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Given a commutative ring $R$, there is a category whose objects are ~~epimorphisms~~ surjective ring homomorphisms $R \to S$ and whose morphisms are commutative triangles making two such ~~epimorphisms~~ surjections compatible, and the skeleton of this category is a partial order that can be identified with the lattice of ideals of $R$. Now, I have always been under the impression that anything one can say about ideals one can phrase in this purely arrow-theoretic language: most importantly, the intersection of ideals is the product in this category and the sum of ideals is the coproduct. (Since we're working in a partial order, product and coproduct are fancy ways to say supremum and infimum. The direction of the implied ordering on ideals may differ here from the one you're used to, but that's not important.)

However, Harry's made some comments recently that made me realize I don't know how to define the product of two ideals purely in terms of this category, that is, via a universal construction like the above. It would be really surprising to me if this were not possible, so maybe I'm missing something obvious. Does anyone know how to do this?

edited Feb 9 at 6:05

asked Feb 9 at 5:15

Qiaochu Yuan
11.7k●18●97

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Epis or surjections? – Mariano Suárez-Alvarez Feb 9 at 5:19

Surely you mean surjections rather than epimorphisms (see mathoverflow.net/questions/109/…), but then you may as well just say you're working with the category of ideals in R where the morphisms are inclusions. – Anton Geraschenko Feb 9 at 5:51

You want them to be epis of the category R-Mod. – fpqc Feb 9 at 6:03

That's what I get for not checking things thoroughly. There are a few other things I have to change now, too... – Qiaochu Yuan Feb 9 at 6:05

opposite category of surjective map(strict epimorphism) of commutative rings is just category of closed subvariery of Spec(R), morphism is injective map as topological space. – Shizhuo Zhang Feb 9 at 6:24

Bjorn Poonen · Accepted Answer · 2010-02-09 06:12:46Z UTC

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Nice question! The answer is that it's not possible! Let $R=\mathbb{F}_3[x,y]/(x^2,y^2)$. The lattice of ideals consists of the eight ideals

$(1)$

$(x,y)$

$(x)$ $(y)$ $(x+y)$ $(x-y)$

$(xy)$

$(0)$,

in which each ideal contains all ideals at lower levels. In the middle level, some of the ideals have square zero, and some don't, but you can't tell which ones just from looking at the (unlabeled) lattice.

answered Feb 9 at 6:12

Bjorn Poonen
6,617●1●18●46

	There is more data in a ring than just the lattice of ideals... I am pretty sure this is false, because Toen and Vezzosi manage to define the Zariski topology on a much more general category than CRing without relying on set-theoretic arguments. – fpqc Feb 9 at 6:28
	I wanted to ignore that extra data; the point of the question is that I wanted to know whether the product could be defined using only arrows and Bjorn gave a great answer of "no." That actually supports the point you're trying to make! – Qiaochu Yuan Feb 9 at 6:41
	Ah, you didn't want to consider all of the other stuff that's attached to R. Then I retract my statement and downvote. – fpqc Feb 9 at 6:45
	And give a +1 – fpqc Feb 9 at 6:46

How can I define the product of two ideals categorically?

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