The rank of the kth exterior power of a free module

Let R be a commutative ring with 1 and let M be a free unital (R,R)-bimodule such that rm = mr. Prove that the kth exterior power \bigwedge^k(M) is a free R-module of rank {n \choose k}.

Suppose E = \{e_i\}_{i=1}^n is a free generating set of M.

Recall from this previous exercise that the kth tensor power is free with free generating set \{e_{i_1} \otimes \cdots \otimes e_{i_k} \ |\ i : k \rightarrow n\}.

First we will construct a generating set for \bigwedge^k(M). Certainly \bigwedge^k(M) is generated by the set of all simple k-tensors m_1 \wedge \cdots \wedge m_k. Note that, by multilinearity, we may assume that each m_i comes from the free generating set E: say e_{i_1} \wedge \cdots \wedge e_{i_k}. Moreover, up to a possible multiplication by -1, we can say that i_1 < i_2 < \cdots < i_k. We claim that these are all distinct. To see this, for each increasing choice function \lambda : k \rightarrow n, define \varphi_\lambda : \mathcal{T}^k(M) \rightarrow R by letting \varphi_\lambda(e_{j_k} \otimes \cdots \otimes e_{j_k}) = \epsilon(\sigma) if \mathsf{im}\ j is a permutation of \mathsf{im}\ \lambda (namely the unique permutation \sigma), and 0 otherwise, and extend linearly. Certianly for increasing choice functions \lambda, \mathcal{A}^k(M) \subseteq \varphi_\lambda, so that we have an induced homomorphism \bigwedge^k(M) \rightarrow R. Moreover, \varphi_\lambda is zero on all of the e_{i_1} \wedge \cdots \wedge e_{i_k} except the one whose indices are given by \lambda. Thus the elemens of our generating set are distinct. There are {n \choose k} distinct ways to choose the indices i_j, and so we have a generating set B of \bigwedge^k(M) of order {n \choose k}.

Note that any nontrivial linear combination \sum \alpha_i z_i = 0 in \bigwedge^k(M) induces a nontrivial linear combination in the free basis on \mathcal{T}^k(M); thus our basis is free.

Hence \bigwedge^k(M) is a free R-module having free rank {n \choose k}.

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