# Prove: In a semisimple ring with a minimal left ideal decomposition, sums of isotypic components are direct.

This problem comes from Lam's A First Course in Non-commutative Rings, the argument after Lemma 3.9. (But I will explain all of the notation.)

$$R$$ is a ring with $$1$$, but not necessarily commutative. A minimal left ideal is a non-zero left ideal not containing any other non-zero left ideal strictly. If $$A$$ is a minimal left ideal, then we define $$B_A$$ to be the sum of all minimal left ideals which are module isomorphic to $$A$$ as left $$R$$-modules.

Assume $$R$$ is semi-simple. It has been proved previously that if $$R$$ is semi-simple, then it is direct sum of finitely many minimal left ideals $$R=\bigoplus\limits_{i=1}^n{A}_i$$. We may assume for the first $$r$$ ideals $$A_1,\cdots, A_r$$, none of them are module-isomorphic to another, while for $$r+1\le i\le n$$, $$A_i$$ is module-isomorphic to one of $$A_1,\cdots, A_r$$. Let $$B_i\overset{\text{def}}{=}B_{A_i}$$. Then the author claimed $$R=\bigoplus\limits_{i=1}^r B_i$$.

Question: Why this sum is direct?

My attempt: Lemma 3.9 says: (1) $$B_i$$s are two-side ideals. (2) $$B_iB_j=\{0\}$$ for $$i\ne j$$.

I have tried to show that if $$0=b_1+\cdots+b_r\cdots(\star),$$ $$b_r\in B_r$$, then $$b_i=0$$ for every $$i$$. For a fixed $$i$$, multiplying both side of eq$$(\star)$$ (at either side is both acceptable), since lemma 3.9 says $$B_iB_j=0$$ ($$i\ne j$$,) we infer $$b_i^2=0$$. But how does this imply $$b_i=0$$?

The title is not equivalent to the question and is false (there are a lot of minimal left ideals).

If $$1_R = \sum\limits_{i=1}^r \beta_i$$, where $$\beta_i \in B_i$$, then $$\beta_i$$ are unities in the subrings $$B_i$$ (as, multiplying by $$1$$ and expanding, $$b_i = b_i \beta_i = \beta_i b_i$$). Then if $$0 = \sum\limits_{i=1}^r b_i$$, we have that $$0 \beta_j = \sum\limits_{i=1}^r b_i \beta_j = b_j \beta_j = b_j$$ is also $$0$$.

• Thanks for your answer. I have to say the title is ambiguous, but I did not find a concise way to summarize. Dec 2, 2023 at 17:04
• @Asigan One can say "sums of isotypic components are direct" (the term is introduced in Ex 2.8). Dec 2, 2023 at 17:11
• Thanks. I have edited the title. Dec 3, 2023 at 5:55