Let a ring $R$ with identity element be such that the category of left $R$-modules has a simple generator $T$.
My question: "Is $T$ isomorphic with any simple left $R$-module $M$?"
I tried the meaning of generation by $T$, i.e. there exists an $R$-epimorphism $f$ from a direct sum $⊕T$ to $M$, but to no avail. Also, I know that any simple left $R$-module is isomorphic with $R/K$ for some maximal left ideal $K$ of $R$.
On
If I understand your question correctly, then the question is whether it follows that all simple modules are isomorphic provided that one has a simple module which is a generator.
I think the answer is yes:
Suppose $S$ is another simple module. Then there is an epi $T^{(I)}\twoheadrightarrow S$.
However, one has the following isomorphism (using universal property of direct sum, basically): $$\mathrm{Hom}_R(T^{(I)}, S)\simeq \mathrm{Hom}_R(T, S)^I.$$
However, it is easy to observe that for simples $T, S$, $\mathrm{Hom}_R(T, S) \neq 0$ iff $T \simeq S$ (take nonzero morphism $T \rightarrow S$ and observe that it is an isomorphism: $\mathrm{Ker}\,f, \mathrm{Im}\,f$ are sumodules of simples $T, S$, hence either $0$ or the whole module).
Thus, since $\mathrm{Hom}_R(T^{(I)}, S) \neq 0$, it follows that $\mathrm{Hom}_R(T, S) \neq 0$ and $T$ and $S$ are isomorphic.
Does this not just follow from the properties of semisimple modules?
If $S$ and $M$ are simple, and $(\oplus_{i\in I} S)/K\cong M$, then $M'\oplus K=(\oplus_{i\in I} S)$ where $M'\cong M$. So $M$ must be isomorphic to a direct sum of copies of $S$, but since it is simple it is just isomorphic to $S$.