$\newcommand{\Gal}{\operatorname{Gal}}
\newcommand{\Aut}{\operatorname{Aut}}$
Suppose we have field extensions $F < K_1, K_2 < K$. It is known that if $K_1/F$ and $K_2/F$ are both Galois and $K_1 \cap K_2 = F$, then the Galois group of the compositum is given by:
$$
\Gal(K_1K_2/F) \cong \Gal(K_1/F) \times \Gal(K_2/F)
$$
If we don't assume $K_1 \cap K_2 = F$, we can still characterize $\Gal(K_1K_2/F)$ as a subgroup of this product. See Dummit and Foote, chapter 14, proposition 21.
My question is: what if we don't assume that $K_1/F$ and $K_2/F$ are Galois? Do we still have that:
$$
\Aut(K_1K_2/F) \cong \Aut(K_1/F) \times \Aut(K_2/F)?
$$
If not, what would be an easy counter-example?
Best Answer
I’m going to interpret your question as asking whether $\mathrm{Aut}(K_1K_2/F) = \mathrm{Aut}(K_1/F)\times\mathrm{Aut}(K_2/F)$ if $K_1$ and $K_2$ are extensions of $F$ such that $K_1\cap K_2=F$, where for fields $L\subseteq M$, $$\mathrm{Aut}(M/L) = \{\sigma\in\mathrm{Aut}(M)\mid \sigma|_L=\mathrm{id}_L\},$$ and hence equals the Galois group in the special case where $M$ is Galois over $L$.
Consider $F=\mathbb{Q}$, $K_1=\mathbb{Q}(\sqrt[3]{2})$, and $K_2=\mathbb{Q}(\omega)$, where $\omega$ is a complex cubic root of unity; that is, a root of $x^2+x+1$.
Then $K_2$ is Galois over $\mathbb{Q}$ and $\mathrm{Gal}(K_2/F)$ is cyclic of order $2$. On the other hand, $\mathrm{Aut}(K_1/F)$ is trivial, because $\sqrt[3]{2}$ is the only real cubic root of $2$, $K_1$ is contained in $\mathbb{R}$, and an automorphism of $K_1$ must send $\sqrt[3]{2}$ to a cubic root of $2$. Thus, $$\mathrm{Aut}(K_1/F)\times\mathrm{Aut}(K_2/F)\cong\{e\}\times C_2 \cong C_2$$ is cyclic of order $2$.
However, $K_1K_2$ is the splitting field of $x^3-2$, which has degree $6$ over $F$. In particular, $\mathrm{Gal}(K_1K_2/F)$ has order $6$ (and is isomorphic to $S_3$) so we get neither an isomorphism with, nor a realization as a subgroup of the direct product.