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Let $R$ be a ring and let $S$ be a subring of R. If $R$ is a semi-local ring and $R$ is integral over $S$, why $S$ is semi-local as well?

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up vote 5 down vote accepted

Because $R$ is integral over $S$, given any maximal ideal of $\mathfrak n$ of $S$, there is a maximal ideal $\mathfrak m$ of $R$ lying over $\mathfrak m$. This shows that the number of maximal ideals of $S$ is bounded above by the number of maximal ideals of $R$. In particular, if $R$ has only finitely many such, then the same is true of $S$.

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Thanks! It is clear now. Problem is, which answer shall I choose? heh, both are great. – user6495 May 16 '11 at 2:55

A ring is semi-local when it has finitely many maximal ideals. By Corollary 5.8 of Atiyah-Macdonald, for every prime ideal $\frak{q}$ of $R$, the ideal ${\frak{p}}={\frak{q}}\cap S$ is maximal if and only if $\frak{q}$ is maximal. By Theorem 5.10 of Atiyah-Macdonald, for every prime $\frak{p}$ of $S$ there is some prime $\frak{q}$ of $R$ with ${\frak{q}}\cap S=\frak{p}$. Thus, because $R$ has finitely many maximal ideals, so will $S$.

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Dear Zev, You also need to note that every maximal ideal $\mathfrak p$ of $S$ does arise from some $\mathfrak q$ in this way. Of course, this follows from integrality of the extension. Regards, – Matt E May 16 '11 at 2:25
@Matt E: You are correct, I'd realized this and added it in just now. Thanks for the heads up though. – Zev Chonoles May 16 '11 at 2:26
Thanks, but why is there an injection from Specmax(R) to Specmax(S) ? – user6495 May 16 '11 at 3:27
@user6495: Ah... there isn't. Sorry, I wasn't thinking clearly about that. You should choose Matt E's answer :) – Zev Chonoles May 16 '11 at 3:40
yeah, I didn't see it either. Thank you a lot for this answer and the previous ones, you have been very helpful! – user6495 May 16 '11 at 3:42

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