Tor$^{R}_{n+1}(k,k)=0$ implies Pd$_RM\leq n$ for any $R$-module $M$?

Question

Let $R$ be a Noetherian local ring with maximal ideal $m$, $k$ its residue field $R/m$.

Then Tor$^{R}_{n+1}(k,k)=0$ implies Pd$_RM\leq n$ for any $R$-module $M$?(Pd$_RM$ is the projective dimension of $M$ over $R$.)

What I know is that since $k$ is a finitely generated $R$-module, Tor$^{R}_{n+1}(k,M)=0$ for any module $M$. Then Tor$^{R}_{n+1}(M,k)=0$ for any module $M$. So for any finitely generated module $M$, Pd$_RM\leq n$.

I am wondering if this can be extended to the case where $M$ is an arbitrary $R$-module.

Answer 1

Note that every module $M$ has projective dimension $\leq n$ iff $\operatorname{Ext}^{n+1}(M,N)=0$ for all $M$ and $N$ iff every module $N$ has injective dimension $\leq n$. Now $N$ has injective dimension $\leq n$ iff the $n$th cokernel $K$ of any injective resolution of $n$ is injective. By Baer's criterion, this is true iff $\operatorname{Ext}^1(R/I,K)=0$ for any ideal $I\subseteq R$. But now $R/I$ is finitely generated and hence has projective dimension $\leq n$, and so $\operatorname{Ext}^1(R/I,K)\cong \operatorname{Ext}^{n+1}(R/I,N)=0$. Thus $K$ is injective, and $N$ has injective dimension $\leq n$ for arbitrary $N$, and thus $M$ has projective dimension $\leq n$ for arbitrary $M$.

More generally, this shows that if every cyclic module over some ring has projective dimension $\leq n$, then every module has projective dimension $\leq n$.

Answer 2

It seems that what you are asking more specifically is

If every fin.-gen. $R$-module $M$ has $\mathrm{pd_R}M\leq n,$ does every $R$-module $M$ have $\mathrm{pd_R}M\leq n$?

The answer is "Yes".

The idea is: Filter your module by an increasing well-ordered chain $(M_{\alpha})_{\alpha < \kappa}$ of submodules such that $M_{0}=0$ and $M_{\alpha}=\bigcup_{\beta < \alpha}M_{\beta}$ at limit ordinals, and such that all the consecutive quotients $M_{\alpha+1}/M_{\alpha}$ are of $\mathrm{pd}_R \leq n$ (in our case finitely generated). Then by induction (on $n$ and $\alpha$), one shows that each $M_{\alpha}$ is of $\mathrm{pd}_R \leq n$, and, by the same argument, that $M$ is.

The induction step for non-limit ordinals is easy, it's just the fact that modules of $\mathrm{pd}_R \leq n$ are closed under extensions (by the long exact sequence for $\mathrm{Ext}_R$), but the limit case requires more trickery: Namely, given $\alpha$ a limit ordinal, for each $\beta < \alpha$ consider the "canonical" epimorphism from free module $$0 \rightarrow K_\beta \rightarrow R^{\oplus M_{\beta}} \rightarrow M_{\beta}\rightarrow 0,$$ Then observe that for each $\beta< \beta' < \alpha,$ there are induced compatible injective maps $K_{\beta} \rightarrow K_{\beta'} $ and split-injective maps $R^{\oplus M_{\beta}} \rightarrow R^{\oplus M_{\beta'}}$. Taking directed union (i.e. applying $\varinjlim_{\beta < \alpha}$), one obtains again the canonical epimorphism from free module $$0 \rightarrow \bigcup_{\beta<\alpha}K_{\beta} \rightarrow R^{\oplus M_{\alpha}} \rightarrow M_{\alpha}\rightarrow 0.$$ But now we expressed $K_{\alpha}=\bigcup_{\beta<\alpha}K_{\beta}$ again in the form of "continuous" filtration with consecutive quotients of $\mathrm{pd}_R\leq n-1$. So from induction on $n$, $\mathrm{pd}_RK_{\alpha}\leq n-1,$ hence $\mathrm{pd}_RM_{\alpha}\leq n.$