I want to prove Lemma 2.5.1 of Silverman's Advanced Topics in The Arithmetic of Elliptic Curves (whose proof is left to the reader):

Let $R$ be a Dedekind domain, let $\mathfrak{a}$ be a fractional ideal of $R$ and let $M$ be a torsion-free $R$-module. Then the natural map $$ \varphi \colon \mathfrak{a}^{-1}M \rightarrow \operatorname{Hom}_R(\mathfrak{a},M) $$ which sends $x \in \mathfrak{a}^{-1}M$ to the $R$-module homomorphism $\varphi_x \colon \alpha \mapsto \alpha x$ is an isomorphism.

This statement looks very similar to the standard isomorphism between a unital module $N$ over a unital ring $S$ and $\operatorname{Hom}_S(S,N)$, but there are some issues. In particular, $\mathfrak{a}^{-1}M$ doesn't make any apparent sense, since $M$ is an $R$ module, but in general is not true that $\mathfrak{a}^{-1} \subseteq R$. Also, the standard proof of $\operatorname{Hom}_S(S,N) \simeq N$ definitely requires that $1 \in S$.

Thank you very much for any help.

EDIT: According to Mohan, I have to prove that the map $\varphi \colon \mathfrak{a}^{-1} \otimes_R M \rightarrow \operatorname{Hom}_R(\mathfrak{a},M)$ given by $$ \varphi \left( \sum_{i} \beta_i \otimes m_i \right) \alpha := \sum_{i} \left( \alpha \beta_i \right)m_i \quad \forall \alpha \in \mathfrak{a} $$ is an isomorphism of $R$-modules (it is well defined since $\alpha \beta_i \in \mathfrak{a} \mathfrak{a}^{-1} = R$). It is quite obvious that $\varphi$ is a homomorphism of $R$-modules, but I can't prove neither injectivity nor surjectivity.

Can someone help me?

  • $\begingroup$ Since $M$ is torsion free, it sits naturally inside $M\otimes_R K$, where $K$ is the fraction field. $\mathfrak{a}^{-1}\subset K$ and thus $M\otimes\mathfrak{a}^{-1}$ makes sense and this is $\mathfrak{a}^{-1}M$. $\endgroup$ – Mohan Jun 30 '15 at 16:17
  • $\begingroup$ Sorry, I've mistakenly deleted my previous comment. Right, I've understood how does $\varphi$ work. Thank you $\endgroup$ – Frankenstein Jun 30 '15 at 21:43
  • $\begingroup$ I have edited my question. $\endgroup$ – Frankenstein Jul 1 '15 at 6:24

If $P$ is a finitely generated and projective $R$-module, then there is a canonical isomorphism $$\zeta:P^*\otimes_RM\simeq\operatorname{Hom}_R(P,M)$$ given by $\zeta(f\otimes m)(x)=f(x)m$.

(For a proof see Proposition 6 from these notes.)

Now, coming back to the question we have $\mathfrak a^*\otimes_RM\simeq\operatorname{Hom}_R(\mathfrak a,M)$. But $\mathfrak a^*\simeq\mathfrak a^{-1}$ (why?) and we are done.

Remark. Since $M$ is supposed to be torsion-free (which is actually not necessary), one can use another approach via the isomorphism mentioned here: Hom and tensor with a flat module.


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