Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $U$ be a universal algebra of type $T$, and denote $\mathrm{Con}(U)\!=\!\{\text{congruence relations on }U\}$ and $\mathrm{Sub}(U)\!=\!\{\text{subalgebras of }U\}$. Let "$\leq$" mean "subalgebra".

The correspondence theorem (2.6.20, p. 54) says: $$\mathrm{Con}(U/\vartheta)\!=\!\{\alpha/\vartheta\,;\, \alpha\!\in\!\mathrm{Con}(U),\vartheta\subseteq\!\alpha\},$$ where $\alpha/\vartheta\!=\!\{(a/\vartheta,b/\vartheta)\!\in\!(U/\vartheta)^2;(a,b)\!\in\!\alpha\}$, and also $$(\alpha\wedge\beta)/\vartheta=(\alpha/\vartheta)\wedge(\beta/\vartheta)\;\;\text{ and }\;\; (\alpha\vee\beta)/\vartheta=(\alpha/\vartheta)\vee(\beta/\vartheta).$$ In particular, for a group $G$ and ring $R$ and module $M$ and algebra $A$, we have $$\{\text{normal subgroups of }G/H\}\!=\!\{H'/H;\,H'\!\unlhd\!G,H\!\subseteq\!H'\},$$ $$\{\text{ideals of }R/I\}\!=\!\{I'/I;\,I'\!\unlhd\!R,I\!\subseteq\!I'\},$$ $$\{\text{submodules of }M/N\}\!=\!\{N'/N;\,N'\!\leq\!M,N\!\subseteq\!N'\},$$ $$\{\text{algebra ideals of }A/I\}\!=\!\{I'/I;\,I'\!\unlhd\!A,I\!\subseteq\!I'\}.$$ But we know that we also have $$\{\text{subgroups of }G/H\}\!=\!\{G'/H;\,G'\!\leq\!G,H\!\subseteq\!G'\},$$ $$\{\text{subrings of }R/I\}\!=\!\{R'/I;\,R'\!\leq\!R,I\!\subseteq\!R'\},$$ $$\{\text{submodules of }M/N\}\!=\!\{M'/N;\,M'\!\leq\!M,N\!\subseteq\!M'\},$$ $$\{\text{subalgebras of }A/I\}\!=\!\{A'/I;\,A'\!\leq\!A,I\!\subseteq\!A'\}.$$

Question: Is there some nice correspondence between $\mathrm{Sub}(U/\vartheta)$ and $\mathrm{Sub}(U)$?

share|cite|improve this question
This isn't very precise but I would expect what we're seeing are artifacts of the fact that, in all of the algebras you list, congruences are represented by a distinguished subalgebra. I'm not sure but I would think the phenomenon should already disappear in lattices. – Miha Habič Jun 2 '12 at 10:37

Note that if $\mathbf{A}$ is any one of the special (congruence modular) algebras you mentioned, then each congruence $\theta \in \mathrm{Con}(\mathbf{A})$ is uniquely determined by a single congruence class, say, $a/\theta$, and each $\theta$ has exactly one class that is a subalgebra of $\mathbf{A}$. For example, a congruence relation of a group is uniquely determined by the congruence class containing 1 (which is a normal subgroup).

It is because of this nice property that subalgebras can be compared with "congruences," as in $H\leq G' \leq G$, where $H$ is a normal subgroup (corresponding to some congruence of $G$). In general, this is not the case.

However, for each subalgebra $\mathbf{B} \leq \mathbf{A}$ we can define $\pi_\theta (B) = \{b/\theta \mid b \in B\}$, and this is a subalgebra of $\mathbf{A}/\theta$. (N.B. we can't write $\mathbf{B}/\theta \leq \mathbf{A}/\theta$, since the expression $\mathbf{B}/\theta$ doesn't make sense if we don't restrict $\theta$ to $B^2$.) But we don't get a nice correspondence like the ones in the examples you mentioned.

As a final remark, to avoid the pitfall of thinking that a normal subgroup is a congruence relation, keep in mind that the set of congruence relations is $\mathrm{Con}(\mathbf{A}) = \mathrm{Eq}(A) \cap \mathrm{Sub}(\mathbf{A}^2)$, where $\mathrm{Eq}(A)$ denotes the set of equivalence relations on $A$. Thus, each congruence relation is, in particular, a subalgebra of $\mathbf{A}^2$ (not of $\mathbf{A}$).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.