# Tor functor and invertible elements

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.

• 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 $A$. – SashaP Nov 3 '19 at 13:51
• @SashaP, 0th Tor (just tensor product) counts too. In case of $M=A$ we have 0th Tor is isomorphic to $A/a$. – Boris Bilich Nov 3 '19 at 14:23
• Got it, just the vanishing of the tensor product already implies that some element of $a$ is invertible if $M$ is finitely generated. – SashaP Nov 3 '19 at 15:59

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 $$f\in S\cap a$$.