# Why is there no isomorphism between exterior powers of a free module?

The following question is related to this post Showing an isomorphism between exterior products of a free module about the existence of a canonical isomorphism of the exterior product of free modules and in particular details the pathological behavior of certain isomorphisms given conditions on the rank of the module.

Let $F$ be a free $R$-module of rank $n\geq 3$ where $R$ is a commutative ring with identity.

Let $T(F) = \oplus_{k=0}^{\infty} T^{k}(F)$ where $T^k(F) = F \otimes F \otimes \ldots \otimes F$ is tensor product of $k$ modules.

Let $\wedge F$ denote the exterior algebra of the $R$-module $F$, that is the quotient of the tensor algebra $T(F)$ by the ideal $A(F)$ generated by all $x \otimes x$ for $x \in F$.

Suppose $1\leq p \leq n-1$ and that $n \neq 2p$

How do we show there cannot exist a bijective module homomorphism $f: \wedge^p F \rightarrow \wedge^{n-p} F$ such that $f \circ (\wedge^p \psi) = (\wedge^{n-p} \psi ) \circ f$ for all $\psi \in Aut(F)$?

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I think that you mean "$f\circ (\wedge^p\psi) = (\wedge^{n-p} \psi)\circ f$" in you last line. – Matt E Oct 21 '11 at 4:31
thank you I just made the correction. – user7980 Oct 21 '11 at 4:54
The last sentence is ungrammatical -- do you mean "such that" (without "if")? – joriki Oct 21 '11 at 7:02
What's the counterexample? I can't find one in the other question you link to. – joriki Oct 21 '11 at 7:07
You should first answer by yourself a similar but simpler problem: prove that we can't find an isomorphism $f:F \to F^\vee$ (the dual of $F$) such that $f \circ \psi = (\psi^\vee) \circ f$ for every $\psi \in GL(F)$. Hint: chose a basis write things down in terms of matrices. – AFK Oct 22 '11 at 21:25