What is the Unique $R$-algebra Homomorphism $\phi: \bigwedge(M) \to A$?

[Chris L T521]

Here's this week's problem!

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Problem: If $A$ is any $R$-algebra in which $a^2 = 0$ for all $a \in A$ and $\varphi: M \to A$ is an $R$-module homomorphism, prove there is a unique $R$-algebra homomorphism $\phi: \bigwedge(M) \to A$ such that $\phi \vert_M = \varphi$

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[Chris L T521]

No one answered this week's problem. You can find the solution below.

[sp]Proof: By the universal property of tensor algebras there exists a unique $R$-module homomorphism $\overline{\phi}: \mathcal{T}(M) \rightarrow A,$ such that $\overline{\phi} \vert_M = \varphi.$ Since $\bigwedge(M) := \mathcal{T}(M) / \mathcal{A}(M)$ with $\mathcal{A}(M)$ generated by the simple tensor $m \otimes m$ for all $m \in M,$ it follows that $\mathcal{A}(M) \subseteq ker \left(\overline{\phi} \right).$ This follows since $\overline{\phi}(m \otimes m) = \varphi(m) \varphi(m) = \varphi(m)^2 = 0$ by the property $a^2 = 0$ for all $a \in A,$ and $\varphi(m) \in A.$ Thus we obtain a well-defined $R$-algebra homomorphism $\phi: \bigwedge(M) \rightarrow A,$ with $\phi(\overline{n}) = \overline{\phi}(n)$ which certainly satisfies $\phi \vert_M = \varphi.$ Uniqueness follows by supposing there exist another such $R$-algebra homomorphism $\omega : \bigwedge (M) \rightarrow A.$ Then $$\omega \left( \bigwedge (m_i) \right) = \prod \omega (m_i) = \prod \varphi(m_i) = \prod \phi(m_i) = \phi \left( \bigwedge(m_i) \right).$$[/sp]