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Jian
Jian
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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,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!

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 $Hom_R(M,r)$ 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!

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!

Source Link
Jian
Jian
  • 496
  • 2
  • 11

How to prove this by extension group

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 $Hom_R(M,r)$ 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!