Suppose that $L/K$ is a normal extension. In his book Commutative Algebra and Algebraic Geometry, Bosch uses the following fact in a proof of the going-down theorem (p.99-100, lemma 5): if $L/K$ is a normal field extension,and $K \subseteq E \subseteq L$ is a field extension such that $E/K$ is normal, then $E \neq L$ implies that there is a finite extension $E'/E, E' \subseteq L$ such that $E'/K$ (not just $E'/E$) is normal. I don't understand why this is true. The idea seems to work with extensions $E(a)/E$ and $E(a)/K$ where $a \in L\setminus E$. By definition, $E(a)/E$ is finite.
I know that every finite field extension is contained is a finite normal field extension, but I don't see how this helps here. Indeed, we could take a normal $E'$ closure of $E(a)/E$, but it wouldn't necessarily hold that $E'/K$ is normal, for normal extensions are not transitive. On the other hand, we could take normal closure $E''$ of $E(a)/K$, but why would $E''/E$ be finite in this case?
I doubt there is a mistake, since Eisenbud in Commutative Algebra with a view toward Algebraic Geometry (p.291) gives a similar proof.
Best Answer
$E''/E$ is finite because it is generated over $E$ by the roots of the $K$-minimal polynomial of $a$.
More generally, we are given a tower $L/M/E/K$ with $L/K$ and $E/K$ normal (possibly infinite) and $M/E$ finite.
Write $M = E(a_1,\ldots,a_n)$. Let $f_j\in K[x]$ be the minimal polynomial of $a_j$.
Then the normal closure $E'$ of $M/K$ is $E(\{ a_{ij}\})$ where $a_{ij}$ is the $i$-th root of $f_j$.
(the proof is relatively immediate, take an element $b$ of $E'=E(\{ a_{ij}\})$, so it is a rational function in the $a_{ij}$ with coefficients in $E$, and apply a $K$-homomorphism $\sigma:E'\to \overline{L}$, you'll get that $\sigma(b)$ is a rational function in the $\sigma(a_{ij})$ with coefficients in $\sigma(E)=E$, where $\sigma(a_{ij})$ must be a root of $f_j$, so $\sigma(b)\in E'$)
So $E'/E$ is a finite extension.
Each $f_j$ splits completely in $L$ so $E'\subset L$.