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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?

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Do you mean splits completely? Presumably you require the field extension to be non-trivial. –  user29743 Oct 7 '12 at 4:51
    
@countinghaus: I mean the prime ideal is NOT nonsplit. Isn't a non-trivial extension necessary from the question? –  Khan Oct 7 '12 at 5:05
    
@Khan: You can take the trivial extension of your Dedekind domain. You certainly what some more conditions. It would also be nice if you could say what means nonsplit. –  user18119 Oct 7 '12 at 6:56
    
@Qil: These are the definitions I was using (page 1) maths.bris.ac.uk/~malab/PDFs/Algae_are_more_numb_9.pdf. In other words, $r>1$. I thought in a trivial extension every prime ideal would be nonsplit. Am I missing something? –  Khan Oct 7 '12 at 7:53
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up vote 1 down vote accepted

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

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Thank you, QiL. –  Khan Oct 7 '12 at 17:18
    
@Khan: now I think it is not necessary to prove that $C$ coincides with $B$ over $A_f$. We see that there are two maximal ideals of $B$ lying over $\mathfrak p$. As $C$ is finite over $B$, each of these ideals lifts to a maximal ideal of $C$. So there are at least two maximal ideals of $C$ lying over $\mathfrak p$. –  user18119 Oct 7 '12 at 18:49
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