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Let $A$ be a valuation ring of a field $K$. Show that every subring of $K$ which contains $A$ is a local ring of $A$.

This problem is already asked and answered at mathoverflow. But I can't understand why $PA_P \subset M_B$ at step (b) of the answer. Or there will be another way to show $B \subset A_P$. How can I prove it?

Community
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Gobi
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1 Answers1

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Recall that $P=A\cap M_B$. Since $M_B$ is an ideal of $B$, and $A_P\subset B$, every element of $PA_P$ is an element of $M_B$.

Zev Chonoles
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  • Oh, I understood $PA_P$ as $P_P$, which is same but gave me an confusion. I tried to localize $P=A \cap M_B$ at $P$, so that $P_P=A_P \cap (M_B)_P \supset A_P \cap M_B$. – Gobi May 19 '11 at 10:12