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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 $\mathrm{Sub}(U)$?

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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

1 Answer 1

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 $\mathbf{A}$).

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