Compositum $FL/F$ Galois if $L/K$ Galois This question arises from following thread of mine: Solvable Field Extension Let $K$ a field, $F /K$ an algebraic and $L/K$ a Galois extension of $K$. Why is the compositum $FL/F$ Galois? My attempts: I have only solved the case: $L= K(a)$. In this case I claim that $FL=F(a)$: "$\supseteq$" is obvious and "$\subseteq$" because $a \in F(A)$ implies $L \subset F(a)$ and $F \subset F(a)$ is obvious. Remark: this argument can be extended to the case that $L/K$ is **_finite_** Galois: Then if $L= K(a,b,..., i)$ then $FL= F(a,b,..., i)$ similary,right? Denote $p_a ^K, p_a ^F$ the minimal polynomials of $a$ in $K[X]$ and $F[X]$. Obvoiusly $p_a ^F \vert p_a ^K$ so $a$ is sparable over $F$ and $p_a ^F $ splits in $FL$ since $ p_a ^K$ splits in $L$. How to handle the general case $L /K$ Galois? By the way: is there a more conventional way to argue than the way I described above?

Take an automorphism $\alpha$ of $FL$ that fixes $F$. Then it fixes $K$ in particular, so as $L/K$ is normal, $\alpha(L)=L$. From this you conclude that $\alpha(FL)=FL$ and so $FL/F$ is normal.

Regarding separability, you sketched the argument yourself.