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I have a theorem and a proof but there's one detail I cannot understand. The theorem is

Let $R$ be a commutative ring and $M$ is a $R$-module. Let $L\leq M$ be a submodule and $N=M/L$. Prove that $\operatorname{Ass}_R M\subset \operatorname{Ass}_R L \cup \operatorname{Ass}_RN$.

Here $\operatorname{Ass}_R M$ denotes the set of all prime ideals $Q$ such that there's a injective homomorphism $R/Q\to M$ as modules.

And this is the proof: "Let $Q\in\operatorname{Ass}_R M$ and consider an embedding $R/Q\to M$. Assume that $Q\notin \operatorname{Ass}_R N$. Then $X=(R/Q) \cap L\neq (0)$. So $\operatorname{Ass}_R X=\{Q\}$ and $Q\in \operatorname{Ass}_R L$ as $X\subset L$."

The detail I don't understand is this: "Then $X=(R/Q) \cap L\neq (0)$". Why is it?

Thank you for your help.

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Let $Q\in \operatorname{Ass}_RM$, $i:R/Q\rightarrow M$ and $p: M\rightarrow M/L=N$. If $Q\notin\operatorname{Ass}_RN$, the morphism $p\circ i:R/Q\rightarrow N=M/L$ is not injective. Let $x\in \ker(p\circ i)$, $p(i(x))=0$. Since $i$ is injective, we deduce that $i (x)\neq 0$ and $p(i(x))=0$. But $i(x)\in L$ since the kernel of $p$ is $L$.

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  • $\begingroup$ I have to say it's not evident to me at all. Thank you for your help $\endgroup$ – chí trung châu Mar 9 '17 at 15:32

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