# Why is it that the congruence relations usually correspond to some type of subobject?

From the perspective of universal algebra, quotient structures of algebraic structures are built using congruence relations. If $$A$$ is an algebraic structure (a set with a bunch of operations on the set) und $$R$$ congruence relation on a set, then the quotient $$A/R$$ is well-defined and it will be an algebraic structure of the same type.

Now, as it turns out, in particular algebraic categories, these congruence relations on $$A$$ correspond exactly to some type of subobject of $$A$$. For instance, the congruence relations on a ring correspond precisely to the ideals of that ring; the congruence relations on a group correspond precisely to the normal subgroups of that group; the congruence relations on a module correspond precisely to the submodules of that module.

Why is it that the congruence relations usually correspond to some type of subobject? Is this a general phenomenon that can be generalized to all algebraic structures (as studied in this generality by universal algebra)?

Recall that congruences on $$A$$ can be viewed as certain subalgebras of its square $$A^2,\,$$ e.g. see here.

In algebras like groups and rings, where we can normalize $$\,a = b\,$$ to $$\,a\!-\!b = \color{#c00}0\,$$ congruences are determined by a single congruence class (e.g. an ideal in a ring). This has the effect of collapsing said relationship between congruences with subalgebras from $$A^2$$ down to $$A.\,$$ Such algebras are called ideal determined varieties and they have been much studied.

One answer to your question is that ideal-determined varieties are characterized by two properties of their congruences, namely being $$\,\rm\color{#c00}{0\text{-regular}}\,$$ and $$\rm\color{#c00}{0\text{-permutable}}$$. Below is an excerpt of one paper on related topics that yields a nice entry point into literature on this and related topics.

On subtractive varieties iv: Definability of principal ideals.

Paolo Agliano and Aldo Ursini

1. Foreword

We have been asked the following questions:

• (a) What are ideals in universal algebra good for?
• (b) What are subtractive varieties good for?
• (c) Is there a reason to study definability of principal ideals?

Being in the middle of a project in subtractive varieties, this seems the right place to address them.

To (a). The notion of ideal in general algebra [13], [17], [22] aims at recapturing some essential properties of the congruence classes of $$0$$, for some given constant $$0$$. It encompasses: normal subgroups, ideals in rings or operator groups, filters in Boolean or Heyting algebras, ideals in Banach algebra, in l-groups and in many more classical settings. In a sense it is a luxury, if one is satisfied with the notion of "congruence class of $$0$$". Thus in part this question might become: Why ideals in rings? Why normal subgroups in groups? Why filters in Boolean algebras?, and many more. We do not feel like attempting any answer to those questions. In another sense, question (a) suggests similar questions: What are subalgebras in universal algebra good for? and many more. Possibly, the whole enterprise called "universal algebra" is there to answer such questions?

Having said that, it is clear that the most proper setting for a theory of ideals is that of ideal determined classes (namely, when mapping a congruence E to its $$0$$-class $$\,0/E$$ establishes a lattice isomorphism between the congruence lattice and the ideal lattice). The first paper in this direction [22] bore that in its title.

It comes out that -- for a variety V -- being ideal determined is the conjunction of two independent features:

1. V has $$\,\rm\color{#c00}{0\text{-regular}}\,$$ congruences, namely for any congruences $$\rm\,E,E'$$ of any member of $$V,$$ from $$\,\rm 0/E = 0/E'$$ it follows $$\rm\,E = E'$$.

2. V has $$\,\rm\color{#c00}{0\text{-permutable}}\,$$ congruences, namely for any congruences $$\,\rm E,E'$$ of any member of $$V,$$ if $$\,\rm 0 \ E\ y \ E'\, x,\,$$ then for some $$\rm z,\ 0\ E'\, z\ E\ x.$$

This is not true "usually"; for example, it's not true for monoids or semirings. It's a very special fact that it's true for groups and rings, and in both cases it's true for the same reason: the presence of inverses allows you to replace thinking about an equivalence relation $$a \equiv b$$ with thinking about $$b^{-1} a \equiv 1$$ for groups (producing normal subgroups) and $$a - b \equiv 0$$ for rings (producing ideals).

Note also that, strictly speaking, ideals are not subobjects in the category of rings (with identity).

This is generally false. The fact that it holds in groups and rings is due to the fact that in each case we have an operation with identity and inverses, namely the group operation or the ring addition respectively.

As long as we have such an operation, every congruence is indeed determined by a single class. To see this, suppose $$S$$ is a structure with an invertible operation $$*$$, $$a\in S$$, and $$\sim,\approx$$ are congruences on $$S$$ with $$[a]_\sim=[a]_\approx$$. Fix $$b\in S$$; we want to show $$[b]_\sim\subseteq[b]_\approx$$ (from which by symmetry and universal generalization we'll get $$\sim=\approx$$).

Suppose $$b\sim c$$. Then $$b b'a\sim cb'a$$, where $$x'$$ denotes the $$*$$-inverse of $$x$$. This means $$a\sim cb'a$$, and so since $$[a]_\sim=[a]_\approx$$ we get $$a\approx cb'a$$. Now we undo the previous step: we multiply on the right by $$a'b$$ to get $$b\approx c$$ as desired.

• Why do you think that this single class that sometimes determines the whole congruence is a subalgebra? – user854455 Nov 27 '20 at 18:07
• @user854455 No idea, that's an interesting question. – Noah Schweber Nov 27 '20 at 18:12
• Thanks. Should I ask it in a new thread? – user854455 Nov 27 '20 at 18:23
• But I realised in the case of ideals it isn't even a subalgebra. ^^ – user854455 Nov 27 '20 at 18:25
• @user854455 Well, sometimes it's a subalgebra. :P But yes, this might make a good further MSE question. – Noah Schweber Nov 27 '20 at 18:25