Why is the left-multiplication map a surjection in this context?

Question

I will be awarding a bounty for this question as soon as I can, because this is a problem that is bothering me for quite a while now.

Here is a theorem that I know is true.

Let $S$ be a simple ring and $L$ be a non-zero left ideal of $S$. Consider $D:= \operatorname{End}({ }_S L)$. Note that $D$ acts on $L$ via evaluation. Then the natural map

$$f: S \to \operatorname{End}_D(L_D): r \mapsto (f_r: L \to L= l \mapsto rl)$$

is a ring isomorphism.

and I want to prove the following:

Let $R$ be a $k$-algebra where $k$ is an algebraically closed field. Suppose that $V:= k^n$ is a simple left $R$-module. Then the map

$$f: R \to \operatorname{End}(V_k): r \mapsto (f_r : V \to V: v \mapsto r.v)$$

is a surjection.

In my notes, a short proof is sketched. I will post it here:

Consider the ringhomomorphism $f: R \to \operatorname{End}(V_k): r \mapsto f_r$. This induces a well-defined ring monomorphism

$$\overline{R}:=R/\ker f \hookrightarrow \operatorname{End}(V_k): \overline{r} \mapsto f_r$$

Note that $\overline{R}$ is finite-dimensional as $k$-algebra, because $\overline{R}$ embeds in $\operatorname{End}(V_k) \cong M_n(k)$. Thus $\overline{R}$ is Artinian, and hence the quotient ring $\overline{\overline{R}}:=\overline{R}/J(\overline{R})$ ($J$ is the jacobson radical) is Artinian as well. Thus $\overline{\overline{R}}$ is a semisimple ring.

By Wedderburn-Artin, there are simple rings $B_1, \dots, B_m$ such that $$\overline{\overline{R}} \cong B_1 \times \dots \times B_m$$

Now, note that $V$ is a simple left $\overline{R}$ -module for the action induced by the action of $R$ on $V$. Since $V$ is simple, we have $J(\overline{R})V = 0$ and hence we can view $V$ as a simple $\overline{\overline{R}}$-module, essentially with an equivalent module structure as the one induced by the $R$-action on $V$.

Going further, we can view $V$ as a simple $B_1 \times \dots \times B_m$-module, and we get

$$V= (B_1 \times \dots \times B_m) V = B_1 V \oplus \dots \oplus B_m V$$

where we have identified $B_i$ with natural ideal of $B_1 \times \dots \times B_m$. Since $V$ is simple, there is a unique $i \in \{1, \dots, m\}$ such that

$$B_i V \ne 0; \quad V = B_i V$$

Employing the theorem above, we get an isomorphism

$$h:B_i \to \operatorname{End}(V_k): b \mapsto h_{b}$$

Composing with the obvious surjections

$$R \to \overline{R} \to \overline{\overline{R}} \to B_1 \times \dots \times B_m \to B_i \to \operatorname{End}(V_k)$$

the result follows. $\quad \square$

Question:

How exactly is the theorem above applied to get that $$h:B_i \to \operatorname{End}(V_k): b \mapsto h_{b}$$ is an isomorphism (and thus a surjection)?

I.e. what ideal $L$ is used? It seems that more is going on because I really can't understand this step.

For anyone trying to help, thousand times thanks!

Answer 1

You're right: a bit more is going on here. Let's denote $B:=B_i$ for the sake of brevity.

As $V$ is simple, we can choose a nonzero element $v_1\in V$ such that $B\cdot v_1 = V$ and then extend it to a $k$-basis $v_1,\dots,v_n$ of $V$. Then we can consider the chain of left ideals $$ L_m:= \{b\in B\mid b\cdot v_j =0,\quad j=2,\dots,m\},\quad m\leqslant n. $$ Obviously, $L_1=B$. If now for some $m\leqslant n$ we have $L_m\neq 0$, but $L_{m+1}=0$, we get that $b\cdot v_{m+1} \neq 0$ for all $b\in L_m$, and so the map $\mu\colon L_m\to V$, $b\mapsto b\cdot v_{m+1}$ is

• injective, because if $b\in L_m$ and $b\cdot v_{m+1}=0$, then $b\in L_{m+1}=0$;
• surjective, because $L_{m}\cdot v_{m+1}$ is a non-zero $B$-submodule of $V$, and the latter is simple;
• a map of $B$-modules by construction.

Thus, $\mu$ an isomorphism of $B$-modules, and we can apply the theorem to $L=L_m$.