Let $A$ be an integrally closed domain whose quotient field is $K$, $L$ be a finite Galois extension of $K$, and $B$ be the integral closure of $A$ in $L$. Let $M\_A$ be a maximal ideal of $A$, and $M\_B$ be a maximal ideal of $B$ that lies above $M\_A$ (that is, $M\_B\cap A = M\_A$). Denote by $F\_A$ the field $A/M\_A$, and by $F\_B$ the field $B/M\_B$; so $F\_B/F\_A$ is an algebraic field extension, and it is normal. 


1) Must $F\_B/F\_A$ be finite ? (this is the case if, say $A$ is Noetherian, and I can show that this is also true if $F\_B/F\_A$ is separable)


2) The most important question for me : Assuming $F\_B/F\_A$ finite, must the residual degree $[F\_B : F\_A]$ divide $[L : K]$ (this is actually the case if $A$ is a Dedekind ring, or $A$ and $B$ are discrete valuation rings). 



Divisibility of the degree of an extension by the degree of its residual field
Let $k$ be the field with $2$ elements. Let $R = k[t\_i, x\_i; i \in \mathbf{N}]$. Let $\sigma : R \to R$ be the order $2$ automorphism sending $t\_i$ to $t\_i$ and $x\_i$ to $x\_i + t\_i$. Let $S \subset R$ be the fixed elements under the action. Then $R$ and $S$ are normal domains and $R$ is integral over $S$. In particular, $R$ is the integral closure of $S$ in the fraction field extension and the fraction field extension has degree $2$ by Galois theory.


Let $\mathfrak p \subset R$ be the prime ideal $\mathfrak p = (t\_i)$. This lies over the prime ideal $\mathfrak q = S \cap \mathfrak p$. We are going to show that $\kappa(\mathfrak q) \subset \kappa(\mathfrak p)$ is infinite. This will give an example where 1) fails, because we can then consider $A = S\_{\mathfrak q} \subset B = R\_{\mathfrak p}$. I omit the verification that $B$ is the integral closure of $A$ in the fraction field extension (this relies on the fact that $\mathfrak p$ is the only prime of $R$ lying over $\mathfrak q$). By using finitely many variables you can find an example where 2) fails.


OK, so let us consider an element $f \in R$. We can write $f$ as
$$
f\_0 + \sum t\_i f\_i + h.o.t.
$$
where $f\_0, f\_i$ depend only on the $x\_j$ and all other terms have more $t\_k$'s in them. Now if $f \in S$, then $\sigma(f) = f$. Note that
$$
\sigma(f) = f\_0 + \sum t\_i(f\_i + \partial f\_0/\partial x\_i) + h.o.t.
$$
This means that $\partial f\_0/\partial x\_i$ is identically zero. In other words, we see that $f\_0 \in k[x\_i^2]$. Hence we see that $\kappa(\mathfrak q) = k(x\_i^2)$. Since $\kappa(\mathfrak p) = k(x\_i)$ we are done.


