Exercise in Weibel's book Lie cohomology, Homology Can you please help me out to prove the following (rather easy (but not for me obviously)) exercise which I didn't manage to solve (Exercise 7.4.2, page 229)?

Assume $\mathfrak{g}$ is the free $\mathbb{k}$-module on basis $\{ e_1, e_2,..., e_n \}$, made it into abelian Lie algebra. Show that $H_p(\mathfrak{g}, \mathbb{k}) = \bigwedge^k (\mathfrak{g})$, the $p^{th}$ exterior power of the $\mathbb{k}$-module $\mathfrak{g}$.

Thank you!

let $L$ be a Lie algebra over $K$. The complex $(C_{\bullet},\partial)$ is given by $C_p(L,V)=\Lambda^pL\otimes_KV$, where $V$ is an $L$-module, and $\partial_p$ is a boundary operator. For the trivial module $V=K$ these space of $p$-chains is just $\Lambda^pL$, and since $L$ is abelian, the boundary map $\partial_p$ is zero, so that we have
$$
H_p(L,K)=\ker(\partial_p)/im (\partial_p)=\ker(\partial_p)=C_p(L,K)=\Lambda^pL.
$$
