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For a finite dimensional vector space $V$ over a field $k$, $\Lambda^n (V)$ is a vector space of dimension $1$, where $n$ is the dimension of $V$. Moreover, $n$ is the highest integer for which $\Lambda^n (V)$ is nonzero (and in general, $\dim ( \Lambda^k (V)) = {n \choose k}$ where $n = \dim(V)$). But most importantly for our purposes, $\dim (V)$ is the highest integer $n$ for which $\text{dim}(\Lambda^n (V))$ is nonzero. This shows that an alternative, basis free definition of dimension of a vector space is $$\dim (V) = \text{max} \{ n \in \mathbb{N} : \Lambda^n_{k} (V) \neq 0 \}$$ Note that this is defined only when $\Lambda^n_k (V)$ vanishes for high enough $n$. I am wondering, may we define the rank of a free module $M$ over a commutative ring $R$ in this way? More precisely, does the definition $$\text{rank}'(M) := \text{max} \{ n \in \mathbb{N} : \Lambda^n_R (M) \neq 0 \}$$ match the rank $\text{rank} (M)$ of a free module $M$ over a ring $R$? If this is true, then could $\text{rank}'$ also be used as a more general definition of rank of a module? Of course this would only be useful when $\text{rank}'(M)$ is finite.

I am thinking that $\text{rank}'(M)$ should be the maximal length of a $R$-linearly independent list of elements in $M$.

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