$R ≅ R/I$ prove that for any two-sided ideals $A$ and $B$ we have $A⊆B $ or $B⊆A$

Question

Assume that $R$ is a ring such that for any two-sided ideal $I$ of $R$ we have $R ≅ R/I$. Prove that for any two-sided ideals $A$ and $B$ we have $A⊆B$ or $B⊆A$ .

I try to solve with proof by contradiction and this property that $A∩B$ is also an ideal but i could not find any thing useful.

Answer 1

Let me state a more general problem and then explain the solution.

Problem. Show that if $A$ is a nontrivial algebraic structure that is isomorphic to each of its nontrivial quotients, then the lattice of congruences on $A$ is a well order.

[Here $A$ is nontrivial means $|A|>1$. A congruence on $A$ is a kernel of a homomorphism.]

[Before giving the solution, here is the terminology: an algebraic structure is simple if its lattice of congruences is a 2-element chain. It is pseudosimple if it is not simple, but it is isomorphic to each of its nontrivial quotients. So the problem is to show that if an algebra is simple or pseudosimple, then its lattice of congruences is a well order.]

Solution. Choose $a\neq b$ in $A$ and let $\alpha$ be a congruence on $A$ maximal for separating these elements. Then $A/\alpha$ is a nontrivial quotient of $A$, since it contains the distinct elements $a/\alpha$ and $b/\alpha$, hence $A\cong A/\alpha$. By the maximality of $\alpha$, $A/\alpha$ has a smallest nonzero congruence, namely the one that identifies $a/\alpha$ and $b/\alpha$. This implies that the zero congruence of $A/\alpha$ is completely meet irreducible. Since $A\cong A/\alpha$, the zero congruence of $A$ is also completely meet irreducible. Since $A\cong A/\theta$ whenever $|A/\theta|>1$, it follows that the zero congruence on any nontrivial quotient of $A$ is completely meet irreducible. By the Correspondence Theorem, every proper congruence $\theta$ on $A$ is completely meet irreducible.

This already implies that the lattice of congruences of $A$ is a well order. First, it must be a linear order, since if it had incomparable elements $\beta$ and $\gamma$, then $\theta=\beta\cap\gamma$ would be a congruence that is not completely meet reducible. Next, it has DCC, since if $(\delta_n)_{n\in\omega}$ is a strictly decreasing $\omega$-chain in the lattice of congruences, then $\theta=\bigcap_{n\in\omega} \delta_n$ is not completely meet irreducible.

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Let me respond to the questions of Batominovski:

1. (Would $A\cap B$ be the maximal congruence $C$ that separates $A$ and $B$?) No, we are separating elements of $R$, not congruences/ideals of $R$. In ring notation, choose $a\neq 0$ in $R$, and then let $A$ be an ideal of $R$ that is maximal for the property that $a\notin A$. Such $A$ exists by Zorn's Lemma. [Now I am taking $R$ for my algebra, $a$ and $0$ for my two elements to be separated, and congruence modulo $A$ for the separating congruence.]

2. (How do we show that the zero congruence of $R$ is completely meet irreducible?) It suffices to show that $R$ has a nontrivial quotient that has completely meet irreducible zero congruence, since $R$ is isomorphic to any such ring. So take a subdirectly irreducible quotient. This is what is being described in the first 2-3 sentences of the Solution. If $a\neq 0$ in $R$ and $A$ is maximal in $R$ for the property $a\notin A$, then $R/A$ is nontrivial, since $\bar{a}\neq \bar{0}$ in this quotient, yet any nonzero ideal of $R/A$ contains $\bar{a}$ by the maximality of $A$. Thus the complete meet of nonzero ideals of $R/A$ also contains $\bar{a}$, which means this complete meet is not zero. This shows that the zero ideal of $R/A$ is completely meet irreducible.

3. (Do you know any ring $R$ with the required property?) As Tsemo Aristide pointed out, if $R$ has a maximal ideal and the required property, then $R$ must be simple, not pseudosimple. This will happen if $R$ has an identity element, as rschwieb has already mentioned. Hence any simple ring has the required property, and these are the only examples when $R$ has a maximal ideal. As for pseudosimple rings, I do not know if there are any noncommutative examples. But it is a not-too-difficult exercise to show that a commutative pseudosimple ring must be a ring with zero multiplication and pseudosimple underlying additive group, and therefore any commutative pseudosimple ring is isomorphic to a Prufer group $\mathbb Z_{p^\infty}$ equipped with zero multiplication.

The pseudosimple commutative semigroups are classified in

Boris M. Schein, Pseudosimple commutative semigroups, Monatshefte für Mathematik March 1981, Volume 91, Issue 1, pp 77-78.

and pseudosimple lattices with congruence lattices of every allowable order type are constructed in

Graetzer, G.; Schmidt, E. T. Two notes on lattice-congruences. Ann. Univ. Sci. Budapest. Eotvos. Sect. Math. 1 1958 83-87.

Answer 2

Let $A$ be an be ideal, we suppose that $A$ is a maximal ideal. We denote by $p_A:R\rightarrow R/A$ the quotient map. There exists an isomorphism $g:R/A\rightarrow R$, $p_A\circ g:R/A\rightarrow R/A$ is a morphism of rings whose kernel contains $g^{-1}(A)$. Since $A$ is maximal, $g^{-1}(A)=0$. We deduce that $A=0$ since $g$ is an isomorphism. This implies that the only non trivial ideal of $R$ is $\{0\}$.