2

I'm having trouble understanding the proof, for the following.

Let $R$ be ring, $M$ an $R$-module. If $M$ is flat over $R$ and for any $m$ maximal ideal we get $mM\neq M$, then for any $R$-module $N$, if $N\otimes_R M=0$, then $N=0$.

In the proof we let $N\neq 0$ be a $R$-module and choose $x\in N-\{{0}\}$. We have a surjection $f: R\rightarrow Rx$ so we get $R/Ker(f)\cong Rx$, which means we have a exact sequence $$0\longrightarrow Ker(f)\longrightarrow R\longrightarrow Rx\longrightarrow 0$$ In the proof they then state that since $M$ is flat and $Rx\subset N$ then $R/Ker(f)\otimes_R M\cong Rx\otimes_R M$ with a injection $Rx\otimes_R M\rightarrow N\otimes_R M$.

I'd say that $R/Ker(f)\otimes_R M\cong Rx\otimes_R M$ since $$0\longrightarrow Ker(f)\otimes_R M\longrightarrow R\otimes_R M\longrightarrow Rx\otimes_R M\longrightarrow 0$$ is exact by using that $M$ is flat (is this correct?), but I don't see how we get the injection $Rx\otimes_R M\rightarrow N\otimes_R M$?

njlieta
  • 411

1 Answers1

2

You get the embedding $Rx\otimes_RM\to N\otimes_RM$ by applying flatness, but not to $0\to\ker f\to R\to Rx\to 0$.

Since $Rx$ is a submodule of $N$, flatness of $M$ implies the exact sequence $0\to Rx\to N\to N/Rx\to0$ remains exact upon tensoring: $$ 0\to Rx\otimes_R M\to N\otimes_R M\to (N/Rx)\otimes_R M\to 0 $$ If $N\otimes_RM=0$, we conclude that $Rx\otimes_RM=0$ and therefore that $$ \ker f\otimes_R M\to R\otimes_RM $$ is an isomorphism. If $x\ne0$, $\ker f$ is contained in a maximal ideal $m$; then we have the commutative diagram $$\require{AMScd} \begin{CD} \ker f\otimes_R M @>>> M \\ @VVV @VV\mathit{id_M}V \\ m\otimes_RM @>>> M \\ \end{CD} $$ that implies $m\otimes_RM\to M$ is surjective, that is, $mM=M$. Contradiction. Therefore $x=0$.

egreg
  • 238,574
  • Thanks! Was it correct that $R/Ker(f)\otimes_R M\cong Rx\otimes_R M$ since $M$ is flat and we have an isomorphism $R/Ker(f)\cong Rx$? – njlieta May 28 '18 at 19:12
  • 1
    @Jta This is true independently of flatness of $M$. Tensoring an isomorphism yields an isomorphism. More generally, applying a functor to an isomorphism yields an isomorphism. – egreg May 28 '18 at 19:15