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.


  • $\begingroup$ 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


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

  • $\begingroup$ 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

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