Dedekind Domain

**particlejohn** · October 25th 2008, 10:06 PM

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

**NonCommAlg** · October 25th 2008, 11:39 PM

Originally Posted by particlejohn

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

$\Longrightarrow$ : let $0 \neq r \in R, \ x_0 \in M,$ and $rx_0=0.$ consider the R-homomorphism $f: R \longrightarrow R,$ defined by $f(s)=rs.$ since R is an integral domain, $f$ is injective.

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

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

$\Longleftarrow$ : 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 $\Longrightarrow$ 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.

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