Can I find a Galois extension which contains a finite set of algebraic elements?

Question

Suppose $K/F$ is an algebraic extension of fields. Pick a finite collection $a_1,...,a_n \in K$. Can I find a Galois extension of $F$ which contains these $n$ elements of $K$?

Answer 1

Yes, this is possible if $K/F$ is a separable extension. Let $P_i \in K[X]$ the minimal polynomial of $a_i$ and define $P$ to be the product of all the $P_i$.

Moreover, let $L$ the splitting field of $P$ over $F$. Then $L/F$ is a Galois extension (it is clearly normal, ans also separable because it is generated over $F$ by separable elements), and $L$ contains all the $a_i$'s (since they are roots of $P$).

But if $K/F$ is not a separable extension, this can be wrong. Let $F = \mathbb F_p(t)$ for some prime $p$, and $K = F(\sqrt[p]{t})$ a rupture field of $X^p-t$ over $F$. If $L/F$ is an extension containing $a=\sqrt[p]{t}$, then the minimal polynomial of $a$ over $F$ is $X^p-t = (X-a)^p$ which is not separable. Then $L/K$ is not a Galois extension.