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I am using Atiyah MacDonald to study commutative algebra and in Chapter 9, it says the following:

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If you look at the paragraph after the proof of 9.1, you can see that to get unique factorization, we must also have that given a nonzero prime ideal $\mathfrak p$ and $n \neq m$, we have $\mathfrak p^n \neq\mathfrak p^m$. I was not able to see why this was the case. Any help will be appreciated.

Kenta S
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LoneStar
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  • What would Nakayama's lemma give you? – Angina Seng Jul 23 '20 at 11:50
  • @AnginaSeng To invoke Nakayama's Lemma, we would need that the ideal $p$ is contained in the Jacobson Nilradical of $A$, which is only true if $p$ is the only maximal ideal in $A$ – LoneStar Jul 23 '20 at 11:56

1 Answers1

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It suffices to show that the chain $$\mathfrak p\supseteq\mathfrak p^2\supseteq\mathfrak p^3 \supseteq \cdots$$ is strictly descending.

Suppose instead that $\mathfrak p^{n+1}=\mathfrak p^n$ for some positive integer $n$.

Let $M=\mathfrak p^n$, regarded as an $A$-module.

Since $A$ is a domain and $\mathfrak p\ne 0$, we get $M\ne 0$.

Since $A$ is Noetherian, $M$ is finitely generated.

From $\mathfrak p^{n+1}=\mathfrak p^n$ we get $\mathfrak pM=M$, hence by Nakayama's lemma

$\;\;\;\;$ https://en.wikipedia.org/wiki/Nakayama%27s_lemma#Statement

we get $aM=0$ for some $a\in A$ with $a\equiv 1\;(\text{mod}\;\mathfrak p)$.

But then $a\mathfrak p^n=0$, contradiction, since $a\ne 0$,$\;\mathfrak p^n\ne 0$, and $A$ is a domain.

quasi
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