Tensor on Exterior Algebra I hope someone can help me on showing that $(\wedge^k(M))^*\otimes \wedge^n(M) = \wedge^{n-k}(M)$, where $M$ is a free $R$-module of rank $n$, and $^*$ is the dual.
From what I know, $(\wedge^k(M))^* = \mbox{Hom}(\wedge^k(M), R)$. But how do I incorporate it with the tensor product, or am I even on the right track? My intuition tells me that if $\varphi \in \mbox{Hom}(\wedge^k(M), R)$, then it somehow "reduces" an element of $\wedge^{k}(M)$ to $R$, and so explains the $\wedge^{n-k}(M)$ part of the equality. Though I am not entirely sure of this.
Any help is much appreciated. Thanks!
 A: First of all, there is a natural pairing $\wedge^kM\times\wedge^{n-k}M\to\wedge^nM$ given by the exterior product. So for any element $v\in\wedge^{n-k}M$, you get a homomorphism $\phi_v:\wedge^kM\to\wedge^nM$ given by $\phi_v(u)=u\wedge v$. This gives you a linear map $\phi:\wedge^{n-k}M\to\mathrm{Hom}(\wedge^kM,\wedge^nM)$. You can prove that this is an isomorphism.
Next, we can prove that $\mathrm{Hom}(\wedge^kM,\wedge^nM)$ is isomorphic to $(\wedge^kM)^*\otimes\wedge^nM$. The isomorphism is the map $\psi:(\wedge^kM)^*\otimes\wedge^nM\to\mathrm{Hom}(\wedge^kM,\wedge^nM)$ that sends $\alpha\otimes q$ to the homomorphism $p\mapsto\alpha(p)\cdot q$.
The isomorphism you're looking for is $\phi^{-1}\circ\psi$.
A: If $M$ is a free $R$-module of rank $n$, then $\wedge^n(M)\cong R$. So $(\wedge^k M)^*\cong \wedge^{n-k}M$ is ultimately what you are trying to show. Then, $(\wedge^k M)^*:=\hom_R(\wedge^k M,R)\cong\hom_R(\wedge^kM,\wedge^nM)$, so we have an isomorphism $\wedge^{n-k}M\to(\wedge^kM)^*$ sending a $(n-k)$-vector $v$ to the map $w\mapsto v\wedge w$ sending $k$-vectors $w$ to $n$-vectors $v\wedge w$.
