Prove that a left semi-simple ring $R$ is both left noetherian and left-artinian.
I am following the proof given in pg 27,A first course in non-commutative rings (T.Y.Lam). Its strategy is to show that R has a finite composition series from which the result will follow.
It starts out by observing :-
$R=\bigoplus_{i \in I} I_{i}$ where each $I_{i}$ is simple and hence a minimal left ideal in $R$. Then it says that $1$ belongs to $R$,hence it is easily seen to be FINITE direct sum. I am not able to figure out this easy part.
An infinite direct sum of nonzero modules is never finitely generated. Since $R$, as a left module over itself, is finitely generated, the result follows.
However, in this case it's even easier: you can write $1=\sum_{\lambda\in\Lambda}x_\lambda$, with $x_\lambda\in I_\lambda$ (I changed slightly the notation, with $\Lambda$ as index set) and all but a finite number of $x_\lambda$ different from $0$. Then, for $r\in R$, $$ r=r1=\sum_{\lambda\in\Lambda}rx_\lambda $$ which shows that $r\in\bigoplus_{\lambda\in\Lambda_0}I_\lambda$, where $\Lambda_0=\{\lambda\in\Lambda:x_\lambda\ne0\}$.
Thus $\bigoplus_{\lambda\in\Lambda}I_\lambda\subseteq\bigoplus_{\lambda\in\Lambda_0}I_\lambda$ and we are done.