I am trying to pin down my lack of understanding in the following case, I know it must be something easy I am missing, so any pointers would be really appreciated.
I have a (finite) Galois field extension $K / F$. I also have another (finite) Galois extension $L / F$. I want to show that if $K \neq L$, then there is some non identity element in the Galois group of $L / F$ that acts trivially on $K$ (obviously assuming that $K,L$ are both contained in some field, and taking a lift of the automorphism to this field).
It seems straight forward, I want to say that we take some element $a \in L$ that is not in $K \cap L$. $L$ is Galois over $K \cap L$, so taking the min poly of $a$ over $K\cap L$ all the conjugates of $a$ are in $K$, but not in $L$. Take an element $\sigma$ of the Galois group of $L / L \cap K$ that acts on $a$, then this is non-trivial on $L$, but trivial on $K$. The nagging doubt I have is how to justify that it is trivial on $K$, i.e. does a lift of $\sigma$ necessarily fix $K$?
Another way of phrasing this (using the notation above) is to consider the compositum $KL$, as both $K$ and $L$ are Galois over $F$ then the Galois group of $KL / F$ is the direct product $Gal(K / (K \cap L) \times Gal(L / (K \cap L))$. This implies that we can take $(id, \tau)$ for any non-trivial $\tau \in Gal(L / (K \cap L))$ and this element satisfies the condition of acting non-trivially on $L$, and as the identity on $K$.
The second argument seems sound, but from the first argument I am worried that I am missing some basic understanding that may come back to bite me. I would appreciate any suggestions as to what I am missing.
Thanks in advance
$\newcommand{\QQ}{\mathbb{Q}}$ Your confusion stems from the fact that your $\sigma$ is an automorphism of $L$, and if $K$ is not a subfield of $L$, then you can't ask if $\sigma$ fixes $K$. If you instead consider a lift of $\sigma$ to an automorphism of say $KL$, then the answer may be yes or no, depending on your choice of lift. As you say in your "second argument", taking any nontrivial $\tau\in\text{Gal}(L/L\cap K)$, the automorphism $(\text{id},\tau)$ is a lift of $\tau$ that fixes all of $K$. However, if $\rho$ is any automorphism of $K/L\cap K$, then $(\rho,\tau)$ is also a lift of $\tau$, which does not fix all of $K$.
Edit: For example, consider $F = \QQ, K = \QQ(\sqrt{2},\sqrt{3})$, $L = \QQ(\sqrt{2},\sqrt{5})$, then any automorphism of $KL = \QQ(\sqrt{2},\sqrt{3},\sqrt{5})$ is determined by how it acts on each of the three generators. For each one it can either fix it, or negate it, so you can easily see that the structure of the galois group of $KL/F$ is just $(\mathbb{Z}/2\mathbb{Z})^3$
In this situation $K\cap L = \QQ(\sqrt{2})$. You can consider the automorphism of $L$ sending $\sqrt{2}\mapsto\sqrt{2}, \sqrt{5}\mapsto-\sqrt{5}$. Lifting this to an automorphism of $KL$ is the same as choosing whether to send $\sqrt{3}$ to $\sqrt{3}$ or to $-\sqrt{3}$. This of course affects whether the lift acts trivially on $K$.