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.


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 $L$.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.