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I was reading through Atiyah-MacDonald and they mention that if a ring $A$ is a Noetherian domain of dimension 1 has the property that every primary ideal is equal to the product of a prime ideal (i.e. if A is a Dedekind domain), then every ideal in A can be written uniquely as the product of powers of prime ideals.

My question concerns the uniqueness of these primary decompositions - one need to check that if $\frak p$ and $\frak q$ are prime ideals in $A$ and ${\frak p}^n = {\frak q}^m$ then $\frak p = \frak q$ and $n=m$.

I see why $\frak p = \frak q$ but it is unclear to me why $n=m$. Is this a general property of powers of prime ideals in (commutative) rings?

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    In a Dedekind domain, the intersection of all powers of a given prime ideal is empty, thus the powers of this prime ideal get strictly smaller and smaller. – Amitai Yuval Aug 06 '14 at 18:58
  • In a general commutative ring, though, you can find examples for a prime ideal which is equal to all its powers. – Amitai Yuval Aug 06 '14 at 19:12
  • @AmitaiYuval. Can you please give me a proof of your claim "In a Dedekind domain, the intersection of all powers of a given prime ideal is empty"? I need this to prove something else. – Barbara Jan 22 '18 at 16:44

2 Answers2

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If $\mathfrak p^m=\mathfrak p^n$ for $m>n$, then $\mathfrak p^n=\mathfrak p^{n+1}$. Localizing we get $\mathfrak p^nR_{\mathfrak p}=\mathfrak p^{n+1}R_{\mathfrak p}$. If $\mathfrak p$ is finitely generated (e.g. when $R$ is noetherian) we can apply Nakayama's lemma and get $\mathfrak p^nR_{\mathfrak p}=0$. Therefore there exists $s\in R-\mathfrak p$ such that $s\mathfrak p^n=0$. If $R$ is an integral domain, then $\mathfrak p=0$.

Conclusion: if $R$ is a noetherian integral domain, then every non-zero prime ideal has different powers.

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No, it is not a general property of commutative rings that different powers of a prime ideal are different.

If you allow zero-divisors there is a simple example: for some field $k$, take $R = k[X]/(X^2)$ and the ideal $I =(X)$.

The ideal $I$ is a prime ideal, since $R/I$ is isomorphic to $k$. Yet $I^2 = (0)$. So all higher powers of $I$ are $(0)$ too and thus equal.

If you do not allow zero-divisors you can get examples via considering for example a domain that has an idempotent prime ideal, that is one with $I^2 = I$. Such domains exist, see the question Can an ideal in a commutative integral domain be its own square? (note that the example in Qiaochu Yuan's answer is in fact shown to be prime, maximal even, while it is not stated).

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