2

Chinese remainder theorem is extremely important in the theory of rings, it is stated that there is a canonical isomorphism between $R/\bigcap I_i$ and $\prod R/I_i$. However, from the proof, neither did I see the reason for canonical, nor feel the 'beauty' of this theorem. That's to say, the proof doesn't adopt any categorical method. What I saw is checking the element and do the induction. I'd like a proof more advanced.

I found this question, however I'm not satisfied with this, since I don't understand why $L$, clearly not some sort of adjoint, keeps pullback, which is a type of limit.

Someone told me that is more or less nature from the sheaf perspective and $\operatorname{Spec}$. Nevertheless I believe that a categorical description will be more effective and clean.

I change my words in order to make myself more clear.

Liam
  • 313
  • 1
  • 8
  • Sure it's canonical. It's the unique homomorphism induced by the pairing of all quotient maps $R\to R/I_i$, which by universal property assemble to a map $R\to\prod_i R/I_i$ – FShrike Dec 03 '23 at 13:17
  • Yes, you are right. I miss this evident fact. However, my problem is more or less why it's isomorphism, and how to convert $I_i$ is coprime into a categorical language. – Liam Dec 03 '23 at 13:23
  • There won't be a purely categorical proof which gives you more insight than the usual proof. CRT is not formal. – Martin Brandenburg Dec 03 '23 at 14:38
  • Thanks. But what's formal? – Liam Dec 03 '23 at 14:40
  • 1
    @WilliamHan "the point of category theory is to make that which is formal, formally formal". Read into that what you will. Formal is one of those words used and learned by assimilation but I'd be hard pressed to exactly define what is meant by it – FShrike Dec 03 '23 at 15:36

1 Answers1

1

Ideals are just subobjects of $R$ in the category of $R$-modules (and $R$ itself is just the representing object for the forgetful functor to Abelian groups or to sets and can be described categorically).

This theorem is saying if you have a finite family of subobjects of $R$ which, pairwise, have categorical union (pushout from their intersection) $R$, then the canonical map $R\to\prod R/I$ induces an canonical isomorphism between $R$ quotient the intersection (pullback from $R$) of these subobjects - which is just $R/\bigcap I$ - and $\prod R/I$. Quotienting itself is just taking a certain coequaliser and is categorical.

This was phrased entirely element-free and in universal properties alone.

FShrike
  • 40,125
  • Well, thanks. However, I didn't make myself clear. I'm seeking for a proof by categorical method, rather than a categorical description of this problem. – Liam Dec 03 '23 at 13:33
  • Meanwhile, I am a bit confused about your sentence 'then the canonical map $R\rightarrow \prod R/I$ induces an canonical isomorphism between $R$ quotient the intersection (pullback from $R$) of these subobjects'. What did you mean by saying isomorphism between ... ? Evidently you didn't show it's surjective. And where did you use the coprime condition? – Liam Dec 03 '23 at 14:25
  • @WilliamHan I didn't prove it's an isomorphism, no, but we knew it's an isomorphism already. Isomorphism "between" was a poor choice of words; I should have said "with". Or I should have just continued my sentence instead of forgetting to say all the words I wanted to say. – FShrike Dec 03 '23 at 15:15
  • @WilliamHan The coprime condition appears where I write: "which, pairwise, have categorical union ... $R$" – FShrike Dec 03 '23 at 15:16
  • Thanks a lot. I think I understand better. But I don't follow which theorem assures us that there is an isomorphism ('We knew it's an isomorphism already'). Is this because the 'categorical union'? Can you explain a little bit? Thanks. – Liam Dec 03 '23 at 15:43
  • @WilliamHan The CRT is just the statement that $R/\bigcap I\to\prod R/I$ is an isomorphism. The categorical union of two subobjects of $R$, ideals $I$ and $J$, is the pushout of $I\leftarrow I\cap J\rightarrow J$ and you can check that that's the ideal $I+J$. So by saying the categorical union of the subobjects is (pairwise) equal to $R$, I'm saying the ideals are pairwise coprime – FShrike Dec 03 '23 at 15:46
  • So you mean that this is because of CRT, thus is not proved by other categorical consturctions or theorems? Am I right? – Liam Dec 03 '23 at 15:48
  • 2
    @WilliamHan I'm saying it's because of the CRT yeah. There is maybe a proof that uses more category theory - I have no idea - but it seems unlikely and I'm convinced you can't find a purely categorical proof – FShrike Dec 03 '23 at 15:49
  • I agree with FShrike. You can interpret it for sure (even saying that a certain functor on closed subschemes is a sheaf), but very unlikely there will be a more direct proof using that. – Martin Brandenburg Dec 03 '23 at 16:06