Tensoring with fraction fields kills the torsion

Question

Assume $R$ is a PID. And $M$ is finitely generated $R$-module. So we have the classification theorem: $M\cong R^r\oplus T(M)$. Is it true that $M\otimes_{R}\mathrm{Frac}(R)=\mathrm{Frac}(R)^r$? At least for $\mathbb{Z}$ it is true.

Answer 1

$\DeclareMathOperator{\Fr}{Frac}\DeclareMathOperator{\Tor}{Tor}$ Yes, it is true that for a finitely generated $R$-module, with a decomposition $\;M=L\oplus T$, where $T$ is the torsion submodule of $M$ and $L$ is free of rank $r$, we have $$M\otimes _R\Fr(R)\simeq \Fr(R)^r.$$

Indeed, the tensor product commutes with direct sums, so $$M\otimes _R\Fr(R)\simeq \bigl(L\otimes _R\Fr(R)\bigr)\oplus\bigl(T\otimes _R\Fr(R)\bigr)\simeq\Fr(R)^r\oplus\bigl(T\otimes _R\Fr(R)\bigr),$$ and as all elements of $T$ are killed by a nonzero element of $R$, which is invertible in $\Fr(R)$, there results that $\;T\otimes _R\Fr(R)=\{0\}$.

As to your question in the comments, it is in general false that $\;\Tor^R_1(M,R)=T$, because $R$ is a flat $R$-module, so $\;\Tor^R_1(M, R)=0$ for any $R$-module $M$.

What is true,if $Rˆ$ is a P.I.D., is that $$\Tor^R_1(M,M)\simeq \Tor^R_1(T,T)\simeq T.$$