The central reduction of the enveloping algebra at a central character

Let $\mathfrak{g}$ be a complex Lie algebra, let $Z(U(\mathfrak{g}))$ be the center of its enveloping algebra (The universal enveloping algebra as a tensor quotient), and let $\chi \text{\hspace{0.17em}⁣}:Z(U(\mathfrak{g}))\to \mathbb{C}$ be a unital $\mathbb{C}$-algebra homomorphism, that is, a central character (Central character of a Lie algebra module). Write $\ker \chi =\{z\in Z(U(\mathfrak{g})):\chi (z)=0\}$ and let

$U(\mathfrak{g})\ker \chi =\{{\sum }_{k=1}^m{u}_k{z}_k:m\ge 0,u_k\in U(\mathfrak{g}),z_k\in \ker \chi \}$

be the two-sided ideal of $U(\mathfrak{g})$ generated by $\ker \chi$ (The sum $I+J$ and product $IJ$ of two-sided ideals, The sum and product of two-sided ideals are two-sided ideals). The central reduction of $U(\mathfrak{g})$ at $\chi$ is the quotient algebra

$U_{\chi }:=U(\mathfrak{g})/U(\mathfrak{g})\ker \chi \mathrm{.}$

It is a unital associative $\mathbb{C}$-algebra, the natural map $U(\mathfrak{g})\to U_{\chi }$ is a surjective algebra homomorphism with kernel $U(\mathfrak{g})\ker \chi$, the image of $Z(U(\mathfrak{g}))$ in $U_{\chi }$ is $\mathbb{C}\cdot 1$, and for a $U(\mathfrak{g})$-module $M$ the following are equivalent:

• (i) $M$ has central character $\chi$, i.e. every $z\in Z(U(\mathfrak{g}))$ acts on $M$ by the scalar $\chi (z)$;
• (ii) $\ker \chi$ annihilates $M$;
• (iii) the action of $U(\mathfrak{g})$ on $M$ factors uniquely through $U_{\chi }$.

Thus the $U(\mathfrak{g})$-modules with central character $\chi$ are exactly the $U_{\chi }$-modules.

• The generating set is central, so the ideal is two-sided. Every $z\in \ker \chi$ commutes with $U(\mathfrak{g})$, hence $u_k{z}_k=z_k{u}_k$ and the set displayed above is already closed under left and right multiplication; it is an additive subgroup because finite sums of such terms are such terms, and it contains $0$ as the empty sum. This is the ideal generated by $\ker \chi$ (Left, right and two-sided ideals).
• Every central element is scalar modulo the ideal. For $z\in Z(U(\mathfrak{g}))$ one has $z-\chi (z)\cdot 1\in \ker \chi \subseteq U(\mathfrak{g})\ker \chi$, so $z$ and the scalar $\chi (z)$ have the same image in $U_{\chi }$. Thus the image of the center consists of scalar classes; this does not imply that every element of $U_{\chi }$ is scalar.
• The equivalences. Condition (i) says $z$ acts by $\chi (z)$ for every central $z$, which is equivalent to $(z-\chi (z))m=0$ for all such $z$ and all $m$, hence to $\ker \chi$ annihilating $M$ because $\ker \chi$ consists exactly of the elements $z$ with $\chi (z)=0$. If $\rho \text{\hspace{0.17em}⁣}:U(\mathfrak{g})\to {\mathrm{End}}_{\mathbb{C}}(M)$ denotes the action, then (ii) says $\ker \chi \subseteq \ker \rho$, so $U(\mathfrak{g})\ker \chi \subseteq \ker \rho$ because $\ker \rho$ is a two-sided ideal; the quotient universal property (A ring homomorphism whose kernel contains a two-sided ideal factors uniquely through the quotient ring) then produces a unique factorization of $\rho$ through $U_{\chi }$, which is (iii). Conversely a factorization through $U_{\chi }$ kills $U(\mathfrak{g})\ker \chi$ and hence $\ker \chi$, which is (ii).