Why is $L'=Lb'$?

I'm reading Rotman- Advanced Modern Algebra p.566, and I got stuck in the proof for Lemma 8.61-(iii)

Here are some lemmas the author introduced to prove lemma 8.61: (Here, all rings are assumed to have a unity)

Lemma 8.53 -(i)

Let $R$ be a ring and $I,J$ be minimal left-ideals of $R$. If $I^2\neq 0$ and $I\cong J$ as $R$-modules, then $IJ\neq 0$.

Lemma 8.53 -(ii)

Let $R$ be a ring and $I,J$ be minimal left-ideals of $R$ such that $IJ\neq 0$. Then, $Hom_R(I,J)\neq 0$ and there exists $b'\in J$ such thay $J=Ib'$.

Now here is a part of the proof for Lemma8.61-(iii) where I got stuck:

Let $R$ be a semisimple ring and $B_1,...,B_n$ be the simple components of $R$. Let $D$ be a nonzero two-sided ideal in $R$. Since $R$ is semisimple, $R$ is Artinian. Thus, there exists a minimal left-ideal $L$ of $R$ such that $L\subset D$. Since $L$ is minimal, $L\subset B_i$ for some $i$. We claim that $B_i\subset D$. Let $L'$ be another minimal-left ideal of $R$ such that $L'\subset B_i$. (Note that $L\cong L'$ as $R$-modules in this case) Then, there exists $b'\in L'$ such that $L'=Lb'$.

I cannot figure out how this last sentence holds true. Using lemma 8.53-(i),(ii), it suffices to prove that $L^2\neq 0$. But how?

Of course $L^2\neq 0$, because it is a nonzero left ideal of a simple ring. (The left annihilator of $L$ in $B_i$ is an ideal of $B_i$.)

$LL'$ is a nonzero submodule of $L'$. It cannot be that $Lx=\{0\}$ for all $x\in L'$. Therefore there exists an $x$ such that $Lx$ is nonzero. Since it is a nonzero left ideal contained in $L'$, it is equal to $L'$.

• Why it cannot happen that $Lx=0$ for all $x\in L'$? – Rubertos Mar 13 '17 at 13:20
• Well, the lemma is given in the text to prove that $B_i$'s are simple rings! So we cannot assume that here.. – Rubertos Mar 13 '17 at 13:23
• @Rubertos That would imply $LL'=\{0\}$! – rschwieb Mar 13 '17 at 13:27
• Yeah I know that. I'm asking why is it $LL'\neq 0$? Actually this is equivalent to my original question.. – Rubertos Mar 13 '17 at 13:28
• @Rubertos Well, it is a bit unfair to declare my work circular when you think there is circularity is in Rotman's presentation. I don't have very good access to the text (just what I can find online). The simplicity of $B_i$ is extremely elementary: for a simple submodule $S$, $End(\oplus_{i-1}^n (S))\cong M_n(End(S))$, and the $End_S$ are division rings by Schur's lemma. The proof that $M_n(D)$ is simple is well-known and does not have any dependencies. – rschwieb Mar 13 '17 at 14:05