Galois closure of a finite extension is finite 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 $[E : F] < \infty$.
 A: 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.)
A: I'll give a quick sketch of my idea here.
Since $K/F$ is a finite field extension, there exists a $p \in F[x]$ such that $K \cong \frac {F[x]}{(p)}$. Thus, in $K$, $p$ factors: $p=p_1 \cdots p_n$. Note that $\deg{p_i} < \deg{p}$.
Define $K_1$ a field extension by $K_1 \cong \frac{K[x]} {(p_1)}$. Then in $K_1$, $p_1$ factors. After a finite number of iterations, $p_1$ will factor into linear terms.
Continuing this process for the other $p_i$, we will eventually have that $p$ factors into linear terms over some field extension $K_l$. That is, in $K_l$, $p=(x-\alpha_1) \cdots (x-\alpha_n)$. Then $K_l$ is a splitting field for $p$ over $F$. Since splitting fields are Galois (and vice versa), we conclude that $K_l$ is a Galois extension of $K$.
