Dual of a vector space coming from a lattice

Question

Let $M$ be a lattice i.e., $M$ is an abelian group and $M\cong\mathbb Z^n$ for some $n$. Let $N:=\hom_{\mathbb Z}(M,\mathbb Z)$. Let $M_{\mathbb R}$ denotes the $\mathbb R$-vector space $M\otimes_{\mathbb Z}\mathbb R$. Then clearly $M_{\mathbb R}\cong\mathbb R^n$.

I need to show that $N_{\mathbb R}:=N\otimes_{\mathbb Z}\mathbb R$ is the dual of $M_{\mathbb R}$, i.e., I need to show that $N_{\mathbb R}:=\hom_{\mathbb Z}(M,\mathbb Z)\otimes_{\mathbb Z}\mathbb R\cong\hom_{\mathbb R}(M_{\mathbb R,}\mathbb R)$.

Any hint will be helpful.

Thank you.

Answer 1

To construct such an isomorphism, you could start by constructing an $\Bbb{R}$-linear map $M_{\Bbb{R}}\ \longrightarrow\ \Bbb{R}$ for every pure tensor $f\otimes c\in\operatorname{Hom}_{\Bbb{Z}}(M,\Bbb{Z})\otimes_{\Bbb{Z}}\Bbb{R}$, where $f\in\operatorname{Hom}_{\Bbb{Z}}(M,\Bbb{Z})$ and $c\in\Bbb{R}$. There are not many sensible options for such a construction. Then show that this extends to an $\Bbb{R}$-linear map $$\operatorname{Hom}_{\Bbb{Z}}(M,\Bbb{Z}),\otimes_{\Bbb{Z}}\Bbb{R}\ \longrightarrow\ \operatorname{Hom}_{\Bbb{R}}(M_{\Bbb{R}},\Bbb{R}).$$ Then show that it is injective and surjective; it might help to note that the two vector spaces are of the same dimension over $\Bbb{R}$.

Answer 2

Note that $M \cong \mathbb{Z}^n$ is isomorphic to $N \cong \hom_\mathbb{Z}(\mathbb{Z}^n,\mathbb{Z})$ as $\mathbb{Z}$-modules, and $M_{\mathbb{R}} \cong \mathbb{R}^n$ is isomorphic to $\hom_\mathbb{R}(\mathbb{R}^n,\mathbb{R})$ as $\mathbb{R}$-modules. So, the result follows.