4
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 $L'$.

$\endgroup$
  • $\begingroup$ Why it cannot happen that $Lx=0$ for all $x\in L'$? $\endgroup$ – Rubertos Mar 13 '17 at 13:20
  • $\begingroup$ Well, the lemma is given in the text to prove that $B_i$'s are simple rings! So we cannot assume that here.. $\endgroup$ – Rubertos Mar 13 '17 at 13:23
  • $\begingroup$ @Rubertos That would imply $LL'=\{0\}$! $\endgroup$ – rschwieb Mar 13 '17 at 13:27
  • $\begingroup$ Yeah I know that. I'm asking why is it $LL'\neq 0$? Actually this is equivalent to my original question.. $\endgroup$ – Rubertos Mar 13 '17 at 13:28
  • 1
    $\begingroup$ @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. $\endgroup$ – rschwieb Mar 13 '17 at 14:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.