Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

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 $M_B$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.