Dimensionality theorem irreps finite dimensional algebra

If $A$ is any ring (and in particular any finite-dimensional complex algebra) then the regular representation of $A$ on itself by left multiplication has submodules consisting of exactly the left ideals, more or less by definition. It is completely reducible iff $A$ is semisimple, again more or less by definition, and for $A$ finite-dimensional over a field (or more generally for $A$ Artinian) this occurs iff $A$ has vanishing Jacobson radical.

If $A$ is Artinian but does not have vanishing Jacobson radical then $A/J(A)$ is semisimple and is the maximal quotient with this property, and by the Artin-Wedderburn theorem it is a finite direct product of matrix rings $M_n(D)$ over division rings. In particular if $A$ is finite-dimensional over a field $k$ then $A/J(A)$ is a finite direct product of matrix rings over division $k$-algebras, so if $k$ is algebraically closed $A/J(A)$ is a finite direct product of matrix rings $M_n(k)$ over $k$.

In the case of semisimple algebras over an algebraically closed field $k$, each $d$-dimensional simple module $S$ corresponds to a matrix ring factor $M_d(k) = text{End}_k(S)$ which, in terms of the left action of $A$ on itself consists of $d$ copies of $S$. A conceptual way to see this is that when $A$ is a semisimple module then the multiplicity of $S$ in $A$ is $dim_k text{Hom}_A(A, S)$, and we have $text{Hom}_A(A, S) cong S$ since $A$ is the free $A$-module. More generally, if $A$ is a semisimple ring and $S$ a simple module with endomorphism ring $text{End}_A(S) cong D$, corresponding to a matrix ring factor $M_d(D)$ in $A$, then the multiplicity of $S$ in $A$ is

$$dim_D text{Hom}_A(A, S) = dim_D S = d.$$

So $S$ still occurs with multiplicity equal to its dimension but its dimension needs to be understood as a module over its endomorphism ring $D$ (which is a division ring by Schur's lemma).