I'm trying to show that when we decompose a semisimple ring $R$ into isotypic components $$ R \overset{_R\mathsf{Mod}}{\cong}\bigoplus_{j=1}^{k_1}{I^{(1)}_j} \bigoplus \dotsb \bigoplus \left( \bigoplus_{j=1}^{k_n}I_j^{(n)} \right) \\ = R^{(1)} \oplus \dotsb \oplus R^{(n)} $$ we have actually decomposed it into the simple two-sided ideals. (I.e. the simple two-sided ideals of $R$ are precisely the sums of all simple left ideals of the same isomorphism type.)
I understand why the isotypic components are two-sided ideals. To show that any two-sided ideal is of this form, I was looking on this page, and their explanation is
...characteristic left ideals are precisely two-sided ideals. Thus two-sided ideals are sums of isotypic components.
I don't understand what this means, nor do I know what a "characteristic ideal" is (is it an ideal $I$ such that any $R$ left-linear automorphism acts trivially on $I$?). Can someone help, or just point me to another proof of what I'm trying to prove?
Remember that given $\varphi \in \text{End}_{_R\textsf{Mod}}(R)$, there is some $a \in R$ such that $\varphi$ is just right multiplication by $a$, because $\varphi(r) = \varphi(r \cdot 1)= r \cdot \varphi(1)$.
If $M \in {_R\textsf{Mod}}$, define $N$ to be a characteristic submodule if it is preserved as a set by every (left) endomorphism of $M$. A characteristic left ideal is defined the same.
Proposition 1: Let $I \subset R$ be a left ideal. Then $I$ is two-sided iff $I$ is a characteristic left ideal of $R$.
Proof: $\text{End}_{_R\textsf{Mod}}(R)$ is exactly those maps consisting of right multiplication by some element of $R$. $\square$
Proposition 2: Suppose $M$ is a left module over a semisimple ring. Then $N < M$ is characteristic iff $N$ is a sum of entire isomorphism classes of simple submodules of $M$.
Proof: Write $M = N \oplus N'$ by semisimplicity, and split $N$ and $N'$ into finitely many simple submodules. $N$ is a sum of entire isomorphism classes of simple submodules of $M$ iff for all $i$ the canonical projection onto the $i$th isotypic component (AKA multiplication by $R^{(i)}$) is zero either on $N$ or on $N'$. Suppose that $S_1, S_2$ are isomorphic simple submodules such that $S_i$ appears in the sum for $N$ and $S_2$ appears in the sum for $N'$. Then there is an endomorphism which swaps them, and so does not preserve $N$ as a set.
Conversely, it is clear that an isotypic component is characteristic, since a map between simple submodules must either be zero or an isomorphism. $\,\,\square$
Corollary: If $R$ is semisimple, $I$ is a simple two-sided ideal iff $I$ is an isotypic component of left ideals.