This is probably an easy question, but I don't know the answer:

Let $R \subset S$ be an integral ring extension, and let $p$ be a prime ideal of $R$. Then $p^{ec} = pS \cap R$ can be different from $p$.

We always have $p \subseteq pS \cap R$, so if $\dim(R)=0$ (or if $R$ is Dedekind; the case $p=(0)$ being trivial), then we have equality. So, what would be a counter-example?

  • 1
    $\begingroup$ Why do you think that $p^{ec}\ne p$? AFAIK, $p=P^c$ and $P^{cec}=P^c$. $\endgroup$
    – user26857
    Commented Nov 19, 2016 at 11:26
  • $\begingroup$ @user26857 : Thank you for your comment. Maybe $p^{ec}=p$, but I wasn't able to show it. I know that there is a prime $q$ of $S$ such that $q^{c}=p$, but it didn't help me. $\endgroup$
    – Watson
    Commented Nov 19, 2016 at 11:29

1 Answer 1


Thanks to user26857's helpful comment above, I think that I'm able to provide an answer.

By the lying-over property, there is a prime ideal $q$ of $S$ such that $q^c=p$. Then: $$p^{ec} = q^{cec}=q^c=p.$$

This is wrong is $S$ is not integral over $R$, for instance $p=2\Bbb Z \triangleleft R=\Bbb Z \subset S=\Bbb Q$.

  • 3
    $\begingroup$ In fact, if $R\subset S$ is a ring extension and $p\subset R$ a prime ideal, then $p^{ec}=p\iff$ there is a prime $P\subset S$ such that $p=P^c$. $\endgroup$
    – user26857
    Commented Nov 19, 2016 at 13:49

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