Homological criterion for $A(B \cap C) = AB \cap AC$?

Question

Is there a homological criterion for the condition $A(B \cap C) = AB \cap AC$ for ideals in a ring $R$? I mean a statement such as "the given equation holds if and only if (some $\operatorname{Tor}$, $\operatorname{Ext}$, local cohomology, etc) group vanishes/does not vanish".

Note that this is a local question, since we are asking when the inclusion $A(B \cap C) \to AB \cap AC$ is surjective. So I'm happy to assume $R$ local.

This is coming from the similar-looking fact that $AB = A \cap B$ if and only if $\operatorname{Tor}^1(R/A,R/B) = 0$.

(I asked this on StackExchange and did not receive a response.)

Answer 1

Ah, here it is: we have $A(B\cap C) = AB \cap AC$ if and only if the natural map

$$\operatorname{Tor}_1^R(R/A,B) \oplus \operatorname{Tor}_1^R(R/A,C) \to \operatorname{Tor}_1^R(R/A,B+C)$$

is surjective.

This is not very pretty, but it's still entirely homological (for example, it can be computed in the completion.) For a more 'geometric' formulation, we may use the natural isomorphism $$\operatorname{Tor}_2^R(M,R/I) \to \operatorname{Tor}_1^R(M,I)$$ to express this in terms of $R/A$, $R/B$, $R/C$, and $R/(B+C)$, so this does indeed relate (somehow) to how $A$ intersects with $B$ and $C$.

Proof: Take $0 \to B \cap C \to B \oplus C \to B + C \to 0$ and tensor with $R/A$. Note that

$$\frac{AB \cap AC}{A(B \cap C)} = \ker\big(\frac{B \cap C}{A(B \cap C)} \to \frac{B}{AB} \oplus \frac{C}{AC} \big)$$

and so is the cokernel of the map above. (In particular, we are asking for the first boundary map in the long exact sequence for $\operatorname{Tor}$ to be zero.)

Answer 2

An obvious sufficient condition would be $$\operatorname{Tor}_1^R(R/A,R/B)=\operatorname{Tor}_1^R(R/A,R/C)=\operatorname{Tor}_1^R(R/A,R/(B\cap C))=0.$$ Unless for some reason you don't want this condition.