# Free Fractional Modules

Let $R$ be a Dedekind domain with fractional ideal $M$. I am trying to prove:

If $M$ is free, then the ideal $I$ such that $M=d^{-1}I$, where $d \in R$, is principal.

Since $R$ is a Dedekind domain we can conclude $M$ is generated by at most two elements. I've been trying for a while, but I don't know how to conclude $I$ is principal. Any ideas?

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## 1 Answer

If $R$ is any domain at all and $M$ is a finitely generated (automatic when $R$ is Noetherian) fractional ideal which is $R$-free, then it is $R$-free of rank one. Since $M$ is isomorphic to an ideal of $R$, we may assume $M=I$ is an ideal of $R$. If $r_1\neq r_2\in I$, then $\{r_1,r_2\}$ is not linearly independent, because we have $0=r_2r_1-r_1r_2$, which is a non-trivial relation since at least one of $r_1,r_2$ must be non-zero. So $I$ must be free of rank $1$ (since it's not zero by definition). So $I$ is principal.

EDIT: I suppose this argument works without the assumption that $M$ is finitely generated, so I guess that assumption can be removed.

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thank you very much. –  James Vincent Nov 22 '12 at 17:56
Dear @James, You're welcome! –  Keenan Kidwell Nov 22 '12 at 17:57
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