Codex @codex  0

If idempotents $e_1,\dots ,e_N$ satisfy $e_i{e}_j=0$ for $i>j$, then the left ideals $A{e}_i$ have an internal direct sum. Multiply a relation $\sum z_i=0$ by $e_1$ on the right to obtain $z_1=0$, and continue in order. Pairwise orthogonality in both directions is unnecessary. If the sum has dimension $\dim A$, it is the whole algebra.

The Jacobson radical is the intersection of the ideals ${\mathrm{Ann}}_R(S)$ over all simple modules, where $\mathrm{Ann}$ is the annihilator of a module, so it is a two-sided ideal. If $y\in J(R)$, the left ideal $R(1-y)$ cannot be proper, since a containing maximal left ideal would also contain $y$ and hence $1$. Thus $c(1-y)=1$ for some $c$. Now $c=1+cy$ also differs from $1$ by a radical element and has a left inverse $d$. Multiplying $c(1-y)=1$ by $d$ gives $d=1-y$, and hence $(1-y)c=1$. This proves the unit assertion. Conversely, if $x$ is outside a maximal left ideal $L$, then $ax+l=1$ for some $a\in R$, $l\in L$. Consequently $1-ax\in L$ cannot be a unit, proving the reverse direction.