Let $(R,m)$ be a Noetherian local commutative ring with a unit. $M$ is a finitely generated module. In the article "Minimal Injective Resolutions" by Robert Fossum, he mentioned it is standard that
$$\operatorname{id}_R(M)=\sup\{i\mid \operatorname{Ext}^i_R(R/m,M)\neq 0 \}$$
why is it the case? I mean $\operatorname{id}_R(M)$ is definitely $\sup\{i\mid\operatorname{Ext}^i(N,M)\}$ in which $N$ ranges through all modules. How can we reduce it to the case $R/m$?
Does it mean $M$ is injective if and only if $\operatorname{Ext}^1(R/m,M)=0\,$? I mean if this is true, the original problem is trivial.
Thank you
(R,m) is Noetherian local ring.
I am very sorry for the wrong argument.thanks for the comment above.it shows that $M$ may not injective if $Ext^1(k,M)=0$.We only can reduce to the case $M$ is injective iff $Ext^i(k,M)=0.\forall i>0$.
Definition:If $M$ is an $R$ module,$p\in Spec(R)$ is called associate prime if $\exists x\in M$ such that $ann(x)=p$.
Denote $ass(M)$ is the set of associated prime ideals.
lemma1:for any nonzero module M,$ass(M)$ is not empty.
corollary1:for any finite generated module $M$,there is a filtration:$M_0=0\subset M_1\subset M_2\subset …\subset M_n=M$ such that $\forall i=0,1,2,…,n-1.\exists p_i\in Spec R$ such that $M_{i+1}/M_i\cong R/p_i$.(In fact,$p_i\in Supp(M)$)
Lemma2:if $N$ is finite generated and $Ext^1(R/p,M)=0 ,\forall p\in Supp(N)$,then $Ext^1(N,M)=0$.
Lemma3:Let $M$ be a finite generated $R$ module.if $p$ is a prime ideal different from $m$,and $Ext^{n+1}(R/q,M)=0，\forall q\in V(p)-p$,then $Ext^n(R/p,M)=0$. (Hint:select $x\in m-p$，then there is a short exact sequence $0\rightarrow R/p \xrightarrow x R/p \rightarrow R/(x,p)\rightarrow 0$. Remark $Supp(R/(x,p)\subset V(p)-p$.
Then you can prove $id M=sup\{i\mid Ext^i(k,M)\ncong 0\}$ by using Lemma 3 repeatedly.