Finite separable field extensions such that $KL/L$ and $K/K\cap L$ have non-isomorphic automorphism groups

Question

If $K/F$, $L/F$ are finite separable extensions (not necessarily finite Galois extensions), then it seems clear that $KL/F$ is also a finite separable extension.

However, in this case, is $\operatorname{Aut}(KL/L)$ isomorphic to $\operatorname{Aut}(K/K\cap L)$? I think there must be some counterexample...but can't find one. Could anyone help me?

Answer 1

Let $F = \mathbf{Q}$, $K = \mathbf{Q}(\sqrt[3]{2})$, and $L = \mathbf{Q}(\zeta_3)$.

Then $KL$ is Galois over $\mathbf{Q}$, hence a fortiori Galois over $L$. Thus $\mathrm{Aut}(KL/L)$ has order $[KL:L] = 3$. On the other hand $\mathrm{Aut}(K/K \cap L) = \mathrm{Aut}(\mathbf{Q}(\sqrt[3]{2}/\mathbf{Q})$ is trivial (there is nowhere for $\sqrt[3]{2}$ to go except to $\sqrt[3]{2}$).