TITLE: Galois closure of a finite extension is finite
QUESTION [5 upvotes]: We proved the fundamental theorem of algebra in my field theory class the other day, but the professor glossed over an (imo) important step which I have found myself unable to prove. Can someone help me out?
If $K/F$ be a finite field extension and $E/F$ is the Galois closure of $K$, then $[E : F] < \infty$.

REPLY [4 votes]: Here’s one of several possible ways of doing it:
Let $\{b_1,\cdots,b_n\}$ be a basis for $K$ as an $F$-vector space. Each $b_i$ has an $F$-minimal polynomial $P_i(X)\in F[X]$. Adjoin all roots of all the polynomials $P_i$. It’s clearly a normal extension of $F$, and pretty clearly the smallest normal extension of $F$ containing $E$.
(If $E$ wasn’t separable over $F$ to start with, I don’t know what a Galois closure of $E$ over $F$ could be.)