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.

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