(Ex 3.13 Lam) Suppose R is a simple, infinite dimensional algebra over a field k. If V is left R module, then V as vsp over k is infinite dimensional.

Question

My approach: Since R is infinite dimensional over k, therefore the infinite basis are $(r_1,...r_i..)$. Let $r_i.v = w_i\in V$ (Since $V$ is a left $R$ module). Now I prove that $w_i$ are independent. Choose finite $w_i$ and let $c_1.w_1+ ...c_n.w_n = 0$. Then taking $v$ common, we have $(c_1.r_1 + ..c_nr_n).v = 0 \implies (c_1r_1 +...+c_nr_n)=0$. Since by choice $r_i$ are independent, therefore $c_i=0$ $\forall i$. Hence $w_i$ are all independent.

Nowhere in the proof I have used the fact that $R$ is simple, so I suspect my proof is wrong. Can someone help me criticize my proof and also provide some hint?

UPDATE: I think not all $c_i= 0$ is implied because there can exist zero divisior of $r_i$. Is this a valid objection

Answer 1

Note: it must be assumed that $V$ is a nonzero module.

Suppose $V$ is a (nonzero) left $R$-module. Consider the map $$ R\to\operatorname{End}_k(V) $$ defined by the left multiplications, which is a $k$-algebra homomorphism and is injective because $R$ is simple. What would be the consequence if $V$ happens to be finite dimensional over $k$?