1

This question was left as an exercise in my class of Algebraic Geometry and unfortunately I am not very good in solving problems of Localization. I have been following my class notes.

So, I am posting it here in hope of getting some guidance.

Question: Let $P\in Spec A$ , where $A$ is a ring and $Spec A$ is Spectrum of $A$, $S= A\setminus P$ is multiplicatively closed. Define , $S^{-1} A= A_P$ . Prove that $\{P A_P\} = Spm A_P$ (where $Spm A_P$ is the max spectrum) and $A_P / P A_P = Q(A/P)$, where $Q(A/P)$ is ring of quotients of $A/P$.

Attempt: For 1st, 1 way is proving by taking an element from set in LHS and proving it into RHS and also converse. But I am looking for an elegant method for both parts and unfortunately I am totally struck on them.

I understand the theory taught in class, but not able to make progress. Can you please give a few hints?

Edit: I found solution of 1st part here:Why is the localization at a prime ideal a local ring?

  • 1
    No problem: but there were still a few corrections to be made. You must really distinguish between $A/P$ and $A\setminus P$. I've set them straight now. – rschwieb May 27 '22 at 12:52
  • The fact that localizing at the complement of a prime ideal always produces a ring with a single maximal ideal is probably in multiple places on the site and in every commutative algebra reference. The other half might be more novel. – rschwieb May 27 '22 at 18:05
  • @rschwieb So, can you please write an answer for this? –  Jun 02 '22 at 10:12
  • I haven’t really seen sufficient attempt on your part yet to know how to help you. I would help you get unstuck with the second half, if I knew where you were stuck. – rschwieb Jun 02 '22 at 11:59
  • @rschwieb I am struck on 2nd part also, can you please help with that? –  Jun 03 '22 at 14:04

0 Answers0