Skip to main content
added 166 characters in body
Source Link

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.

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.

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.

Source Link

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.