# The ring $R/I$ is a total quotient ring iff $I$ is... what?

Conventions.

1. All my rings are commutative.
2. By a total quotient ring (TQR) , I mean a ring whose every regular element is a unit.

Now let $R$ denote a ring and $I$ denote an ideal of $R$. The following fact is well-known:

Proposition. The ring $R/I$ is a field iff $I$ is maximal among proper ideals.

I'm wondering if there is a similar theorem for total quotient rings. Therefore, I ask:

Original Question. The ring $R/I$ is a TQR iff $I$ is... what?

Edit. I'm also interested in the following, related issue:

Second Question. Is there some weakening of freeness that holds for modules over a TQR?

If so, does this help us answer the previous question?

## 1 Answer

There doesn't seem to be any strong connection with any standard ring theoretic conditions.

To give you an idea of the difficulty in pinning down a condition, consider this. For a von Neumann regular ring, every ideal has the property you speak of (that is, $R/I$ is a total quotient ring for every ideal $I$. What standard properties to the ideals have? They are all idempotent, semiprime and flat and divisible (in the sense given by Lam in Lectures on modules and rings). Unfortunately, the zero ideal in a nonfield domain also has all of these properties, and yet such a domain is not a total quotient ring.

So, it seems rather hard to see a common thread here beyond the literal translation of your condition:

$R/I$ is total quotient iff for every $x\in R$ with $(I:x)= I$, there exists a $y$ such that $1-xy\in I$.

Incidentally, I have also seen the rings you describe called "classical rings" and "cohopfian rings" in case that leads to more literature for you.

• I happened on this post by accident a couple of weeks ago. Independen tly, K. Varadarajan (Hopfian and co-Hopfian objects, Publ. Mat. vol 36 (1992), 293-317) and I (Dedekind finite objects in module categories, J. Pure Appl. Algebra 82 (1992), 71-80) noted that a commutative ring $R$ is co-Hopfian as an $R$-module if and only if $R$ is a TQR. Later (Thm 2.3, co-Hopfian modules, arXiv:2201.09961v1) I showed that if $M$ is nonzero finitely generated faithful co-Hopfian module over a commutative Noetherian ring $R$, then $R$ must be semilocal, and its maximal ideals are precisely Commented Aug 30, 2022 at 16:06
• those ideals maximal in Ass(M). This has relevance to the question above. The ring $R/I$ is a f. g. faithful $R/I$-module whose endomorphism ring is isomorphic to $R/I.$ In this context, $R/I$ is co-Hopfian if and only if $R/I$ is semilocal and its maximal ideals are the only maximal associated primes of $R/I$ (as $R/I$-module). In essence, $I$ must be contained in a finite number of maximal ideals of $R$ and these contain all other prime divisors of $I$. Commented Aug 30, 2022 at 16:43
• @ChrisLeary neat! Thanks for the followup Commented Aug 30, 2022 at 17:05
• @ChrisLeary Huh, do we perhaps have a common connection through Todd?. Commented Aug 30, 2022 at 17:08
• @rscwhieb - No common connection. Although I did have a brief correspondence with your thesis advisor, Sergio, when submitting a paper to Journal of Algebra and its Applications. The paper didn't work out, but I broke it into two parts and did some revisions. They're both on arXiv, one is the one I mentioned in my comment, with author name F. C. Leary. Interstingly enough, there is another Chris Leary floating around in mathematics. He's at SUNY Geneseo. If I recall correctly, he does do some work in set theory. Commented Aug 30, 2022 at 17:24