Generalization of normal subgroups and ideals

Question

I have seen that it is possible in universal algebra to generalize the concept of congruence and of quotient algebra.

If $A$ is an algebra (a set endowed with some operations), a congruence is an equivalence relation on $A$ which preserves those operations (a subset of $A \times A$ which is an equivalence relation and a subalgebra). Therefore, in the quotient set, operations via representatives are well-defined.

Given an equivalence relation we obtain a partition of the set $A$. In groups, this partition is a congruence if and only if it has been "generated by a normal subgroup via product". i.e:

Let $G$ be a group and $N \lhd G$ a normal subgroup. Then $x Ry$, if $xy^{-1} \in N$ is a congruence. The equivalence classes are $[x]=xN=\{xn : n \in N\}$ and $G/R=G/N=\{xN : x\in G\}$.

Similarly, in commutative rings a partition is a congruence if an only if it has been "generated by" an ideal.

Is there any way to formalise this concept of a subset of an algebra that provides a partition making use of some operation, being this equivalence relation a congruence?

Could this concept be generalized to any algebra, or at least to some collection of algebras?

Answer 1

Is there any way to formalise this concept of a subset of an algebra that provides a partition making use of some operation, being this equivalence relation a congruence?

Here is the way it has been done. If A is an algebra and $\theta$ is a congruence, then $\theta$ is called regular if it is generated as a congruence by any one of its classes. That is, if $C$ is any $\theta$-class, then $\theta$ is the least congruence containing $C\times C$. A is congruence regular if all of its congruences are regular. A variety of algebras is congruence regular if all of its members are.

Notes.

(1) The term regular was introduced by Mal'cev in

A.I. Mal'cev, On the general theory of algebraic systems, Mat. Sb., 35 (77) (1954), pp. 3-20.

(2) Congruence regular varieties were characterized by Csakany in

B. Csakany, Characterization of regular varieties, Acta Sci. Math. (Szeged), 31 (1970), pp. 187-189.

(3) In unpublished notes from the 1970's, J. Hagemann proved that a congruence regular variety must be congruence modular and congruence $n$-permutable for some $n$. It is a consequence of this that any variety containing an algebra that has a non-discrete compatible partial order must fail to be congruence regular. So, for example, any unary variety, the variety of all semigroups, any variety of lattices, ETC, will fail to be congruence regular.

Answer 2

For any algebra, the answer is no.
Consider, for example, lattices (or semi-lattices). Among these, consider chains. If two elements in a chain are related, that tells you nothing about all other elements but the ones that are in the interval whose least and greatest elements are the given elements.

This works, for example, for Boolean Algebras, where congruences are given by the class of $0$ (equivalently, by the class of $1$), which is an ideal (the class of $1$ is a filter).
But in this special case, we also have that Boolean Algebras are term-equivalent to Boolean Rings, and as you say, it works for rings, with the ideals.

It also works for Heyting Algebras (but I think it is only via filters, not ideals), and for many other varieties, but right now I don't know if there is a characterization of these.

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Edit: Even for Boolean Algebras (and thus, also for Heyting Algebras) this doesn't work completely.
If $\mathbf{B}$ is a infinite Boolean Algebra, then $\mathbf{B}$ has at least one ultra-filter $U$ which doesn't have a minimum element, and so this subset cannot be seen as a subalgebra. Principal filters don't have this limitation and so if $\mathbf{B}$ is a finite Boolean Algebra, then you have the desired property; for infinite ones, no (only for the congruences associated with principal filters).
Of course you always have homomorphic images, but that is obvious, and you're asking for subalgebras...

Edit 2: While what I wrote in the previous paragraph is true, it is perhaps a bit misleading.
The reason is that, the relevant question is not if a congruence is determined by a subalgebra, but if it is determined by a single congruence class, that is associated with some special subset of the original algebra.
Actually, if you have a ring $\mathbf{R}$ with unit $1$, then the only ideal that is a subalgebra of the ring is $R$ itself. The reason is that if it is a subalgebra, it must contain the nullary operations, such as $1$, and an ideal that contains $1$ is the whole ring.
So in the sense of Universal Algebra, ideals are not, in general, subalgebras of the rings (although normal subgroups are subalgebras of their groups).
That said, in a Heyting Algebra, you have a correspondence between congruences and filters: if $\mathbf{H}$ is a Heyting Algebra, $\theta$ is a congruence and $F$ is a filter, then $$F_{\theta} = \{ x \in H : x \theta 1 \}$$ is a filter and $$\theta_F = \{ (a,b) \in H^2 : (a \to b) \wedge (b \to a) \in F \}$$ is a congruence. Moreover, $\theta_{F_{\theta}} = \theta$ and $F_{\theta_F} = F$.
Boolean Algebras can be seen as Heyting Algebras, if we define $a \to b = a' \vee b$.
Again, this works for many other cases.