Exercise 17, § 1

We retain the notations of the preceding exercise. Show that, for all $n\geq 1$, the algebra $U(\mathfrak{g})$ has a unique two-sided ideal $\mathfrak{m}_n$ such that $U(\mathfrak{g})/\mathfrak{m}_n$ is a central simple algebra of dimension $n^2$ (extend scalars to reduce to the case in which $\mathfrak{g}=\mathfrak{s}$ and show that $\mathfrak{m}_n$ is then the kernel of the homomorphism $U(\mathfrak{g})\rightarrow$ End(V($n-1$))). Every two-sided ideal of $U(\mathfrak{g})$ of finite codimension is of the form $\mathfrak{m}_{n_1}\cap \mathfrak{m}_{n_2}\cap \cdots \cap \mathfrak{m}_{n_h}$, where $n_1, . . . , n_h$ are distinct; its codimension is $n^2_1+\cdots +n^2_h$ (apply the density theorem). The $\mathfrak{m}_n$ are the only maximal two-sided ideals of $U(\mathfrak{g})$ of finite codimension.

Show that $\mathfrak{m}_2$ is generated (as a two-sided ideal) by the elements $x^2-\frac{1}{8}\Phi (x, x) (x\in \mathfrak{g})$, and that the quotient $U(\mathfrak{g})/\mathfrak{m}_2$ can be identified with the quaternion algebra A of Exerc. 16.

When $\mathfrak{g}$ is isomorphic to $\mathfrak{s}, U(\mathfrak{g})/\mathfrak{m}_n$ is isomorphic to $\mathbf{M}_n(k)$. Show that, when $\mathfrak{g}$ is not isomorphic to $\mathfrak{s}, U(\mathfrak{g})/\mathfrak{m}_n$ is isomorphic to $\mathbf{M}_n(k)$ if and only if $n$ is odd (use Exerc. $16g$)$).^9$

$^9$ It can be shown that, when $n$ is even, $U(\mathfrak{g})/\mathfrak{m}_n$ is isomorphic to $\mathbf{M}_{n/2}(A)$.