# Why assume $R$ Noetherian for the primary decomposition of $R$-modules?

When considering primary decomposition in modules, why is it often assumed that $M$ is a finitely generated module over a Noetherian ring $R$ and not only that $M$ is Noetherian? As I understand this extra assumption is not needed to prove the existence or the uniqueness properties of the primary decomposition of submodules.

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@Manny: Isn't this an answer? –  Martin Brandenburg Feb 21 '13 at 19:43
@MartinBrandenburg, you're probably right. I've reposted below. –  Manny Reyes Feb 21 '13 at 20:30
Why some authors (like Kaplansky) prefer not to use the primary decomposition and others (like Atiyah & MacDonald) prefer to use it? Is this a real question? –  user26857 Feb 21 '13 at 22:04

I think that's because in a lot of proofs the people use the fact that $M\neq 0 \Rightarrow \mbox{Ass}(M)\neq \varnothing$ and this is true only for Noetherian rings.

Idea: Let $\mathcal{S}=\{\mbox{Ann}(m)\subset R,\;m\neq 0 \}$. If $R$ is Noetherian then $\mathcal{S}$ has a maximal element, say $\mathfrak{p}=\mbox{Ann}(m)$ with $m\neq 0$. Then $1\notin \mathfrak{p}$ as $m\neq 0$ and if $xy\in \mathfrak{p}$ but $y\notin \mathfrak{p}$ then $xym=0$, but $ym\neq 0$ and therefore $x\in \mbox{Ann}(ym)$. It's clear from the definition that $\mathfrak{p}=\mbox{Ann}(m)\subseteq \mbox{Ann}(ym)$, so $\mathfrak{p}=\mbox{Ann}(ym)$ by maximality. Thus $x\in \mathfrak{p}$ and $\mathfrak{p}$ is a (associated) prime.

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I agree with you that it seems slightly more clean to simply assume that M is Noetherian. However, if M is a Noetherian module over a commutative ring R, then M is a faithful module over R/ann(M), and the latter ring is Noetherian! So in a sense, there's very little distinction between the two cases.

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