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Edit: (Short version over Noetherian Domain) Exactly when M is projective One "extension" of your statement is that if $M$ has rank $r$, then $\mbox{rank}\wedge^k M = {r \choose k}$. In particular, when $k > r$ the module $\wedge^k M$ has rank $0$. This holds in any commutative ring for any finite rank module. (It comes considering $R^r \to M$, and ...

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I think the solution is not possible, though with you may have a solution to similar systems: By 1), $A$ must be a function, i.e., a $0$-form, and , since the wedge of functions is their product, you must have $A==\pm 1$, But then, substituting in 2), you will have $\pm\eta+\pm\eta=\eta$, which is not possible.

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The exterior derivative is a sort of infinitesimal operation. See for example Bachman's notes on differential forms. (See also comments to the question, above).

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