Let $A$ be a commutative ring, $a \subset A$ be an ideal. For $A$-module $M$ let $S \subset A$ be the set of elements, which are invertible in $M$, so $M$ is actually a $S^{-1}A$-module. It is not hard to show, that 



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> If $S\cap a \neq \emptyset$, then $\operatorname{Tor}\_\*^A(A/a, M) = 0$. 
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Under what conditions the converse is also true? I can prove it for PID, so I am interested if it can be extended to wider classes of rings, Noetherian for example. 


Also, I would be glad to receive some references on relations between Tor functors, quotient rings and multiplicative sets.


**UPD** I’m concerned about special case, when $a$ is prime, or even maximal (because of some geometric interpretations), but the general case is interesting too.



Tor functor and invertible elements
Without any finiteness assumptions on $M$, the converse fails already for $A=k[x, y]$ ($k$ is a field). 


Take $M=k[x,y^{\pm 1}]\oplus k[x^{\pm 1},y]$ and $a=(x,y)$. The module $M$ is flat over $A$ so $M\otimes^{\mathbb{L}}\_A A/a=M\otimes\_{A}A/a=M/aM=0$. However $S=k\setminus\{0\}$ because the sets of elements invertible on $k[x^{\pm 1},y]$ and $k[x,y^{\pm 1}]$ are $\{ax^n|a\in k\setminus\{0\},n\in\mathbb{N}\}$ and $\{ay^n|a\in k\setminus\{0\},n\in\mathbb{N}\}$ respectively.




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On the positive side, for a finitely generated $M$ the vanishing of the tensor product $A/a\otimes\_A M$ already implies that some element of $a$ is invertible on $M$.


Indeed, consider the annihilator ideal $I=Ann(M)\subset A$ of the module $M$. Since $M$ is finitely generated, the closed subspace $V(I)\subset Spec\, A$ is the support of $M$.


**Claim.** If $A/a\otimes\_A M=0$ then $V(a)\cap Supp\, M=\emptyset$


Proof. This is immediate from Lemma 10.39.9 (1) on Stacks Project <https://stacks.math.columbia.edu/tag/00L3>.


The intersection $V(a)\cap V(I)$ is empty if and only if $a+I=A$. Take $f\in a, g\in I$ such that $f+g=1$. The element $g$ annihilates $M$, so the multiplication by $f$ on $M$ is the identity endomorphism, hence $f\in S\cap a$.


