## Over a PID, flat and torsion free are equivalent

Let $R$ be a principal ideal domain. Prove that a right unital $R$-module $A$ is flat if and only if it is torsion free.

Suppose $A$ is flat as a right $R$-module. Given $r \in R$ nonzero, define $\psi_R : R \rightarrow R$ by $\psi_r(a) = ra$. Because $R$ is a commutative ring, $\psi_r$ is a left module homomorphism. Because $R$ is a domain, $\psi_r$ is injective. Because $A$ is flat, $1 \otimes \psi_r : A \otimes_R R \rightarrow A \otimes_R R$ is injective. Now suppose $ar = 0$, with $r \neq 0$. In $A \otimes_R R$, we have $a \otimes r = 0 \otimes r$, so that $(1 \otimes \psi_r)(a \otimes 1) = (1 \otimes \psi_r)(0 \otimes 1)$. Since $1 \otimes \psi_r$ is injective, we have $a \otimes 1 = 0 \otimes 1$. Recall that $A \otimes_R R \cong A$ via the multiplication map $a \otimes r \mapsto ar$. Thus we have $a = 0$, and so $A$ is torsion free as a right $R$-module.

Conversely, suppose $A$ is torsion free as a right $R$-module, and let $I \subseteq R$ be a nonzero ideal. Since $R$ is a PID, say $I = (r)$ with $r \in R$ nonzero. Note that the map $\theta_r : A \rightarrow A$ given by $\theta_r(a) = ar$ is a module homomorphism. Moreover, $\theta_r$ has a trivial kernel since $A$ is torsion free- so $\theta_r$ is injective. Now consider the map $\rho : R \rightarrow I$ given by $\rho(x) = rx$; this is a module homomorphism. Certainly $\rho$ is surjective, and moreover since $R$ is a domain, $\rho$ is injective. Thus $\rho$ is a module isomorphism. Now the homomorphism $1 \otimes \rho : A \otimes_R R \rightarrow A \otimes_R I$ is also an isomorphism since $(1 \otimes \rho) \circ (1 \otimes \rho^{-1}) = 1 \otimes 1 = 1$ and $(1 \otimes \rho^{-1}) \circ (1 \otimes \rho) = 1 \otimes 1 = 1$. Recall that $A$ and $A \otimes_R R$ are isomorphic as abelian groups via the mappings $\iota : A \rightarrow A \otimes_R R$ and $\tau : A \otimes_R R \rightarrow A$ given by $\iota(a) = a \otimes 1$ and $\tau(a \otimes r) = ar$. Consider the map $1 \otimes \iota : A \otimes_R I \rightarrow A \otimes_R R$.

Now $(\tau \circ (1 \otimes \iota) \circ (1 \otimes \rho) \circ \iota)(a) = \tau((1 \otimes \iota)((1 \otimes \rho)(a \otimes 1)))$ $= \tau((1 \otimes \iota)(a \otimes r))$ $= \tau(a \otimes r)$ $= ar = \theta_r(a)$. Thus we have $\tau \circ (1 \otimes \iota) \circ (1 \otimes \rho) \circ \iota = \theta_r$, and so $1 \otimes \iota = \tau^{-1} \circ \theta_r \circ \iota^{-1} \circ (1 \otimes \rho)$. Thus $1 \otimes \iota$ is injective.

By the flatness criterion for modules (proved here), $A$ is flat as a right $R$-module.