Let \(R\) be a commutative ring. Recall that an \(R\)-module \(Q\) is flat if \(-\otimes_R Q\) is an exact functor; namely, given any short exact sequence of \(R\)-modules \(0 \to M \to N \to P \to 0\), the sequence \[ 0 \to M \otimes_R Q \to N \otimes_R Q \to P \otimes_R Q \to 0 \] remains exact.
Let \(Q\) be any module over an integral domain \(R\). Show that if \(Q\) is flat, then \(Q\) is torsionfree.
In terms of the classification theorem for modules over PIDs, characterize exactly which finitely generated modules over a PID \(R\) are flat.
Solution
1. Let \(R\) be an integral domain and \(0\neq r\in R\). Multiplication by \(r\) gives an injective \(R\)-linear map\[ R\xrightarrow{\cdot r}R. \]
If \(ra=0\) in the domain \(R\) and \(r\neq0\), then \(a=0\). Hence multiplication by \(r\) is injective.
2. If \(Q\) is flat, multiplication by every nonzero \(r\in R\) is injective on \(Q\).
From 1 we have a short exact sequence \[ 0\longrightarrow R\xrightarrow{\cdot r}R\longrightarrow R/rR\longrightarrow0. \] Tensoring with the flat module \(Q\) preserves exactness, so \[ 0\longrightarrow R\otimes_R Q \xrightarrow{(\cdot r)\otimes 1} R\otimes_R Q \] is exact. Under the canonical identification \(R\otimes_RQ\cong Q\), the displayed map is precisely \[ Q\xrightarrow{\cdot r}Q. \] Therefore it is injective.
3. Every flat module over an integral domain is torsionfree.
Suppose \(q\in Q\) and \(0\neq r\in R\) satisfy \(rq=0\). By 2, multiplication by \(r\) on \(Q\) is injective, so \(q=0\). This is exactly torsionfreeness.
4. Let \(R\) be a PID and let \(M\) be finitely generated.By the structure theorem,\[ M\cong R^s\oplus T, \]where \(T\) is a finite direct sum of nonzero cyclic torsion modules \(R/(d_i)\) if the torsion part is nonzero.
This is the classification theorem for finitely generated modules over a PID.
5. A finitely generated \(R\)-module over a PID is flat if and only if its torsion part \(T\) in 4 is zero.
If \(M\) is flat, then 3 implies that \(M\) is torsionfree. Hence the torsion summand \(T\) must vanish.
Conversely, if \(T=0\), then \(M\cong R^s\) is free. Every free module is flat because tensoring with a direct sum of copies of \(R\) is a direct sum of copies of the original exact sequence and therefore preserves exactness.
6. Thus the finitely generated flat modules over a PID are exactly the finitely generated free modules.
This is the equivalence established in 5.