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I like to think in terms of commutative diagrams rather than referring to elements. So to me a group is really a group object, i.e. an object with some maps satisfying certain commutative diagrams. You can define rings and modules similarly. You can define a field object as a commutative ring object with the inverse axiom from a group object applied to the multiplication (alternately as a commutative ring object over which every module is free). Assuming your category is abelian, one would hope that you could define a maximal ideal $M$ by an exact sequence $0 \rightarrow M \rightarrow R \rightarrow k \rightarrow 0$ where $k$ is a field object. I'm thinking ahead to generalizing this to a triangulated category where $M$ could be uniquely defined via a fiber sequence.

I’d like to define an ideal $I$ as a subobject of $R$ which is also an $R$-bimodule. Both of these can be defined by diagrams, without reference to elements. However, I'm a bit afraid that this is completely wrong since I've never heard of an "ideal object", and it seems Google has not either. Is there something I’m missing here that makes my definition fail?

The next object I’d like to define is a prime ideal. I have no idea how to do this without referring to elements. The trick with maximal ideals above doesn’t seem to work unless someone has a way to classify “integral domain objects” via diagrams.

Has anyone ever heard of a way to define prime ideals purely diagrammatically?

Alternately, assuming that I totally botched that attempt to define an "ideal object," one could still ask about prime modules, i.e. nonzero modules $M$ such that $IN = (0)$ implies $N = (0)$ or $IM = (0)$ for any ideal $I$ of $R$ and any submodule $N$ of $M$. Has anyone heard of a diagrammatic way to define these?

I'm tagging this Topoi because an answer of Peter Arndt to a totally different MO question gives me a small amount of hope that some of the mysteries of Topos Theory could help me.

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The category in which you work seems to vary rather wildly. In the title, it's the category of rings (which are apparently not required to have a 1 element, so that ideals can be viewed as objects there). The first couple of sentences of the text seem to be about general categories with finite products; I don't think you want, for example, group objects in the category of rings. A few sentences later, you're looking into an abelian category (therefore definitely not the category of rings). The net effect of this is that I can't really tell what you want. –  Andreas Blass Jun 28 '11 at 4:46
I agree with Andreas Blass. I interpreted the question as following: Find a categorical description of the ring homomorphisms of the form $R \to R/\mathfrak{p}$, where $\mathfrak{p}$ is a prime ideal, within the category of unital, commutative rings. –  Martin Brandenburg Jun 28 '11 at 8:30
If you really want to talk about ideals themselves from a category theory perspective (and really, why would you want to? they don't lend new light on the theory of abelian categories, and Martin's answer probably accomplishes whatever you application is), then their correct categorical home is the category of modules. Note that this category is better than a category: it comes equipped with a distinguished object (the free rank-one module $R$). An ideal is a subobject thereof. But from this perspective, maps to $R$ that need not be mono are also natural, for example, or maps from. –  Theo Johnson-Freyd Jun 28 '11 at 12:35
Thanks all for the comments. I now realize I was far too fast and loose with which category I was thinking about, but I really appreciate that people still gave me great answers. The application I have in mind is a unital (often commutative) ring objects in the category of spectra. I can see already how to make Theo's comment work there, though I'm also going to spend some time trying to make Martin's excellent answer work in that setting. –  David White Jun 28 '11 at 13:11
I have a slight problem with your notion of field object. For example, take k a field. Then, I guess, you want the affine line A to be a field object over k. But if I take any k algebra R which is not a field, then A(R) equals R which is, well, not a field. –  Arno Kret Jun 28 '11 at 13:19
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up vote 6 down vote accepted

The lattice of ideals of $R$ is isomorphic to the lattice of regular quotients of $R$. Here, a regular quotient is an equivalence class of regular epimorphisms $R \to S$ in the category of rings (which are precisely the surjective ring homomorphisms). So this also serves as a categorical definition of an ideal. Also the ideals of $R$ are exactly the subobjects of $R$ in $\text{Mod}(R)$, but I think that you want to keep in the category of rings, right? Now, a prime ideal is characterized by the fact it is a proper ideal $\mathfrak{p}$ and for all ideals $\mathfrak{a}, \mathfrak{b}$ we have that $\mathfrak{a} * \mathfrak{b} \subseteq \mathfrak{p}$ implies $\mathfrak{a} \subseteq \mathfrak{p}$ or $\mathfrak{b} \subseteq \mathfrak{p}$. Unfortunately, there is no lattice-theoetic definition of the product of ideals, see this previous MO question, but there I give a definition which comes from ideas of Rosenberg's noncommutative algebraic geometry and uses subcategories of $\text{Mod}(R)$.

However, you can also give the following characterization within the category of rings: A prime ideal corresponds to a regular quotient $R \to S$ where $S$ is an integral domain. Now an integral domain is characterized by the property that it embeds into a field, where embeddings are given by monomorphisms (which actually coincide with the injective ring homomorphisms as in every algebraic category). A field $K$ is characterized by the following property: $K \neq 0$, where $0$ is the terminal object, and the lattice of ideals has exactly two elements, namely $0$ and $K$.

You may also consult the overlooked book "Categories of commutative algebras" by Yves Diers. Here the author developes all the basic commutative algebra for so-called Zariski categories, which are special locally presentable categories which resemble categories of commutative rings. In the chapter "Classical objects" we defines integral objects etc. in every Zariski category. Namely: First a simple object is defined to have exactly two congruences, and an integral object is defined to be one which embeds into a simple one.

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Nitpick for the second sentence: Isn't the inclusion of $\mathbb Z$ into $\mathbb Q$ a typical example of a ring epimorphism which isn't surjective? –  Peter Samuelson Jun 28 '11 at 2:10
Yes but regular epimorphisms coincide with surjective ring homomorphisms. –  Martin Brandenburg Jun 28 '11 at 2:16
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