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It is well known that if a (unital commutative) ring A has only three ideals ({0}, J, A), then the quotient A/J is a field.

But, what can we conclude about A/J if A is not commutative nor unital but has only 3 ideals ({0}, J, A)?

Let me know if the question is poorly written/explained.

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    $\begingroup$ It would of course be simple by the correspondence theorem, but I doubt one could say more. Here's a "natural" example of a noncommutative unital ring with a unique proper ideal (note that the Jacobson radical is 0): Full linear ring of a countable dimensional right vector space. The quotient is $R_{58}$. $\endgroup$
    – Amateur_Algebraist
    Commented Jul 8 at 20:29
  • $\begingroup$ Of course I can improve the question: does someone know a counter-example? I.e., a nonunital, noncommutative ring with only three ideals ({0}, J, A}, such that the quotient A/J is unital? $\endgroup$
    – Felipe
    Commented Jul 8 at 22:16

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I'd just like to point out the "natural unital" example in the comments can be broadened to a natural nonunital example. It uses the fact that the ring of transformations is von Neumann regular.

Let $V$ be a vector space of dimension $\aleph_1$. It's known the ideals of the full ring of linear transformations $R$ are linearly ordered and are exactly

  1. The two trivial ideals
  2. The subset of transformations with images dimension $<\aleph_0$ (Call it $I_1$.)
  3. The subset of transformations with images dimension $<\aleph_1$ (Call it $I_2$.)

Another lemma:

If $C$ is a (possibly nonunital) von Neumann regular ring, and $B$ is an ideal of $C$ also considered as a (possibly nonunital) ring. Then if $A\lhd B$, it is also true that $B\lhd C$. That is, the ideals of $B$ are ideals of $C$.

Using this, we can say that $I_2$ is a rng contained in $R$ whose ideals must also be ideals of $R$, but that leaves only the zero ideal and $I_1$.

$I_2$ cannot have an identity, because such an identity would have to have to be a transformation with image having dimension at least as large as the dimension of $I_1$, but it is easy to see that the dimension of $I_1$ is $\aleph_1$. This is a problem since $I_2$ only contains transformations with image dimension $<$ $\aleph_1$.

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  • $\begingroup$ Amazing response, and now I ask: what can we say about the quotient I2/I1 ? $\endgroup$
    – Felipe
    Commented Jul 10 at 14:17
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    $\begingroup$ @Felipe I'm don't really have anything to say about it. It is an overly broad thing to ask I think. $\endgroup$
    – rschwieb
    Commented Jul 10 at 16:01

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