Universal property of exterior algebras

Let R be a commutative ring with 1 and let M be a unital (R,R)-bimodule such that rm = mr. Prove that if A is an R-algebra such that a^2 = 0 for all a \in A and if \varphi : M \rightarrow A is an R-module homomorphism, then there exists a unique R-algebra homomorphism \Phi : \bigwedge(M) \rightarrow A such that \Phi|_M = \varphi.

By the universal property of tensor algebras, there exists a unique R-module homomorphism \overline{\Phi} : \mathcal{T}(M) \rightarrow A such that \overline{\Phi}|_M = \varphi. Now recall that \mathcal{A}(M) is the ideal of \mathcal{T}(M) generated by simple tensors of the form m \otimes m. Note that \overline{\Phi}(m \otimes m) = \overline{\Phi}(m)^2 = 0 in A, so that \mathcal{A}(M) \subseteq \mathsf{ker}\ \overline{\Phi}. By the first isomorphism theorem for R-algebras, we have a well-defined R-algebra homomorphism \Phi : \bigwedge(M) \rightarrow A given by \Phi(\overline{t}) = \overline{\Phi}(t). Certainly then \Phi|_M = \varphi.

To see uniqueness, suppose \Psi : \bigwedge(M) \rightarrow A is another such homomorphism. Then \Psi(\bigwedge m_i) = \prod \Psi(m_i) = \prod \varphi(m_i) = \prod \Phi(m_i) = \Phi(\bigwedge m_i).

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