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I have never been a "rings must be unitary as an axiom" kind of person, nor do I harbor any hard feelings against those that are. Just like any religion I have my own reasons (normal subgroups are groups, so ideals should be rings) but I also understand the need for simplicity and smoother assumptions.

My question is honest and possibly naive. In those cases where the existence of a multiplicative identity is built into the definition of ring---and so ideals are no longer rings themselves---how does one invoke any useful category theory machinery into their work on the subject? For instance, I am looking at articles I've written where subobjects and quotient/cofiber sequences make many things work, and I have to believe [after Quillen] that these constructs are useful in ring theory. But now we are in a position where we don't even have meaningful subobjects to serve as the "denominator" in quotient constructions: ideals are not objects in the category.

So as a unitary kind of person, does this somehow not matter? Is there some clever way around this, or is this simply not an issue?

To attempt to make this question answerable and not opinion: how do ideals manifest themselves in the category of rings-which-must-have-identity?

Randall
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    This is not an answer to your question, but rather a comment on the first paragraph: "normal subgroups are groups, so ideals should be rings" is a very group-centric way of looking at algebra. It doesn't generalize at all to wider algebraic contexts (monoids, lattices, etc.) where the idea of taking the quotient of $A$ by a subalgebra of $A$ fails much more badly than it does for rings. In these wider contexts, the correct kind of object to quotient by is a congruence: an equivalence relation on $A$ which, when viewed as a subset of $A\times A$, is a subalgebra. – Alex Kruckman Aug 30 '24 at 16:56
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    ... so maybe this is an answer to your question after all. Given a ring $R$ and an ideal $I\subseteq R$, we can form the congruence ${(r,r+i)\mid r\in R,i\in I}$, which is a subring $C$ of $R\times R$. Composing the inclusion with the two projections $\pi_1$ and $\pi_2$ gives two ring homomorphisms $C\to R$, whose coequalizer is the quotient $R/I$. Conversely, given any congruence on $R$, the equivalence class of $0$ is an ideal, and these constructions are mutually inverse. – Alex Kruckman Aug 30 '24 at 17:01
  • I'm not posting it as an answer because I don't know enough about category theory to know wether this is actually the way out of your dilemma, but I could imagine that since both rings and ideals are modules over the ring, you can work in the category of modules. – Vercassivelaunos Aug 30 '24 at 18:03
  • @Vercassivelaunos that's an interesting approach. – Randall Aug 30 '24 at 18:04
  • @AlexKruckman I like that. – Randall Aug 30 '24 at 18:04
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    See also Why is it that the congruence relations usually correspond to some type of subobject?, esp. the notion of ideal-determined varieties. $\ \ $ – Bill Dubuque Aug 30 '24 at 19:07

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I think your understanding of quotients is not fully correct. We do not quotient out subobjects. We quotient out congruences or, say, group actions. The quotient object $A/R$ is then defined as the coequalizer of two maps $R \rightrightarrows A$. That congruences on a group correspond to special subobjects is a pure coincidence in the landscape of categories. You cannot (reasonably) quotient out a submonoid from a monoid, for example. And you cannot (reasonably) quotient out a subset from a set. (Also the quite common construction of the quotient of a topological space by a subspace is problematic, you should work with pointed spaces here in order to formulate a good universal property. But that just as a side note.)

In the context of universal algebra, a congruence relation on an algebra $A$ is given by an equivalence relation $R$ on the underlying set such that $R \subseteq A \times A$ is a subalgebra. Then we can form the quotient $A/R$ as usual, and it is the coequalizer of the two maps $R \rightrightarrows A$.

This holds for rings and for non-unital rings (and every algebraic structure you can think of). In both cases, there is a bijection between the congruences on $A$ and the ideals of $A$. (The presence of a unit doesn't matter for this since $(1,1) \in R$ follows from reflexivity of $R$.)

On the other hand, this way we only capture two-sided ideals. Left ideals of $A$ are just subobjects with respect to the category of left $A$-modules, and we may regard the construction $A/I$ as a special case of the quotient of left modules. Same for right ideals, and for two-sided ideals we work with $A$-bimodules.

In the category of abelian groups, and more generally any abelian category, there is a correspondence between congruences and subobjects, though. And for this reason one often sees this construction "object modulo a subobject". But that is not the rule.

You were also concerned about applying category theoretic machinery to rings (i.e., rings with unit). Well, the category $\mathbf{Ring}$ is complete, cocomplete, and the forgetful functor $\mathbf{Ring} \to \mathbf{Set}$ is monadic. This is as good as it can get. In particular, all colimits exist. The same holds for the category $\mathbf{Rng}$ of rngs. When it comes to categorical constructions, there is almost no difference. What maybe sets them apart is that there is a fully faithful functor $\mathbf{Ab} \hookrightarrow \mathbf{Rng}$ whose image consists of the rngs with zero multiplication. Also, $\mathbf{Rng}$ has a zero object.

PS: There are a couple of reasons why rings are unital by default, some of them are explained by Bjorn Poonen in Why all rings should have a 1.

Martin Brandenburg
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