Proposition: For a ring $R$ the following statements are equivalent:
(a) $R$ has a simple left generator;
(b) $R$ is simple left artinian;
(c) For some simple $_RT, _RR \cong T^{(n)}$ for some $n$;
(d) $R$ is simple and $_RR$ is semisimple.
My try to understand above proof:
$(a) \Longleftrightarrow (c)$: OK
$(a) \Longrightarrow (d)$: Please explain why $_RT$ is faithful. If i understand it, clearly $I \leq l_R(R/L)=l_R(T)=0$
$(d) \Longrightarrow (b)$: Since $_RR$ is semisimple and finite generate so $_RR= \oplus_{i=1} ^n T_i, T_i$ is simple. So, $_RR$ has composition series of length n:
$0=T_0\leq T_1\leq T_1\oplus T_2 \leq ... \leq \oplus_{i=1} ^n T_i = _RR$
Then $R$ is simple left artinian.
$(b) \Longrightarrow (a)$
- First: If $R$ is left artinian, then $R$ has a minimal non-zero left ideal $T$: OK.
- Please explain: $Tr_R(T) \neq 0$
Thanks for all!
$(a)\implies (d)$ Generators have to be faithful. There exists $I$ and a surjection of $R$ modules $\oplus_I T\twoheadrightarrow R$. What would be a consequence if $T$ wasn't faithful?
$(d)\implies (b)$ No question?
$(b)\implies (a)$ The trace of $T$ in $R$ is the largest submodule of $R$ generated by $T$. You just got done saying that there is a copy of $T$ in $R$, so the trace of $T$ in $R$ is certainly larger than that...