F is flat if and only if every ideal I of R, the R-module $I\otimes _R F$ is torsion-free.

Question

Let $R$ be an integral domain, and $F$ an $R$-module. Prove $F$ is flat if and only if for every ideal $I$ of $R$, the $R$-module $I\otimes _RF$ is torsion-free.

Here's my weak attempt to solve it:

($\Rightarrow$) Assume $F$ is flat then every $I\otimes _RF\rightarrow R\otimes _RF$ is injective. I know $M$ is torsion-free if and only if $M\rightarrow M_Q$ is injective. And $Q$ is flat as an $R$ module. But I can't draw any conclusion from that information.

($\Leftarrow$) The $R$-module is $I\otimes _RF$ is torsion-free. We want to show $I\otimes _RF\rightarrow R\otimes _RF$ is injective which would imply $F$ is flat. But I don't know how to prove it.

Answer 1

Proof: =>] For any $r\ne0\in R$, the multiplication by $r$ map $I\to I$ is injective, so $I\otimes_RF\xrightarrow{r}I\otimes_RF$ is injective.

<=] We hope to check the inclusion $I\otimes_RF\to F$ is injective, so let $x:=\sum_{i=1}^na_i\otimes x_i\in I\otimes_RF$ be such that $\sum_{i=1}^na_ix_i=0$, where all $a_i\ne0$. Then, for any $r\ne0\in I$, we have $$r\cdot x=\sum_{i=1}^nra_i\otimes x_i=\sum_{i=1}^nr\otimes a_ix_i=r\otimes\sum_{i=1}^na_ix_i=0.$$ Since $I\otimes_RF$ is torsion-free this shows $x=0$.

Alternatively, for any $r\ne0\in I$, we have the inclusions $rI\subset (r)\subset I$, which induces morphisms $rI\otimes_RF\to rR\otimes_RF\to I\otimes_RF$. The composition is injective since $I\otimes_RF$ is torsion-free. Thus, $rI\otimes_RF\to rR\otimes_RF$ must be injective. But this is equivalent to $I\otimes_RF\to R\otimes_RF$ being injective.