Projectivity of $U(\mathfrak(g))$ as a $U(\mathfrak(b))$-module

Question

Let $\mathfrak{g}$ be a semisimple Lie algebra, and $U(\mathfrak{g})$ be its enveloping algebra, with a triangular decomposition $$ U(\mathfrak{g}) = U(\mathfrak{n}^-) \otimes_\Bbbk U(\mathfrak{h}) \otimes_\Bbbk U(\mathfrak{n}) = U(\mathfrak{n}^-) \otimes_\Bbbk U(\mathfrak{b}).$$

Is $U(\mathfrak{g})$ a projective (right) $U(\mathfrak{b})$-module?

I think this is true, unless the following is wrong:

Consider any right $U(\mathfrak{b})$-module $M$, and let $n \geqslant 1$. Then

\begin{align*} \text{Ext}_{U(\mathfrak{b})}^n(U(\mathfrak{g}),M) &= \text{Ext}_{U(\mathfrak{b})}^n(U(\mathfrak{n}^-) \otimes_\Bbbk U(\mathfrak{b}) ,M) \\ &\cong \text{Ext}_{\Bbbk}^n(U(\mathfrak{b}), \text{Hom}_{U(\mathfrak{b})}(U(\mathfrak{n}^-),M)) \\ & = 0 \end{align*} where the isomorphism follows since $U(\mathfrak{n}^-)$ is projective as a right $\Bbbk$-module, and the last equality follows since $\Bbbk$ is semisimple. Hence $U(\mathfrak{g})$ a projective (right) $U(\mathfrak{b})$-module.

Assuming this is correct, I feel like there has to be a simpler way to prove this. Is there?

Answer 1

You can take a $\{b_i\}$ basis for $\mathfrak b$ and extend it to $\{c_j,b_i\}$ a basis for $\mathfrak g$ (put the $\mathfrak b$-part on the right side). Now use it to make Poincare-Birkhoff-Witt basis for $U(\mathfrak{g})$. Ordered monomials in $\{c_j\}$ will constitute a free basis for $U(\mathfrak g)$ as right $U(\mathfrak b)$-module. So it is free, and therefore projective.

Edit: As stated by Tobias Kildetoft, it works for any subalgebra of $\mathfrak g$, and $\mathfrak g$ not necessarily semisimple.