$\operatorname{Ext}^{1}(M,R/m)=0$ implies $\operatorname{Tor}_{1}( M,R/m)=0$ 
Let $(R,m)$ be a commutative local Noetherian ring and $M$ a finitely generated $R$-module. I want to show that $\operatorname{Tor}_{n+1}(M,R/m)=0$ if and only if $\operatorname{Ext}^{n+1}(M,R/m)=0$.

I know that $\operatorname{Tor}_{n+1}(M,R/m)=0$ if and only if the projective dimension of $M$ is less than equal to $n$. So $\operatorname{Tor}_{n+1}(M,R/m)=0$ implies $\operatorname{Ext}^{n+1}(M,R/m)=0$. 
By using induction on $n$, the converse reduces to the case $n=0$. But I can't prove it.
Does anyone  have a good idea for this?
 A: Proof for the fact that $M$ is projective, if $\operatorname{Ext}^1_R(M,R/\mathfrak m)=0$ holds:
Let $$\dotsc \to R^{n_2} \to  R^{n_1} \to R^{n_0} \to M \to 0$$ be a minimal free resolution, i.e. the maps between the free modules are matrices with entries in $\mathfrak m$.
Apply $\operatorname{Hom}(-,R/\mathfrak m)$ to obtain
$$0 \to \operatorname{Hom}(M,R/\mathfrak m) \to (R/\mathfrak m)^{n_0} \xrightarrow{\alpha} (R/\mathfrak m)^{n_1} \xrightarrow{\beta} (R/\mathfrak m)^{n_2} \to  \dotsc$$
The property of the resolution being minimal assures $\alpha = \beta =0$, hence
$$\operatorname{Ext}^1_R(M,R/\mathfrak m)=\operatorname{ker} \beta / \operatorname{im} \alpha=(R/\mathfrak m)^{n_1}.$$
The assumption - the ext group vanishes - yields $n_1=0$, hence $0 \to R^{n_0} \to M \to 0$ is exact, which means $M \cong R^{n_0}$.
A: I can elaborate on this answer if you want, but the following facts are useful:


*

*A projective module is always flat.

*A finitely generated flat module over a noetherian ring is projective.

*$\text{Tor}$ (in either component) can be used to detect flatness and $\text{Ext}$ (in the first component) can be used to detect projectivity.

*Over a commutative local ring, a module $M$ is projective iff $\text{Ext}_R^1(M,R/\mathfrak{m}) = 0$. A similar result holds for flatness and Tor.

