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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?

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Epis or surjections? –  Mariano Suárez-Alvarez Feb 9 '10 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 '10 at 5:51
You want them to be epis of the category R-Mod. –  Harry Gindi Feb 9 '10 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 '10 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 '10 at 6:24
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2 Answers

up vote 46 down vote accepted

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



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



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.

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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. –  Harry Gindi Feb 9 '10 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 '10 at 6:41
@Harry: The topological space does not need ideal products since $V(IJ)=V(I \cap J)$ as a set, and the structure sheaf is just defined on the open complements which are also equal as schemes. –  Martin Brandenburg Jun 29 '11 at 0:59
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Bjorn's answer shows that the product of two ideals cannot be defined by means of the partially ordered set of ideals. However, there is a category-theoretic definition of the product of two ideals if we allow us to use the category of modules. This works because of the Rosenberg reconstruction Theorem.

Explicitely, there is a bijection between ideals of $R$ and reflective, topologizing subcategories of $\text{Mod}(R)$, which is given by $I \to \{M \in \text{Mod}(R) : I M = 0\}$. If $T$ is such a subcategory with reflector $F : \text{Mod}(R) \to T$, then the corresponding ideal is $I=\ker(R \to F(R))$. Now, the multiplication of ideals corresponds to the so called Gabriel product: If $S,T$ are subcategories of an abelian category, then $S \bullet T$ is the subcategory which consists of those objects $M$ for which there is an exact sequence $0 \to M' \to M \to M'' \to 0$ with $M' \in T, M'' \in S$. The idea for the notation is $(S \bullet T)/T = S$.

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