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Let $(R,m,k)$ be a complete local Noetherian ring and let $E$ be an $R$-module such that $\operatorname{Ass}E=\left\{m\right\}$. Let $N$ be a proper submodule of $E$.

Question: Is it true that $\operatorname{Ass}E/N=\left\{m\right\}$ and why? If it is not true, how does the answer change when $E$ is the injective hull of $k$?

PS: I am also not sure how the completeness comes into play, if at all.

Manos
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1 Answers1

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There is a short exact sequence $\newcommand{\Supp}{\text{Supp}}$ $\newcommand{\Ass}{\text{Ass}}$ $$0 \to N \to E \to E/N \to 0$$

which gives that $\Supp(E) = \Supp(N) \cup \Supp(E/N)$. Now $\Ass(E) = \{m\}$ implies $\Supp(E) = \{m\}$, since for a prime ideal $p$, $\Ass_{R_p}(E_p) = \Ass_R(E) \cap \text{Spec}(R_p)$ (viewing $\text{Spec}(R_p) \subseteq \text{Spec}(R)$). Since $E/N \ne 0$, $\emptyset \ne \Supp(E/N) \subseteq \{m\}$, so $\Supp(E/N) = \{m\}$. Also $\Ass(E/N) \ne \emptyset$, and $\Ass(E/N) \subseteq \Supp(E/N)$, so $\text{Ass}(E/N) = \{m\}$.

Notice that completeness of $R$ was not used (but Noetherianness was crucial, to bypass finite generation hypotheses on $E$).

zcn
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  • Excellent answer, thanks. PS: I was trying to prove this using only $Ass(E) \subset Ass(N) \cup Ass(E/N)$. – Manos Feb 28 '14 at 19:21
  • Yes, this proof shows that a posteriori, $\text{Ass}(E) = \text{Ass}(N) \cup \text{Ass}(E/N)$ if $E$ is $m$-primary. But a priori this is not so clear. – zcn Feb 28 '14 at 20:24