 Solution

The Jacobson radical $J=J(R)$ is the intersection of all maximal right ideals, equivalently the largest ideal annihilating every simple right module. The Artin–Wedderburn theorem givesbecause $k$ is algebraically closed.

The descending chain $J\supseteq J^2\supseteq \cdots$ stabilizes since $R$ is finite-dimensional. If $J^m=J^{m+1}$, Nakayama lemma applied to the finite right module $J^m$ gives $J^m=0$. Thus $J$ is nilpotent.

For $f\in {\mathrm{End}}_R(P_i)$, the Fitting lemma givesfor large $n$. Indecomposability makes one summand zero, so $f$ is either invertible or nilpotent. In the latter case $1-f$ is invertible. This is the criterion that ${\mathrm{End}}_R(P_i)$ is a local ring.

Let $M={⨁}_i{P}_i$. Reduction modulo $MJ$ definesSince $M/MJ={⨁}_i{S}_i$ and the $S_i$ are pairwise nonisomorphic simples,by Schur lemma. Arbitrary scalars on the direct summands lift to scalar identity maps on the $P_i$, so $\theta$ is surjective.

If $f\in \ker \theta$, then $f(M)\subseteq MJ$. A product of $r$ such maps sends $M$ into $M{J}^r$, so $\ker \theta$ is nilpotent. A nilpotent ideal lies in the Jacobson radical, while the semisimplicity of the quotient $k^n$ gives the reverse inclusion. HenceThis is exactly the definition of a basic algebra.