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This is a problem I asked in SE, but it seems the question is more suitable for MO.

Consider a ring $R$ (not necessary with identity or commutative) such that for any proper two-sided ideal $I$, $R\cong\frac RI$ (as rings), e.g., $\Bbb{Z}_{2^\infty}$ with zero product. Is it true that the set of two-sided ideals of $R$ is a chain ?!

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If there are ideals $I_1,I_2$ not forming a chain, we can assume that $I_1,I_2\ne0=I_1\cap I_2$ just passing to $R\simeq R/(I_1\cap I_2)$. Pick any $0\ne r\in R$. By the Zorn lemma, there is a maximal ideal $I$ in the set of the ideals with the property $r\notin I$. Passing to $R\simeq R/I$, we can assume that every nonnull ideal of $R$ contains $r$. So, $r\in I_1\cap I_2=0$. A contradiction. Therefore, the ideals of $R$ form a chain.

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  • $\begingroup$ Why if $r\neq0$ the Zorn lemma implies that there is a maximal ideal I such that $r\notin I$?, if this is is true then the Jacobson radical is always zero. $\endgroup$
    – Murphy
    Commented Jan 13, 2014 at 4:03
  • $\begingroup$ @Murphy $I$ is not maximal in the absolute sense. It is maximal with the property $r\notin I$. I will edit my answer. $\endgroup$
    – Sasha Anan'in
    Commented Jan 13, 2014 at 11:56

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