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I'm working through a proof and I've got stuck on one detail, it seems like it is supposed to be totally obvious, but need help to figure out why.

Let $D$ be a Noetherian integrally closed domain with unique prime ideal $(\pi)$. For an ideal $I$ of $D$ we have that $I\pi ^{-k} \subseteq D$ for all $k\leq m$ but $I\pi ^{-m-1}\nsubseteq D$. I wonder how to make the conclusion that $I\pi ^{-m}\nsubseteq (\pi )$.

This is used in proving that every ideal in the ring $D$ above is a power of $(\pi )$. I hope I didn't leave out any crucial information...

harajm
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2 Answers2

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Suppose, for the sake of contradiction, that $I\pi^{-m}\subseteq(\pi)$. Then every element of $I\pi^{-m}$ is of the form $\pi a$ for some $a\in D$. Thus, every element of $I\pi^{-m-1}$ is an element of $D$, i.e. $I\pi^{-m-1}\subseteq D$; but we chose $m$ so that this is not the case.

Zev Chonoles
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Since $D$ is a domain, one should say that $(\pi)$ is the unique non-zero prime ideal. If so, $D$ has dimension $1$ and thus is a Dedekind domain (by definition).

One knows that in a Dedekind domain an ideal is always a product of prime ideals. Thus, a fortiori, the products $(\pi)^n=(\pi^n)$ are all the non-zero ideals that there might be.

This is the approach taken, for instance, by P. Samuel in his classic Théorie Algébrique des Nombres

Andrea Mori
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