# Simple rings/algebras fail to be division rings/algebras because of zero-divisors?

Is the following statement true/false?

For a non-trivial ring $$R$$, and a maximal modular ideal $$M\subset R$$, if $$M$$ is not one-sided maximal, then the non-invertible elements of the quotient $$R/M$$ are zero-divisors.

I know if an ideal is left-maximal or right-maximal, then the quotient must be a division ring. But want to understand why this fails when it's only maximal. Do simple rings/algebras fail to be division rings/algebras only because of zero-divisors? I'm aware of this answer, but that only shows an example. Thanks in advance.

• I'm trying to figure out what "maximal ideal not one-sided maximal" is supposed to mean. You're saying it's maximal as a two-sided ideal, not as a one-sided ideal? And you mean it's modular both as a left and a right ideal? Commented Apr 11, 2023 at 14:27
• @rschwieb Yes. That's what I mean. So $M$ is two-sided maximal. But there is a larger proper one-sided ideal containing $M$. And yes, it is two-sided modular, so the quotient is unital. Commented Apr 11, 2023 at 14:30

• Thanks. But then is it possible to modify the OP and makes it correct? Maybe it's true if it's changed to something like, "the non-invertible elements of the quotient is of the form $a+\alpha$, where $a$ is invertible but $\alpha$ is a nonzero zero-divisor"? Or maybe some other form? There must be some relations to zero divisors? Commented Apr 11, 2023 at 14:36
• "$a+\alpha$ where $a$ is invertible and $\alpha$ is a nonzero zero divisor" surely won't work: in the example given above there are no choices for $\alpha$. In a ring where nonzero zero divisors don't exist, your quest to associate them with elements outside the maximal ideal is going to be fruitless. Commented Apr 11, 2023 at 14:56
• Thanks. Can I understand it as this? We can always think of a ring as a free algebra quotient-ing out some relations. So if the ring has a pair of invertible elements $(a,b)$ such that $ab=ba=1$, it is quotient of the free algebra of some generators quotient out that relations. Similarly if there're zero-divisors $ab=0$, then the ring is the quotient of some free algebra over that relation. So it's always possible to start from some free algebra and choose some relations to quotient out, producing invertibles or zero-divisors. And there's no restriction whatsoever? Commented Apr 11, 2023 at 15:28