Isomorphism of modules implies isomorphism of tensor products Let $R$ be a Noetherian ring, $M$ a finitely generated left $R$-module and $I,J$ two sided ideals in $R$ such that $I \subset J$. Suppose that $IM=JM$. What are some conditions that will allow us to conclude that $$ I \otimes_R M \cong J \otimes_R M?$$
I know that $M$ being a flat $R$-module is enough, but I am looking for some weaker conditions. Many thanks. 
 A: A more precise statement is: supposing that $I \subseteq J \subseteq R$ are ideals with $IM=JM$, the canonical map $I \otimes M \to J \otimes M$ is an isomorphism if the following two Tor groups are $0$,
$$\mathrm{Tor}_R^1(J/I,M)=0 \quad \text{and} \quad \mathrm{Tor}_R^1(R/J,M)=0.$$ In particular, this holds if $M$ is acyclic for $\otimes_R$ (that is, flat as an $R$-module).
I don't think the Noetherian hypotheses are necessary.
Proof: There is an exact sequence $0 \to I \to J \to J/I \to 0$ of $R$-modules. Tensoring this with $M$ gives an exact sequence
$$\mathrm{Tor}_R^1(J/I,M) \to I \otimes M \to J \otimes M \to J/I \otimes M \to 0.$$ Thus the canonical map is surjective if and only if $J/I \otimes M=0$, and is injective if and only if the map $\mathrm{Tor}_R^1(J/I,M) \to I \otimes M$ is the zero map. On the other hand, tensoring the exact sequence 
$$0 \to J/I \to R/I \to R/J \to 0$$ by $M$ produces
$$\mathrm{Tor}_R^1(R/J,M) \to J/I \otimes M \to M/IM \to M/JM \to 0,$$ where we have used the canonical isomorphisms $R/I \otimes M \cong M/IM$ and $R/J \otimes M \cong M/JM$. By hypothesis the map $M/IM \to M/JM$ is an isomorphism, so by exactness 
$$\mathrm{Tor}^1_R(R/J,M) \to J/I \otimes M$$ is surjective, finishing the proof.
