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Let $R$ be a Noetherian local domain with unique maximal ideal $M$. Then I want to show that if every $M$-primary ideal is a power of $M$, then $R$ is a Discrete Valuation Ring.

I know I'll be done if I can show that $M$ is principal, or that $M$ is the only prime ideal (since then I can invoke, or that $R$ is integrally closed in its field of fractions, but I'm not sure how to show any of those things. Could I have some hints?

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There is no ideal (properly) between $M^2$ and $M$ (why?). Let $x\in M-M^2$. (What can you say if $M=M^2$?) Then $M^2+(x)=M$, and from Nakayama Lemma get $M=(x)$.

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  • $\begingroup$ Oh, so since any ideal contained between $M$ and $M^2$ has $M$ as its radical, it must then be $M$ or $M^2$? But this ideal between $M$ and $M^2$ doesn't have to be primary? $\endgroup$ – Nishant Jul 4 '14 at 23:05
  • $\begingroup$ Okay, I think I got it: by primary decomposition, such an ideal is contained in the intersection of primary ideals, and thus in the intersection of the associated primes. But since the radical of this ideal is $M$, $M$ is the only prime that contains it, so this ideal is just an $M$-primary ideal, and thus just $M$ or $M^2$. $\endgroup$ – Nishant Jul 4 '14 at 23:42
  • $\begingroup$ @Nishant In general, an ideal whose radical is a maximal ideal $M$ is necessarily $M$-primary. $\endgroup$ – user26857 Jul 5 '14 at 6:47

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