1. Dedekind Domain

Let $\displaystyle M$ be a finitely generated module over a Dedekind domain. Prove that $\displaystyle M$ is flat $\displaystyle \Longleftrightarrow$ $\displaystyle M$ is torsion free.

2. Originally Posted by particlejohn

Let $\displaystyle M$ be a finitely generated module over a Dedekind domain. Prove that $\displaystyle M$ is flat $\displaystyle \Longleftrightarrow$ $\displaystyle M$ is torsion free.
$\displaystyle \Longrightarrow$: let $\displaystyle 0 \neq r \in R, \ x_0 \in M,$ and $\displaystyle rx_0=0.$ consider the R-homomorphism $\displaystyle f: R \longrightarrow R,$ defined by $\displaystyle f(s)=rs.$ since R is an integral domain, $\displaystyle f$ is injective.

thus the map $\displaystyle f \otimes \text{id}_M: R \otimes_R M \longrightarrow R \otimes_R M$ must be injective too because M is assumed to be flat. but $\displaystyle R \otimes_R M \simeq M.$ so the map $\displaystyle f \otimes \text{id}_M$ is basically

the map $\displaystyle f^* : M \longrightarrow M,$ defined by: $\displaystyle f^*(x)=rx, \ \forall x \in M.$ now we have $\displaystyle 0=rx_0=f^*(x_0).$ thus $\displaystyle x_0=0,$ since $\displaystyle f^*$ is injective. therefore M is torsion free.

$\displaystyle \Longleftarrow$: since R is a Dedekind domain and M is finitely generated and torsion free, M is isomorphic to a direct sum of ideals of R. so M is projective, because every

ideal of a Dedekind domain is projective. finally this trivial fact that every projective module is flat completes the proof. Q.E.D.

Remark 1: for $\displaystyle \Longrightarrow$ we only needed R to be a domain.

Remark 2: the equivalence of "torsion free" and "flat" holds for any module (not necessarily finitely generated) over a Dedekind domain.