Multi-pullbacks and the relative chinese remainder theorem

Question

Let $I,J$ be two-sided ideals of a ring $R$. In this question I asked for an "automatic" proof of the fact the natural map $R/(I\cap J)\rightarrow R/I\times _{R/(I+J)} R/J$ is an isomorphism. At any rate, if $I,J$ are comaximal, this immediately yields an isomorphism $R/(I\cap J)\cong R/I\times R/J$.

Now for $n$ ideals. I'll denote the multi-pullback of the arrows $R/I_k\rightarrow R/(\require{amsart}\Large +$$_k I_k)$ by $P$. There is clearly a canonical map $R/\bigcap_ kI_k\rightarrow P$ and by an analogous direct proof it is an iso (Incorrect; see answer). Now, if I try to collapse into the chinese remainder theorem, I find myself merely asking for $I_k$ to jointly cover $R$, i.e $\require{amsart}\Large +$$_k I_k=R$. On the other hand, every source I've seen asks for $I_k$ to be pairwise comaximal.

Why is this?

Added: As mentioned in the comments, the result just ain't true without pairwise comaximality, so I'm missing something in the diagrams. Each of the squares below is both a pullback and a pushout. $$\require{AMScd} \begin{CD} R/(I_i\cap I_j) @>>> R/I_i\\ @VVV @VVV\\ R/I_j @>>> R/(I_i+I_j) \end{CD}$$ Hence, each square from the multi-pullback $$\require{AMScd} \begin{CD} R/\bigcap_kI_k @>>> R/I_i\\ @VVV @VVV\\ R/I_j @>>> R/\sum_kI_k \end{CD}$$ factors through the one above and we get induced maps $R/\bigcap_kI_k\rightarrow R/(I_i\cap I_j)$ and $R/(I_i+I_j)\rightarrow R/\sum_kI_k$. What's the missing link?

Added again: I'm getting increasingly large diagrams and starting to think this needs all kinds of combinatorial stuff related to covers. Hoping there's an elegant argument that cuts through this.

Answer 1

The map $R/\bigcap_ kI_k\rightarrow P$ is not generally an isomorphism. This is clear if one thinks concretely. Too bad I was determined to work with diagrams alone. Comaximality is a convenient condition to make sure this map is an iso.

Already in the case of three ideals $I,J,K$, the problem is that there are no induced maps $P\rightarrow R/(I\cap J)$ because the square below need not commute. $$\require{AMScd} \begin{CD} P @>>> R/I\\ @VVV @VVV\\ R/J @>>> R/(I+J) \end{CD}$$ Commutativity would mean the implication below, which has no reason to hold. $$r+I+J+K=s+J+K+I\implies r+I+J=s+J+I$$ In the binary case the square is a pullback, so commutativity is not an issue.

Now for the comaximality. As this question shows, comaximality is not a necessary condition for the binary CRT. By associativity, $P$ can be calculated via iterated pullbacks. By induction based on the binary case, we can tentatively calculate $$\begin{aligned} P & \;\;= (R/\bigcap _{k=1}^{n-2}I_k\times _{R/\sum_{k=1}^n}R/I_{n-1}) \times_{R/\sum_{k=1}^n}R/I_n \\ & \overset{\text{I wish}}{=} (R/\bigcap _{k=1}^{n-2}I_k\times _{R/\sum_{k=1}^{\color{#c00}{n-1}}}R/I_{n-1}) \times_{R/\sum_{k=1}^n}R/I_n \\ & \;\;=R/\bigcap_{k=1}^{n-1}\times _{R/\sum_{k=1}^nI_k}R/I_n \\ & \overset{\text{I wish}}{=}R/\bigcap_{k=1}^{n-1}\times _{R/(\color{#c00}{\bigcap_{k=1}^{n-1}I_k+I_n)}}R/I_n \\ & \;\;=R/\bigcap_{k=1}^nI_k \end{aligned}$$

Sufficient conditions would be $\sum_{k=1}^{n-1}=\sum_{k=1}^n, \sum_{k=1}^nI_k=\bigcap_{k=1}^{n-1}I_k+I_n$. We want these to hold for all $1\leq k\leq n, n\geq 3$. An easy condition for the first equality is pairwise comaximality. The following fact takes care of the second one.

Fact. Let $I_1,\dots,I_n$ be pairwise comaximal ideals. Then $I_j,\prod_{k\neq j}I_k$ are comaximal. (Proof by induction.) In particular, $I_j,\bigcap_{k\neq j}I_k$ are comaximal since $\prod_kI_k\subset \bigcap _kI_k$.

So for pairwise comaximal ideals the tentative calculation is justified. Since we're pulling back over $R/\sum_k I_k$ and $\sum_kI_k=R$, $P$ is just the product $\prod _kR/I_k$ and we have the Chinese remainder theorem. If the ring is moreover commutative, then multiplication and intersection coincide for pairwise comaximal ideals, and we have $\prod _kR/I_k\cong R/\prod_kI_k$.