Suppose $V^{(n)}$ is the simple $\mathfrak{g}=\mathfrak{sl}_2$-module of dimension $n$, and let $I_n$ be the annihilator, an ideal of the universal enveloping algebra $U(\mathfrak{g})$. Why is $U(I_n)=U(\mathfrak{g})/I_n$ isomorphic to the matrix ring $M_n(\mathbb{C})$? I can identify $U(I_n)$ as a subalgebra of $M_n(\mathbb{C})$, and then $V^{(n)}$ is naturally a simple $n$-dimensional $U(I_n)$-module.
The justification I read is that by Artin-Wedderburn, the only simple finite dimensional complex algebra with a simple $n$-dimensional module is $M_n(\mathbb{C})$, so $M_n(\mathbb{C})$ is a quotient of $U(I_n)$. I think the result then follows just by counting the dimensions as complex vector spaces, to see they're equal. But how does the observation from Artin-Wedderburn imply $M_n(\mathbb{C})$ is a quotient? I don't see any sort of obvious surjective map or anything.
Saying the $\mathfrak{g}$-module $V=V^{(n)}$ is simple is equivalent to saying that the associated representation $$\rho:U(\mathfrak{g})\to\mathrm{End}_{\mathbb{C}}(V)$$ is surjective. Now, $\ker\rho=I_n$, so by the fundamental homomorphism theorem $$ U(\mathfrak{g})/I_n\cong\mathrm{End}_{\mathbb{C}}(V)\cong M_n(\mathbb{C}). $$