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!

The $a$ be a finite dimensional abelian Lie algebra, let $\{x_1,\dots,x_n\}$ be a basis for $a$ and let $U$ be its enveloping algebra. It is easy to see that $U$ is a polinomial ring with generators $x_1,\dots,x_n$. You can check that $x_1,\dots,x_n$ is a regular sequence in $U$, so using the result of Chapter 4 in the book that refer to th Koszul complex, you get a free resolution of the module $U/(x_1,\dots,x_n)$, which is just th trivial module over $a$, as a let $U$-module. Use that resolution to compute $Ext_U(k,k)$, which is the Lie cohomology $H^*(a,k)$.
