Exterior algebra universal property

Question

I am trying to show the following (D&F Ex. 11.5.6):

Let $R$ be a commutative ring with $1$, $A$ an $R$-algebra such that $a^2=0$ for all $a\in A$, and $\varphi:M\to A$ an $R$-module homomorphism. Then there exists a unique $R$-algebra homomorphism $\Phi:\bigwedge(M)\to A$ such that $\Phi|_m=\varphi$.

Idea: The map $$M\times \cdots \times M \to A, \qquad (m_1,\dots,m_k) \mapsto \varphi(m_1)\varphi(m_2)\cdots \varphi(m_k)$$ is an alternating $k$-multilinear map and therefore induces a unique $R$-module homomorphism $\Phi_k:\bigwedge^k(M)\to A$ such that $\Phi_k|_M=\varphi$. The collection $\{\Phi_k\}_{k=0}^{\infty}$ induces a unique map $\Phi:\bigwedge(M)\to A$ such that $\Phi|_M=\varphi$.

Is $\Phi$ an $R$-algebra homomorphism and if so, why?

Answer 1

The universal property of the tensor algebra gives a homomorphism $\Psi:T(M)\to A$ with $\Psi\mid_M=\varphi$. As $\bigwedge(M)$ is the quotient of $T(M)$ by the ideal generated by the $m\otimes m$ then to show $\Psi$ induces $\Phi:\bigwedge(M)\to A$ all one needs to prove is that $\Psi(m\otimes m)=0$.