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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

If $S\cap a \neq \emptyset$, then $\operatorname{Tor}_*^A(A/a, M) = 0$.

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.

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  • $\begingroup$ Apparently I’m misunderstanding something, but the vanishing condition holds for any $M$ flat over $A$, in particular for $M=A$ in which case $S$ is the set of invertible elements in $A$ so it has empty intersection with any non-unital ideal. The converse is thus false for every ring $A$. $\endgroup$ – SashaP Nov 3 '19 at 13:51
  • $\begingroup$ @SashaP, 0th Tor (just tensor product) counts too. In case of $M=A$ we have 0th Tor is isomorphic to $A/a$. $\endgroup$ – Boris Bilich Nov 3 '19 at 14:23
  • $\begingroup$ Got it, just the vanishing of the tensor product already implies that some element of $a$ is invertible if $M$ is finitely generated. $\endgroup$ – SashaP Nov 3 '19 at 15:59
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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.


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 $f\in S\cap a$.

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