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Let R be a commutative ring.$r\in R$.Let M and N are R-modules such that $rM=0$ and there is a short exact sequence $0\rightarrow N\xrightarrow r N\rightarrow N/rN \rightarrow 0$.

if X is arbitrary R-module,$X\xrightarrow r X$ is R module morphism by multiplication,then we can get the induced map $Hom_R(M,r):Hom_R(M,X)\rightarrow Hom_R(M,X)$ is zero.

by above,it is not difficult to see $Hom_R(M,N)=0$.and we have exact sequence:$0\rightarrow Hom_R(M,N/rN)\rightarrow Ext^1_R(M,N)\xrightarrow {Ext^1(M,r)} Ext^1_R(M,N)$.By the same argument,we use the definition of derived functor $Ext^1$,take a injective resolution of $N$,then we can prove $Ext^1(M,r)=0$. So $Hom_R(M,N/rN)\cong Ext^1_R(M,N)$.

My question:How to prove this by extension group?

That is:for any s.e.s. $\delta:0\rightarrow N\rightarrow W\rightarrow M\rightarrow 0$,take pushout with $\delta$ and morphism $N\xrightarrow rN$.We get a new s.e.s.,How to prove this new s.e.s. splits?

Thank you in advance!

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    $\begingroup$ What is $Hom_R(M,r)$? $\endgroup$
    – YCor
    Oct 22, 2017 at 12:05
  • $\begingroup$ @YCor morphism induced by $N\xrightarrow rN$. $\endgroup$
    – Jian
    Oct 22, 2017 at 12:33
  • $\begingroup$ Why the downvote ? $\endgroup$
    – tj_
    Oct 22, 2017 at 17:12
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    $\begingroup$ Doesn't $rM=0$ imply $Ext^n(M,r)=0$ for $n \ge 0$ without any assumptions for $N$ ? $\endgroup$
    – tj_
    Oct 23, 2017 at 22:08
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    $\begingroup$ @tj_ it should not be injective.We can see s.e.s $0\rightarrow Z\rightarrow Q\rightarrow Q/Z\rightarrow 0$.consider injection $Z\xrightarrow 2Z$,the multiplication of Q is injective.but the multiplication of Q/Z is not injective.since $Z\rightarrow Q$is injective envelope,we know we can't get injective map of $I_Z$. $\endgroup$
    – Jian
    Oct 25, 2017 at 1:42

1 Answer 1

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Let $\,\,E: 0 \to N \xrightarrow{i} W \xrightarrow{\varepsilon} M \to 0$ be a short exact sequence and let $r \in R$.

The short exact sequence induced from $E$ by the homomorphism $\mu: N \to N,\,n \mapsto r\cdot n$ is by definition$$\,\,E': 0 \to N \to X \to M \to 0$$ where $X= W \oplus N\,/\,\{(-i(n),rn)\mid n \in N\}$ is the pushout of $(i,\mu)$.

If $rM=0$ then the short exact sequence $E'$ splits. A splitting is defined by $$\gamma: M \to X,\,m \mapsto \overline{(w,n)}$$ where $w\in W, n \in N$ are choosen such that $\varepsilon(w)=m$ and $i(n) = -rw$.

Since the map $X \to M$ is $\overline{(w,n)} \mapsto \varepsilon(w)$, it's obvious that $\gamma$ is a splitting.

I leave it as an exercise to show that $\gamma$ is a well-defined homomrphism of $R$-modules.

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  • $\begingroup$ You prove it by analysis of element.before you answer this.I tty to prove by analysis of morphism,but failed.Maybe analysis of element id very helpful.thank you. $\endgroup$
    – Jian
    Oct 25, 2017 at 0:27

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