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Math Help - Dedekind Domain

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    Dedekind Domain

    Let  M be a finitely generated module over a Dedekind domain. Prove that  M is flat  \Longleftrightarrow  M is torsion free.
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    Quote Originally Posted by particlejohn View Post

    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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