Let $S$ and $R$ be Noetherian integral domains. Suppose $S$ is $R$-flat. Let $M$ be an $R$-module. Is it true that $\text{Torsion}_S(S\otimes_R M)=S\otimes_R (\text{Torsion}_R(M))$?
I can see that $S\otimes_R (\text{Torsion}_R(M))\subseteq\text{Torsion}_S(S\otimes_R M)$ but I can't see the reverse inclusion.
If this statement is false are there any well known conditions we can add so that this holds?
Since $t(M/t(M))=0$ and $$S\otimes_R(M/t(M))=(S\otimes_RM)/(S\otimes_Rt(M))$$ we get $t((S\otimes_RM)/(S\otimes_Rt(M)))=0$, so $t(S\otimes_RM)\subseteq S\otimes_Rt(M)$.
Now let's prove the first claim: if $t(N)=0$ then there is a $K$-vector space $V$ such that $0\to N\to V$. (Here $K=Q(R)$ the field of fractions of $R$.) Then we have an exact sequence $$0\to S\otimes_RN\to S\otimes_RV.$$ But $S\otimes_RV=(S\otimes_RK)\otimes_KV$ is a torsion-free $S$-module hence $S\otimes_RN$ is torsion-free, and we are done.