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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 $b_i=0$?

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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 $0$.

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

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