Homomorphism between exterior powers of a free module of finite rank I´m looking for homomorphisms between  exterior powers of a free module M of rank m

ΛmR M → Λm-1R M
Exactly, I´m looking for an explicit isomorphism

M → Hom R (ΛmR M , Λm-1R M)
I compare the ranks and the things go, but I can not imagine a concrete expression.
Suggestions are welcome
 A: If $R$ is commutative, let $e_1, \ldots, e_m$ be an $R$-basis of $M$. Then $\Lambda^m M$ is a free $R$-module of rank one generated by  the vector $e_1 \wedge e_2 \wedge \ldots \wedge e_m$ and $\Lambda^{m-1}M$ is free of rank $m$ with basis $$\lbrace e_1 \wedge \ldots \wedge \hat{e_i} \wedge \ldots \wedge e_m \rbrace_{i=1}^{m}$$ where the hat denotes an omitted basis vector. Here is an explicit isomorphism: $$e_i \in M\mapsto \Big(e_1 \wedge e_2 \wedge \ldots \wedge e_m \mapsto e_1 \wedge \ldots \wedge \hat{e_i} \wedge \ldots \wedge e_m \Big) \in \mathrm{Hom}_R(\Lambda^m M, \Lambda^{m-1} M)$$
A: Here is a basis free expression. 
Let the rank of $M$ be $r$. Pick an isomorphism $\phi: M \to M^\*=\hom_R(M,R)$. Now, if $m\in M$, define a map $f_m:\Lambda^rM\to\Lambda^{r-1}M$ by contracting with $\phi(m)$, so that if $m_1$, $\dots$, $m_r\in M$, then $$f_m(m_1\wedge\cdots\wedge m_r)=\sum_{i=1}^r\;(-1)^i\;\langle\phi(m),m_i\rangle\; m_1\wedge\cdots\wedge\hat m_i\wedge\cdots\wedge m_r.$$ Here $\langle\mathord-,\mathord-\rangle:M^*\times M\to R$ is the evaluation map. It is not hard to show that $m\in M\mapsto f_m\in\hom(\Lambda^rM, \Lambda^{r-1}M)$ is an isomorphism 
(This isomorphism is not natural, because it depends on the choice of $\phi$. Of course, there is a natural isomorphism $\Psi_M:M^\*\to\hom(\Lambda^rM, \Lambda^{r-1}M)$ given by essentially the formula, and there is no natural isomorphism $M\to\hom(\Lambda^rM, \Lambda^{r-1}M)$, because such a thing would, when composed with the inverse of the natural isomorphism $\Psi_M$, give a natural isomorphism $M\to M^\*$, which does not exist.)
