Does the Künneth theorem hold for (co-)homology with coefficients in a module? Let $R$ be a principal ideal domain, and let $M$ be an $R$-module. To compute $H_*(X \times Y;R)$, one typically applies the Künneth theorem, which tells you that there's a split short exact sequence
$$0 \to \bigoplus_{i + j = k} H_i(X; R) \otimes_R H_j(Y;R) \to H_k(X;R) \to \bigoplus_{i + j = k - 1} \operatorname{Tor}^R_1\big(H_i(X;R), H_j(X;R)\big) \to 0\,\text{.}$$
My question is whether this formula still holds true if we're working with homology $H_*(\,\cdot\,,M)$ with coefficients in an $R$-module $M$. (Side question: I'd also be interested in the case where $R$ is no longer is a PID. Is there some general spectral sequence that covers all relevant cases?)
 A: No, the algebraic Kunneth sequence tells us that if we have free chain complexes C,D over the PID R, then we have the short exact sequence:
$0 \rightarrow \bigoplus\limits_{i+j=k} H_i(C) \otimes_R H_j(D) \rightarrow H_k (C \otimes D) \rightarrow \bigoplus\limits_{i+j=k-1} Tor^R_1 (H_i(C),H_j(D)) \rightarrow 0$.
The Eilenberg-Zilber theorem tells us that $S_*(X \times Y,R)$ and $S_*(X,R) \otimes S_*(Y,R)$ are quasi-isomorphic. From these one can deduce the usual Kunneth formula in topology. To figure out the case for coefficients in a module M we can just tensor each of these free chain complexes by $M$ and we preserve the quasi-isomorphism. So we have $S_*(X \times Y,R) \otimes_R M$ is quasi-isomorphic to $S_*(X,R) \otimes S_*(Y,R) \otimes_R M$. Rewriting, the former is $S_*(X \times Y,M)$ and the latter is $S_*(X,R) \otimes_R S_*(Y,M)$.
Applying the algebraic Kunneth sequence tells us that we have a short exact sequence:
$0 \rightarrow \bigoplus\limits_{i+j=k} H_i(X,R) \otimes_R H_j(Y,M) \rightarrow H_k (X \times Y,M) \rightarrow \bigoplus\limits_{i+j=k-1} Tor^R_1 (H_i(X,R),H_j(Y,M)) \rightarrow 0$
And of course you can swap the roles of X and Y if you'd like to pick which space gets which coefficients. The reason why your proposed sequence does not work is because there are modules such that $M \otimes_R M \neq M$.
In the case R is not a PID, you can use the algebraic Kunneth spectral sequence because I believe the Eilenberg-Zilber theorem holds with no conditions on the ring. This spectral sequence involves the higher Tor terms which vanish for PID's.
