Is $\mathcal{U}(\mathfrak{g})$ semisimple (as a module over itself)? (and related examples) Recall that an object in an abelian category is semisimple if it is a (finite) direct sum of simple objects. An abelian category is semisimple if every object is semisimple.
In studying representations of semisimple lie algebras, one often restricts to the subcategory called $\mathcal{O}_{int}$ inside the category left modules for the universal enveloping algebra, $\mathcal{U}(\mathfrak{g})$.  The main theorem about $\mathcal{O}_{int}$ says it is semisimple (or, at least, that all the objects in $\mathcal{O}_{int}$ are semisimple. Perhaps one needs to check several other things to see that $\mathcal{O}_{int}$ is abelian as a subcategory). 
I am new to this area and would like to motivate this restriction. 
A ring $R$ is semisimple in the category of left $R$ modules iff the category of left $R$ modules is semisimple. Thus,  

Is $\mathcal{U}(\mathfrak{g})$ semisimple as a left $\mathcal{U}(\mathfrak{g})$-module (for $\mathfrak{g}$ a semisimple Lie algebra)?

As a bonus, are there other nice examples of $\mathcal{U}(\mathfrak{g})$-modules that aren't semisimple?
 A: If $U(\mathfrak{g})$ were semi-simple as a $U(\mathfrak{g})$-module, then the same would be true of any quotient. However, $U(\mathfrak{g})/U(\mathfrak{g})\mathfrak{b}\cong M(0)$, where $M(0)$ is the Verma module of highest weight $0$, $\mathfrak{b}=\mathfrak{h}\oplus\mathfrak{n}^+$ and $\mathfrak{g}=\mathfrak{n}^-\oplus\mathfrak{h}\oplus\mathfrak{n}^+$ is a triangular decomposition of $\mathfrak{g}$.
It is well known, and easy to check that $M(0)$ is not semi-simple. Try it for $\mathfrak{sl}_2$.
A: Here is an alternate, somewhat indirect approach. The Artin-Wedderburn theorem completely classifies semisimple rings as finite products $\prod_i M_{n_i}(D_i)$ of matrix rings over division rings. The PBW theorem implies that $U(\mathfrak{g})$ has no nontrivial zero divisors (see, for example, this math.SE answer), but a semisimple ring $\prod_i M_{n_i}(D_i)$ has nontrivial zero divisors unless it's a division ring $D$.
$U(\mathfrak{g})$ is never isomorphic to a division ring unless $\mathfrak{g} = 0$; this is because it has a counit map $U(\mathfrak{g}) \to k$ to the underlying field, so can only be a division ring if this map is an isomorphism.
