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Edit : My question has been linked with this :Why does a minimal prime ideal consist of zerodivisors? but this proof doesn't use localization, while I want to use localization. So, kindly reopen my question.

The following proof was outlined in my online class of commutative algebra and I was not able to understand and I asked to some of my friends and it's not clear with them also.

Question: Let A be a ring. Then every minimal prime ideal in A is contained in Z(A)= the set of all zero divisors of A.

Proof: Let $P \in Spec A$ be a minimal prime ideal in A To show that $P\subseteq Z(A)$. Consider the local ring $A_P = S^{-1} A$, where S=A/P Then {$PA_P$}= $Spm A_P$ ( I understand this) = $Spec A_P$ and $Nil A_P= P A_P$.

Can you please tell how can I deduce that $Spm A_P =Spec A_P$ and $Nil A_P= P A_P$? I am not getting any ideas on how this should be proved.

  • You seem to be asking a lot of standard commutative algebra material: please kindly remember to search as much as possible before posting. Quite often you will find solutions for such questions pretty rapidly. – rschwieb May 27 '22 at 18:32
  • Just to make sure you understand the last line: there is a correspondence of prime ideals in $S^{-1}A$ and primes contained in $P$. If you choose $P$ to be minimal, then there are no other primes, so that's why the spec equals maxspec. The nilradical is the intersection of all primes, and if there is only one prime that is exactly the nilradical. – rschwieb May 27 '22 at 18:34
  • @rschwieb The linked answer doesn't uses localization. –  Jun 03 '22 at 13:31
  • @rschwieb It will really help me if you can vote to reopen it. –  Jun 03 '22 at 13:37
  • @rschwieb I understand your 2nd comment completely .But how does $Spm A_P = Spec A_P$ and $ Nil A_P = PA_P$? –  Jun 03 '22 at 13:48

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