## 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 $\Psi(\bigwedge m_i) = \prod \Psi(m_i)$ $= \prod \varphi(m_i)$ $= \prod \Phi(m_i)$ $= \Phi(\bigwedge m_i)$.