$\operatorname{Tor}_i(M,N)$ is torsion for $i>0$ Let $R$ be a commutative domain. Let $M$, $N$ be $R$-modules. A module is torsion if $M=\tau(M)$, i.e. for any $m$ there exists $r\neq 0$ s.t. $rm=0$.
Then we should have $\operatorname{Tor}_i(M,N)$ being torsion for all $i>0$, for all $M$, $N$.
I guess I should get a short enough projective resolution of $N$ to start with. But I could not think of one.
Any hint would be helpful.
 A: Hint: Let $K$ be the field of fractions of $R$.  A module is torsion iff its tensor product with $K$ is trivial.  Also, $K$ is flat, so tensoring with $K$ is exact.  So try tensoring the entire computation of $\operatorname{Tor}_i(M,N)$ with $K$.
A: I'm expanding on Eric's hint:
I think the idea is like this:
Fix $i>0.$ 
(Following sloppy conventions, various equalities below are actually natural isomorphisms.)


*

*Let $\ldots \to P_1 \to P_0 \to M$ be a projective resolution of $M$. Let $C_* = \ldots \to P_1 \to P_0$, so $C_*$ is quasi-isomoprhic to $M$, and a complex of projectives.

*Then $Tor^i(M,N) = H_i(C_* \otimes N)$.

*We want to show that $Tor^i(M,N)$ is torsion. This is equivalent to $Tor^i(M,N) \otimes_R K = 0$.

*So we wish to show that $0 = Tor^i(M,N) \otimes K$:

*$Tor^i(M,N) \otimes K = H_i(C_* \otimes N) \otimes K = H_i(C_* \otimes N \otimes K)$. Since $K$ is flat, $\otimes K$ is exact, since exact functors "commute" with taking homology, we can bring the tensor product inside.

*Now $C_* \otimes_R N \otimes_R K = (C_* \otimes_R K) \otimes_K (N \otimes_R K)$, by associativity of tensor products and what is usually called cancellation for tensor products: $(A \otimes_R B) \otimes_B C= A \otimes_R C$, for $B$ an $R$ algebra. 

*However, $H_i((C_* \otimes_R K) \otimes_K (N \otimes_R K))= Tor^i_K(M \otimes_R K, N \otimes_R K)$. This is zero because $K$ is a field - in particular to compute this Tor we can take the projective resolution of $M \otimes_R K$ to be the complex consisting of $M \otimes_R K$ in degree $0$.

