by definition, it contains the smallest subfield of $L$ containing $K$ and $F$, that is, $KF$.

If $K$ and $F$ are finite extensions of some base field $E$, you can get an explicit description of the composite as follows (for the general case, refer to J. Darné's answer): $$KF=\left\{\sum_{i=1}^nk_if_i\ \colon n\in\mathbb{N},k_i\in K,f_i\in F\right\}.$$ The explicit description can come in handy at times, but I think it's often more useful to think of $KF$ in terms of its defining property, i.e. being the smallest subfield containing $K$ and $F$.

An automorphism of $\mathbb{C}$ that fixes $\mathbb{Q}(\pi)$ is the complex conjugation. Others (a lot even) exist, but are non-explicit and quite misbehaved.

Answer 2

The answer is yes : elements of $KF$ are all that you can obtained from elements of K and F by applying the field operations to elements of K and of F (they are algebraic fractions in elements of K and F). As a consequence, $\sigma$, which fixes all elements of K and F and preserves the field operations, must preserve all these elements.

If you want to write it down completely, you can take : $$A := \left\{\frac{P(k_1, ..., k_n,f_1, ..., f_m)}{Q(k_1, ..., k_n,f_1, ..., f_m)} \middle| P, Q \in \mathbb C[X_1, ..., X_n,Y_1, ..., Y_m],\ k_1,..., k_n \in K,\ f_1, ..., f_n \in F,\ Q(k_1, ..., k_n,l_1, ..., l_m) \neq 0 \right\},$$ then $A$ is a subfield of $L$, and it is obviously the smallest containing $K$ and $F$, thus $A = KF$. And it is easy to see that $\sigma$ must preserve these elements.