1
$\begingroup$

While studying primary decomposition from Atiyah-Macdonald I came across this problem:

For any prime ideal $p$ in a ring A, let $S_p(0)$ denote the kernel of the homomorphism $A \rightarrow A_p$ Prove that

i) $S_p(0) S \subset p$

ii) $r(S_p(0)) = p \iff p$ is a minimal prime ideal of $A$.

While I was able to solve first one easily, I am having a hard time understanding the meaning of second problem. Specially the quote

$p$ is a minimal prime ideal of $A$.

Here I can't understand, the meaning of minimal prime ideal. What I have read so far says minimal prime ideal over an ideal $a$ of the ring $A$ is the set of minimal prime elements of minimal primary decomposition of the ideal $a$. But in the question no such ideal is given. I have read in some place that they are trying to imply here $p$ is a minimal prime ideal of $A$ means it's a minimal prime ideal over the ideal $(0)$. But in this case that will imply $(0)$ ideal is decomposable. But correct me if I am wrong, $(0)$ ideal is not always decomposable. So it will be great if you can help me understanding the question.

Thanks

$\endgroup$
1
  • $\begingroup$ A minimal prime ideal of $A$ means a minimal prime ideal of $A$ over (0). $\endgroup$
    – user682705
    Aug 17, 2019 at 19:56

1 Answer 1

1
$\begingroup$

Prime ideals form a poset. A minimal prime ideal is a minimal element in that poset.

$\endgroup$
3
  • $\begingroup$ So what you are saying that if $P'$ is any other prime ideal in $A$ then $P \subset P'$ ? $\endgroup$
    – user631697
    Aug 17, 2019 at 19:36
  • $\begingroup$ @user631697 not quite, I am saying that if $P'$ is any prime ideal in $A$ such that $P'\subset P$, then $P'=P$ $\endgroup$ Aug 17, 2019 at 19:38
  • $\begingroup$ Okay okay.. just to clear it completely it's saying that if $S_p(0) \subset P' \subset P$ with $r(S_p(0))=p$ then $P'=P$. Am I correct now? $\endgroup$
    – user631697
    Aug 17, 2019 at 19:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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