Finding separable extensions in which a prime ideal in a Dedekind domain splits Given a prime ideal in a Dedekind domain, can we always find a separable extension in which the prime ideal splits? If the answer is no in general, is it true under mild conditions on the Dedekind domain?
 A: Let $k$ be the residue field at a given maximal ideal $\mathfrak p$ of a Dedekind domain $A$. Suppose $\mathrm{char}(k)\ne 2$. Let $a\in 1+\mathfrak p$ such that $a$ is not a square in $A$. Consider 
$$ B=A[T]/(T^2-a).$$ 
This is an integral domain finite over $A$. Let $C$ be the integral closure of $A$ in $\mathrm{Frac}(A)$. Then $B\subseteq C$. Let $f=2a$. Then 
$$B_f=A_f[T]/(T^2-a)$$
is unramified over $A_f$ because for all $\mathfrak q$ not containing $f$, we have 
$$ B\otimes_A A/\mathfrak q=k(\mathfrak q)[T]/(T^2-\bar{a})$$
is reduced. This implies that $B_f$ is integrally closed, hence $B_f=C_f$. 
Above $\mathfrak p$, 
$$ C\otimes_A A/\mathfrak p=B\otimes_A A/\mathfrak p=k[T]/(T^2-1)$$ 
is direct sum of two copies of $k$, so there at least (hence exactly) two prime ideals of $C$ above $\mathfrak p$. The extension is actually completely split.
If $\mathrm{char}(k)=2$, one can use one equation of the form $T^2+T+a$.
Remark An element $a$ as at the begininning usually exist: if $\mathfrak q$ be another maximal ideal of $A$, use Chine Remainder Theorem to find 
$$a\equiv 1 \mod \mathfrak p, \quad a\equiv 0 \mod \mathfrak q, \quad a\not\equiv 0 \mod \mathfrak q^2.$$
If $A$ is a local ring with uniformizing element $\pi$, then use the equation $(T-1)^2T^2-\pi=0$. 
