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Stream: learning: questions

Topic: Quotient rings

Ryan Schwiebert (Dec 24 2025 at 04:11):

Maybe it's because it's late, or maybe it's because the only exposition I saw so far was in categories with zero morphisms, but I can't seem to figure out for myself if one can express quotients in CRing (or Ring for that matter) as colimits. This is possible, right?!

The only thing I can see is that if I have a ring $R$ and I first take the coslice category $R/Ring$, and then reduce to the subcategory using only objects (arrows leaving $R$) that have $I$ in their kernels, then $R\to R/I$ would be a initial object in that category... I think!

Maybe someone can think of something pithy to de-confuse me about the situation? I think I am more acclimated to the category of modules than the category of Rings or CRings

fosco (Dec 24 2025 at 08:40):

I recently stumbled upon the same problem! One way to argue, that I learned from the categorical algebra people, is to regard the ideal $I$ as a non-unital subring or $R$. Then, the quotient $R/I$ is the pushout of the span

$R \leftarrow I \oplus \mathbb Z \to \mathbb Z$

where the right leg is the projection; one easily checks that a cocone for this diagram is a ring homomorphism $R\to A$ the kernel of which contains $I$.

fosco (Dec 24 2025 at 08:40):

this diagram happens in the category of rings, because $I\oplus \mathbb Z$ is indeed the free unitarization of $I$, regarded as a non-unital ring

Graham Manuell (Dec 24 2025 at 12:28):

I think I must be misunderstanding the question, but in any finitely complete category, a regular epimorphism is the coequaliser of its kernel pair. In any algebraic category, the regular epimorphisms are the surjections. So you can easily express ring quotients as colimits in a way that really has nothing to do with rings in particular.

Ryan Schwiebert (Dec 24 2025 at 13:05):

fosco said:

as a non-unital subring

Yeah, I am finally setting aside a little time to see how that works. I've seen it alluded to before. Thanks for your sketch!

fosco (Dec 24 2025 at 13:07):

I also think Graham's approach is more elegant and portable

Ryan Schwiebert (Dec 24 2025 at 13:10):

Graham Manuell said:

a regular epimorphism is the coequaliser of its kernel pair.

That could be it, it's just that it's a level above the fluency I currently have. I can barely remember that regular epics are a strengthening of epics, and I don't know for sure if I ever learned they coincided in Ring and CRing. It sounds like I should just look at the span with apex $R$ and legs both $f:R\to S$ and see if it starts making sense...

Chris Grossack (she/they) (Dec 24 2025 at 13:26):

A coequalizer of $A \rightrightarrows B$ gives you an epimorphism $B \twoheadrightarrow C$, but not every epi arises in this way -- for instance, $\mathbb{Z} \to \mathbb{Q}$ in CRing. We say that an epi is regular exactly when it has a good reason to be epi, because it's the coequalizer of some pair of arrows. So when you said that you're looking to recognize the quotients (read: coequalizers), it's kind of a linguistic trick to say that you're looking for the regular epis.

BUT, in favorable cases the regular epis admit another, more intrinsic description! (This is what Graham's answer provides). Given an epi $f : B \twoheadrightarrow C$, you can try to build a parallel pair $A \rightrightarrows B$ whose coequalizer is $f$. If you meditate on it, the obvious thing to try is called the [[kernel pair]] of $f$, and in nice cases an epi is regular if and only if it's the coequalizer of its kernel pair! So this lets us recognize the regular epis among the epis, and thus to recognize the quotient objects in CRing (among many other categories)

fosco (Dec 24 2025 at 14:01):

Ryan Schwiebert said:

Graham Manuell said:

a regular epimorphism is the coequaliser of its kernel pair.

That could be it, it's just that it's a level above the fluency I currently have. I can barely remember that regular epics are a strengthening of epics, and I don't know for sure if I ever learned they coincided in Ring and CRing. It sounds like I should just look at the span with apex $R$ and legs both $f:R\to S$ and see if it starts making sense...

Don't worry about your lack of fluency, these things are hard, and require maturity to be understood. At a very layman level, when you move from Set to other categories you notice that there are many inequivalent notions of epimorphism. You can be an epi because you're surjective, or because you're a coequalizer, or because you coincide with your image, or...

fosco (Dec 24 2025 at 14:08):

in different categories, one crafts different notions of epimorphism at different levels of strength

• cancellativity: an arrow is an epi if it is right cancellable
• invertibility: an arrow is a split epi if it admits a right inverse
• coequalizing: an arrow is a regular epi if it is the coequalizer of a pair (every such map must be an epi, for the "is unique" part of the universal property of a coequalizer)
• extremality: an arrow is an extremal epi if "it doesn't factor through any nontrivial subobject"
• strength: an arrow is a strong epi if it has the right lifting property with respect to all monics

In the category of rings, an extremal epi is the correct notion of surjection, because the class coincides precisely with surjective homomorphisms. Part of the proof uses the fact that the forgetful functor into Set is conservative, and thus (I believe) this example together with similar others was a stimulus to generalize the same argument to other algebraic categories (the categories of "algebraic structures", elementarily but broadly intended).

And the epimorphism that embeds the integers in the rationals is not extremal, because it factors through any localization of $\mathbb Z$ at a prime, e.g. $\mathbb Z \to \mathbb Z [\frac 12]\hookrightarrow \mathbb Q$.

fosco (Dec 24 2025 at 14:09):

An unparalleled source to get intuition and many (counter)examples related to this taxonomy is "the joy of cats".

Damiano Mazza (Dec 24 2025 at 22:31):

@Ryan Schwiebert if you want a more concrete description of what was explained above, which makes the ideal $I$ appear (more or less) explicitly, remember that the quotient $R/I$ is just equating to zero every element of $I$. So we need to do this with a coequalizer.

Let $V_I$ be the set containing an element $X_a$ for each $a\in I$. That is, $V_I$ is really just $I$ but in which the names of elements have been changed so that we do not confuse them with elements of $R$. Consider now the polynomial ring $\mathbb Z[V_I]$. Its elements are polynomials in the variables $X_a$ (with $a\in I$) with coefficients in $\mathbb Z$. Observe that, for example, in this ring $X_a+X_b$ has nothing to do with $X_{a+b}$, they are two distinct polynomials (this is why I changed the names, so that we wouldn't be tempted to impose these equalities). In fact, it turns out (in case you do not already know this) that $\mathbb Z[V_I]$ is the free commutative ring on the set $V_I$. This has the important consequence that a ring homomorphism $\mathbb Z[V_I]\to R$ is just a plain function $V_I\to R$, that is, an assignment giving to each variable $X_a$ a value in $R$.

Consider now the homomorphism $i:\mathbb Z[V_I]\to R$ mapping each $X_a$ to $a$, and the homomorphism $z:\mathbb Z[V_I]\to R$ mapping every $X_a$ to $0$. At this point, it shouldn't be too hard to guess that $R/I$ is exactly the coequalizer of $i$ and $z$.

As was pointed out above, this works in great generality (basically, every time you have a way of presenting the objects of your category in some algebro-logical way, by means of generators and equations), but I hope that seeing it in the concrete case of rings will help you understand how it works.

Damiano Mazza (Dec 24 2025 at 22:34):

(Of course, once you understood what's going on, you can really just write $\mathbb Z[I]$ rather than $\mathbb Z[V_I]$, and send every "formal version'" of an element of $I$ to "itself'', or to zero. But perhaps this would have been too confusing if you're seeing this for the first time).

Damiano Mazza (Dec 24 2025 at 22:36):

(Also, I'll let you think about the fact that, if $I$ is generated by a set $S\subseteq R$, then considering $\mathbb Z[S]$ and taking the coequalizer of the homomorphisms sending every "formal version" of $S$ to "itself" or to zero also defines $R/\langle S\rangle=R/I$).

Morgan Rogers (he/him) (Dec 24 2025 at 23:07):

Damiano's construction is nice, but it's not the kernel pair that Graham mentioned! That would be the ring of pairs $(r,s)$ such that $r-s \in I$, with the two apparent projections to $R$.

Morgan Rogers (he/him) (Dec 24 2025 at 23:10):

(The pullback in rings is just the pullback in sets with inherited operations :) )

Ryan Schwiebert (Dec 25 2025 at 04:32):

Hmm, yeah I do feel like I'm getting lost with three competing pictures now.

#1
#2
#3

I think #2 is closest to what I expected, there's just a few nuances I want to get straight.

So I thought about the coequalizer of the kernel pair for $f$, and I think what I concluded is that if $d:S\to C$ is the coequalizer, $f$ factors through that map. That's when I realized a few things I hadn't been thinking hard enough about. I did not think of $f$ as being a surjection, and I have not yet connected the dots from what was said before about this $f$ and $d$.

@Chris Grossack (she/they) said: in nice cases an epi is regular if and only if it's the coequalizer of its kernel pair! So the first dot to connect is: Ring and CRing are examples of this nice case? Regular epis are exaclty the coequalizers of kernel pairs? And that is not the case in some categories?

Morgan Rogers (he/him) (Dec 25 2025 at 08:33):

@Chris Grossack (she/they) made that sound more mysterious than it really is, I think :wink: The condition for this is just the existence of kernel pairs! It's a nice exercise in diagram chasing to show that if a morphism is a coequalizer of any pair of morphisms, then it's the coequalizer of its kernel pair. So yes, Ring and CRing are both categories where this is true.

Chris Grossack (she/they) (Dec 25 2025 at 13:23):

Oh no! I was trying to make it sound less mysterious, haha

Ryan Schwiebert (Dec 25 2025 at 13:24):

Oof, that means the time I spent trying to prove that still isn't over. I thought i misunderstood and that I was spinning my wheels.

Chris Grossack (she/they) (Dec 25 2025 at 13:25):

"Nice cases" is extremely mild, just the existence of finite limits, iirc. So like Morgan says, CRing is definitely enough.

Ryan Schwiebert (Dec 25 2025 at 14:12):

While I plod through that, I have a followup.

In Rng, you have zero morphisms, and now ideals are objects of the category. In this context, do the ring theoretic kernel and category theoretic kernel coincide? Feels like they should...

Morgan Rogers (he/him) (Dec 25 2025 at 22:27):

Beware that "kernel" (equalizer with zero) and "kernel pair" (pullback of morphism along itself) are different concepts with very similar names

Ryan Schwiebert (Dec 25 2025 at 22:34):

Yes, I’m thinking of the former, along with the ring theoretic definition. That’s what the equalizer with a zero morphism amounts to, right?

Morgan Rogers (he/him) (Dec 25 2025 at 23:12):

That's right!

Damiano Mazza (Dec 26 2025 at 08:15):

Morgan Rogers (he/him) said:

Damiano's construction is nice, but it's not the kernel pair that Graham mentioned!

Whooops, my bad. I thought the two should be related at some level, but they actually yield quite different parallel arrows (as we see in this example).

Sorry @Ryan Schwiebert, I didn't mean to make things more confusing!

James Deikun (Dec 28 2025 at 11:57):

I don't know too much about it, but the ability to describe congruences by something like an "ideal" is connected to the tower of notions including [[Malcev category]] and [[protomodular category]] as explicated mostly by Dominique Bourn.

Ryan Schwiebert (Dec 29 2025 at 04:11):

I'm beginning to realize that perhaps my attention was focused on the wrong half of an imbalanced pair.

Just so I'm being clear, Ring is meant to be the category of rings with identity and identity preserving morphisms, and Rng is meant to be the category of rings not necessarily having identity, and whose morphisms are ring morphisms (with nothing required related to identity elements.)

If we consider ring ideals and quotient rings, we apparently have quotient rings in both Ring and Rng, those being the regular epic morphisms (equivalence classes thereof, probably.) Whenever $R\to S$ is surjective, $S$ is the quotient of $R$ by the kernel of that morphism.

But the ideals don't (can't) materialize in Ring, but they do appear as objects in Rng. So maybe I should think more about that. I know not all monics in Ring are regular. (Equivalence classes of) monic morphisms are the initial subrings , are they not?

I'm trying to get my head around what changes for regular monics, if anything. They look dual to the quotients but I’m not sure what they’d mean in Ring.

OTOH in Rng, quotients seem to be straightforwardly coequalizers, and ideals seem to be equalizers.

Ryan Schwiebert (Dec 29 2025 at 04:37):

I wish this page were as fleshed out with information as its Ring counterpart is: https://ncatlab.org/nlab/show/Rng

Ryan Schwiebert (Dec 29 2025 at 04:45):

Ooh, there’s this and it’s follow-up too: https://math.stackexchange.com/questions/695685/regular-monomorphisms-of-commutative-rings