= Solution
Since $A$ is a nonzero <associative algebra> of finite <dimension>, it has a nonzero <left ideal> $I$ of least positive <dimension>. Such a <minimal left ideal> is a <simple module> under left multiplication.
For $a\in A$, right multiplication defines the <module homomorphism>
$$
R_a:I\longrightarrow A,\qquad x\longmapsto xa.
$$
Its <kernel> is a <submodule> of $I$, so its <image> $Ia$ is either zero or a <simple module> <isomorphic> to $I$. Each $Ia$ is a <left ideal>, since $c(xa)=(cx)a$.
Consider
$$
J=\sum_{a\in A}Ia.
$$
It is a <left ideal>. It is also stable under right multiplication, because $(Ia)b=I(ab)$, so it is a <two-sided ideal>. Moreover $I=I1\subseteq J$, giving $J\ne0$. The assumption that $A$ is a <simple ring> therefore forces $J=A$.
Only finitely many summands are needed: express each member of a finite <basis> of $A$ as a finite sum of elements of the $Ia$, and collect the finitely many indices used. After discarding zero summands, $A$ is a finite sum of <simple modules>. Part (a)(ii) makes its <left regular module> a <semisimple module>. This is exactly the definition of a <semisimple algebra>, proving that a <finite-dimensional simple algebra is semisimple>. The finite <dimension> assumption is essential to this assertion about a <simple ring>.
Back to article page
