Prove $\operatorname{pd}_R M\leq n$ iff $\operatorname{Ext}^{n+1}(M,N)=0$

Question

I have a problem concerning Ext and finitely generated module.

Let $R$ be Noetherian ring, $M$ a finitely generated $R$-module. Prove $\operatorname{pd}_R M\leq n$ iff $\operatorname{Ext}^{n+1}(M,N)=0$ for all finitely generated $R$-modules $N$.

I have asked numerous questions concerning this issue and something like this. We have a broad equivalence $\operatorname{pd}_R M\leq n$ iff $\operatorname{Ext}^{n+1}(M,N)=0$ for all $R$-modules $N$. But in some cases we can reduce the conditions like this. And each proof uses a different method.

Can anyone help me with this? Thank you.

Answer 1

we only need to prove if $\forall N\in R-mod,Ext^{n+1}(M,N)=0$,then $pdM\leq n$,where M is finite generated.

First,we recall an isomorphism:$Ext^i(M,\coprod X_k)\cong\coprod Ext^i(M,X_k)$ where M is finite generated and R is Noetherian .

Second,we know every module $X$is direct limit of finite generated submodules $X_k$.So there is a short exact sequence:$0\rightarrow \coprod X_k\xrightarrow{1-shift}\coprod X_k \rightarrow X\rightarrow 0$.

So we have exact sequence $Ext^{n+1}(M,\coprod X_k)\rightarrow Ext^{n+1}(M,X)\rightarrow Ext^{n+2}(M,\coprod X_k)$

So we can get $Ext^{n+1}(M,X)=0$.