Finding separable extensions in which a prime ideal in a Dedekind domain splits

Question

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?

Answer 1

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 $(T-1)^2T^2-\pi=0$.