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Let $A$ be a Noetherian domain of dimension $1$, $\mathfrak a$ is a non-zero ideal in $A$, then $\mathfrak a$ has a minimal primary decomposition $\mathfrak a=\bigcap_{i=1}^n \mathfrak q_i$, where each $\mathfrak q_i$ is $\mathfrak p_i$-primary. Since $\dim A=1$, each non-zero prime ideal of $A$ is maximal. But

Why $\mathfrak p_i$ are distinct maximal ideals?

Why every primary ideal is a prime power?

Michael Hardy
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Peter
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  • Q1 follows from Q2 since $p^n \cap p^m=p^{\max(n,m)}$. Hence Q2 is the interesting one! – tj_ Dec 06 '13 at 16:54
  • @user: The question I'm interested in is: Given a 1-dimensional noetherian domain, is it true that each primary ideal is a power of a prime ideal ? – tj_ Dec 06 '13 at 18:20

1 Answers1

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Minimal primary decomposition implies that $\mathfrak p_i$ are pairwise distinct and since they contain $\mathfrak a$ must be non-zero which in a ring of dimension one is equivalent to maximal.

If every primary ideal is a prime power then $A$ is Dedekind, so in general is not true what you said.

Edit. If $A$ is a Noetherian domain of dimension $1$ and every primary ideal is a prime power then $A$ is Dedekind.

Proof. Every ideal has a primary decomposition which means that, in fact, every ideal can be written as an intersection of prime powers. But different prime powers of different prime ideals are pairwise comaximal (since $\dim A=1$), so every ideal is a product of prime powers.