0
$\begingroup$

Let $R$ be a ring with identity, and $e^2=e\in R$ such that both $eJe$ and $(1-e)J(1-e)$ are nil, where $J=J(R)$ is the Jacobson radical of $R$. When $R$ is commutative, it is easy to see that $J$ is nil. Indeed, the sum $ej+(1-e)j=j$ is nilpotent, for every $j\in J$ in the commutative setting. I guess that if Koethe's Conjecture holds for $R$, then $J$ is nil too. (The conjecture is true for the commutative rings.)

Any suggestion/help is appreciated in advance!

Improve this question
$\endgroup$
4
  • $\begingroup$ Am I guessing right that your question is your "guess"? $\endgroup$
    – rschwieb
    Commented Oct 9, 2017 at 13:45
  • $\begingroup$ @rschwieb: I do not understand the tenor of your comment. What a "doubt" does there exist about the "guess"? Is it presented before by another person? $\endgroup$
    – karparvar
    Commented Oct 9, 2017 at 16:33
  • $\begingroup$ You have written a paragraph of statements without any questions, and I just wanted to make sure I was thinking about your intended problem, that is all. Sometimes questions come out not as they were intended to, and I'm just playing it safe by asking. $\endgroup$
    – rschwieb
    Commented Oct 9, 2017 at 17:08
  • $\begingroup$ @rschwieb: Yes, I meant if Koethe Conjecture holds, could we deduce the assertion I raised, i.e., is the nilness of $eJe$ and $(1-e)J(1-e)$ yield that of $J$? $\endgroup$
    – karparvar
    Commented Oct 14, 2017 at 17:22

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .