# Relative chinese remainder theorem and the lattice of ideals

Let $R$ be a ring with two-sided ideals $I,J$. The proof of the "absolute" chinese remainder theorem revolves around the fact that if $I,J$ cover $R$ in the lattice of ideals, i.e $I+J=R$, then the canonical map $R/(I\cap J)\rightarrow R/I\times R/J$ is an isomorphism.

The "relative" version says that given any ideals $I,J$, the canonical map $R/(I\cap J)\rightarrow R/I\times _{R/(I+J)} R/J$ is an isomorphism. Evidently if $I,J$ cover $R$ the quotient $R/(I+J)$ is the terminal object and we collapse into the "absolute" version. This relative version seems more natural, so I'm trying to understand it better.

In the lattice of ideals, as in any lattice, the square below is both a pullback and a pushout. $$\require{AMScd} \begin{CD} I\cap J @>>> I\\ @VVV @VVV\\ J @>>> I+J \end{CD}$$ The relative version then tells us the square below is a pullback. $$\require{AMScd} \begin{CD} R/(I\cap J) @>>> R/I\\ @VVV @VVV\\ R/J @>>> R/(I+J) \end{CD}$$ Is there some easy way to get this from the correspondence theorem for ideals?

I think the bottom square is the image of the top square by the functor taking an inclusion $I\subset J$ to the map $R/I\rightarrow R/J$ defined by $r+I\mapsto r+J$. Denote by $\mathcal L(R)$ lattice of ideals. This functor seems to be the cokernel over the subcategory of $\mathcal L(R)\downarrow R$ generated by inclusions of ideals. Hence this functor only preserves pushouts. So the bottom square is also a pushout, and the content of the relative version is that this pushout square is also a pullback. Is this at all correct?

• It is trivial that the second square is a pullback and a pushout in the category of quotients of $R$, since this follows from the first square. It is not trivial that it is also a pullback and a pushout in the category of rings. I don't think that there is a purely category-theoretic proof. – HeinrichD Sep 23 '16 at 12:36
• SInce you are interested in generalizations, here is one: Let $A$ be an algebra in the sense of universal algebra. Let $U,V$ be congruences on $A$ with $U \circ V = V \circ U$. (The $\circ$ denotes composition of relations). Then, $U \vee V = U \circ V$ in the lattice of congruences, and we have $$A/(U \wedge V) \cong A/U \times_{A/(U \vee V)} A/V.$$ This gives at once a chinese remainder theorem for groups, rings, Lie algebras; here $U \circ V = V \circ U$ is automatically satisfied. – HeinrichD Sep 23 '16 at 12:43
• @HeinrichD Where's this from? – Arrow Nov 10 '16 at 23:13
• @JoseBrox pullback. – Arrow Jul 12 '17 at 20:41
• @JoseBrox I have never noticed this relative CRT mentioned in a book. I stumbled upon it in an answer by Martin Brandenburg on some MSE question. The result mentioned by HeinrichD is extremely general and probably appears in some paper about Mal'cev categories, since Mal'cev theories are characterized by the commutation of the composition of (internal) equivalence relations. – Arrow Jul 14 '17 at 19:26