# Question regarding the meaning of a problem from Atiyah Macdonald chapter 4

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

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

• So what you are saying that if $P'$ is any other prime ideal in $A$ then $P \subset P'$ ? Aug 17, 2019 at 19:36
• @user631697 not quite, I am saying that if $P'$ is any prime ideal in $A$ such that $P'\subset P$, then $P'=P$ Aug 17, 2019 at 19:38
• 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? Aug 17, 2019 at 19:45