$\varpi$ does not contain non-trivial ideals of $\R$ and so we have the following result.

The additive character $\varpi$ is a generator of the $\R$-module $\R^\wedge$.

The map $$\M^\vee\rightarrow \M^\wedge,\,\,\,f\mapsto \varpi\circ f$$ is an isomorphism of $\R$-modules. \[Pont\]

Let $g\in \M^\wedge$. Consider $\M$ as the quotient of some $\R^n$ and denote by $\pi$ the quotient map $\R^r\rightarrow \M$. Let $(x_1,\dots,x_r)$ be the canonical basis of $\R^r$. Then for each $i=1,\dots,r$, let $\alpha_i\in\R^\wedge$ be defined by $\alpha_i(h)=(g\circ\pi)(h x_i)$ for $h\in\R$. Since $\varpi$ is a generating character of $\R^\wedge$, we get for each $i=1,\dots,r$, an element $\lambda_i\in\R$ such that $\alpha_i=\lambda_i\cdot\varpi$.

We let $\overline{f}:\R^r\rightarrow\R$ be the $\R$-linear form defined by $\overline{f}(x_i)=\lambda_i$ for all $i=1,\dots,r$. We then have $g\circ\pi=\varpi\circ\overline{f}$. We now prove that there exists a unique $\R$-linear form $f:\M\rightarrow\R$ making the following diagram commutative

$$\xymatrix{\R^r\ar[r]^-\pi\ar[rd]_{\overline{f}}&\M\ar[r]^-g\ar[d]^f&\C^\times\\
&\R\ar[ru]_\varpi&}$$ We need to see that ${\rm Ker}(\pi)\subset{\rm Ker}(\overline{f})$. For $x\in{\rm Ker}(\pi)$, we have $(\varpi\circ\overline{f})(x)=(g\circ\pi)(x)=1$ and so $\overline{f}(x)\in{\rm Ker}(\varpi)$. Moreover for all $\lambda\in\R$, we have $\lambda\overline{f}(x)=\overline{f}(\lambda x)$ and $\lambda x\in{\rm Ker}(\pi)$, and so ${\rm Ker}(\varpi)$ contains the ideal generated by $\overline{f}(x)$. Since the kernel of $\varpi$ does not contain non-trivial ideals of $\R$, we must have $\overline{f}(x)=0$ for all $x\in{\rm Ker}(\pi)$.

The $\R$-module $\M$ is reflexive. \[reflexive\]

We 