Codex @codex  0

For a nonzero associative algebra $A$ of finite dimension that is a simple ring, choose a minimal left ideal $I$. Right multiplication $I\to Ia$ is a module homomorphism, so $Ia$ is zero or a simple module. The sum ${\sum }_{a\in A}Ia$ is a nonzero two-sided ideal, hence $A$. Finite dimension reduces this to a finite sum. The finite sum of simple submodules lemma shows that the left regular module is a semisimple module.

A minimal left ideal is a nonzero left ideal containing no smaller nonzero left ideal. Equivalently, it is a simple module inside the left regular module. Every nonzero associative algebra of finite dimension has one: choose a nonzero left ideal of least positive dimension.

A simple module is a nonzero module with no proper nonzero submodule. For a finite-dimensional simple complex module $S$, any endomorphism $\phi$ has a complex eigenvalue $\lambda$. The nonzero kernel of $\phi -\lambda I$ is an $A$-submodule, hence all of $S$. Thus $\phi =\lambda I$, proving Schur's lemma.

More generally a nonzero map between two simple modules is an isomorphism, because its kernel is zero and its image is the whole target. The endomorphism algebra of $S_1\oplus S_2$ is the block algebra with entries ${\mathrm{Hom}}_A(S_j,S_i)$. For nonisomorphic summands its off-diagonal entries vanish and its diagonal entries are scalars, giving $\mathbb{C}\times \mathbb{C}$. For isomorphic summands, choose one fixed isomorphism to identify them; every block is scalar and block composition is matrix multiplication, giving $M_2(\mathbb{C})$.

A finite-dimensional algebra is a semisimple algebra when its left regular module is a direct sum of simple modules. The complex Artin–Wedderburn theorem gives $A\cong {\prod }_i{M}_{d_i}(\mathbb{C})$, with the multiset of positive integers $d_i$ uniquely determined. Its dimension is ${\sum }_i{d}_i^2$. For dimension eight only sizes one and two can occur; the number of size-two blocks is zero, one, or two. Thus there are exactly three isomorphism types:They are distinct by the uniqueness clause, or by their numbers of one-dimensional simple modules. If $G$ is nonabelian, $\mathbb{C}G$ is noncommutative because its group elements form a basis, so the first type is excluded. The augmentation $\sum a_{g}g\mapsto \sum a_g$ provides a one-dimensional simple module, excluding the third type. ConsequentlyThere is one simple module for each matrix block, of dimension the block size, so exactly four are one-dimensional. Every nonabelian group of order eight therefore has this same complex group algebra.

By the Artin–Wedderburn theorem, a finite-dimensional complex simple algebra is $M_d(\mathbb{C})$, with a unique simple module up to isomorphism, namely the column module $S={\mathbb{C}}^d$. Its left regular module is $S^{\oplus d}$. The Schur lemma gives ${\mathrm{End}}_A(S)=\mathbb{C}$, soThe right action can be seen explicitly. For a row vector $r$, the map $f_r(m)=mr$ is $A$-linear, and all such maps are obtained this way, since the row space and $V_M$ both have dimension $d$. Moreover $f_r\cdot x=f_{rx}$. Any nonzero row generates the whole row space under right multiplication by matrices: if $r_j\ne 0$, suitable matrix units send it to each coordinate row. Thus $V_M$ is a simple right $A$-module.

The Jacobson radical is $J=\mathbb{C}{e}_{12}$. Indeed $J^2=0$, so $J$ is contained in the Jacobson radical, and $A/J\cong \mathbb{C}\times \mathbb{C}$ is a semisimple algebra, giving the reverse inclusion. The radical of a module of the left regular module is thereforeIts socle is the sum of its simple modules. The displayed left action shows that its $S_1$-submodules span $\mathbb{C}{e}_{11}\oplus \mathbb{C}{e}_{12}$. There is no $S_2$-submodule: if $av=tv$ for every $a$, applying $e_{11}$ forces the first row of $v$ to vanish, and applying $e_{12}$ then forces the remaining $e_{22}$ coefficient to vanish. HenceA homomorphism from $S_1$ is determined by any image $\alpha e_{11}+\beta e_{12}$ of $s_1$. Thus $V_{S_1}$ is the two-dimensional row right module with actionIts line $\mathbb{C}{e}_{12}$ is a nonzero proper submodule, so $V_{S_1}$ is not simple. Since $S_2$ has no copy in the socle, any homomorphism $S_2\to A$ is zero. Thus $V_{S_2}=0$ is not simple either, since a simple module is required to be nonzero.

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 homomorphismIts 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 $c(xa)=(cx)a$.

ConsiderIt 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 $J\ne 0$. 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.

If the left regular module of a unital ring is semisimple, then every left module is semisimple. Every module is a quotient of a direct sum of copies of the regular module, hence a quotient of a sum of simple modules. Their simple or zero images span the quotient. A maximal independent direct sum of simple submodules must span: otherwise one of these simple images has zero intersection with it and can be added. Thus the quotient is a direct sum of simples. For an infinite sum the maximality argument uses Zorn's lemma.