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I was reading this question on why algebraic geometry looks at prime ideals instead of only maximal ideals, and I understand Anton's answer, but I'm a little confused as to how this fits with Hilbert's Nullstellensatz - affine algebraic sets are in bijection with radical ideals, not prime ideals, and it seems like we'd want the extra information we'd get by looking at "RadSpec(R)" (my own imagined notation). Also, the preimage of a radical ideal is radical, so there isn't the same objection as to maximal ideals - "RadSpec" would also be a contravariant functor.

So, why not radical ideals instead of only prime ideals, and what kinds of things could we say about RadSpec(R) even if they aren't very interesting?

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This is a bit vague. The others have mentioned what they know about the radical ideals, and to what extent they relate to the Zariski topology. But what do you want to know about the radical ideals? – Martin Brandenburg Feb 6 '10 at 0:23
up vote 18 down vote accepted

The short answer is that every radical ideal is the intersection of the prime ideals containing it, so that the pullback map on Specs determines the pullback map on your wouldbe RadSpecs.

Note that a similar reason underlies the success of looking only at maximal ideals in classical algebraic geometry: a finitely generated algebra over a field is a Hilbert-Jacobson ring: every radical ideal is the intersection of the maximal ideals containing it.

I won't rule out the possibility that RadSpec could be useful for something, though...

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Is anything like RadSpec used in noncommutative geometry? – Harry Gindi Feb 5 '10 at 21:26
@HG: In case that question is directed at me: I haven't the faintest idea. If someone else knows, it would be reasonable to put that information in an answer to this question. – Pete L. Clark Feb 5 '10 at 21:29

The space Spec(R) has a universal property:

In the category of sets there is no such thing as the initial local ring into which some given ring R maps, i.e. a local ring L and a map f:R-->L such that any map from R into a local ring factors through f.

But a ring R is a ring object in the topos of Sets. Now if you are willing to let the topos vary in which it should live, such a "free local ring on R" does exist: It is the ring object in the topos of sheaves on Spec(R) which is given by the structure sheaf of Spec(R). So the space you were wondering about is part of the solution of forming a free local ring over a given ring (you can reconstruct the space from the sheaf topos, so you could really say that it "is" the space).

An even nicer reformulation of this is the following (even more high brow, but it nicely singles out the space):

A ring R, i.e. a ring in the topos of sets, is the same as a topos morphism from the topos of sets into the classifying topos T of rings (by definition of classifying topos). There also is a classifying topos of local rings with a map to T (which is given by forgetting that the universal local ring is local). If you form the pullback (in an appropriate topos sense) of these two maps you get the topos of sheaves on Spec(R) (i.e. morally the space Spec(R)). The map from this into the classifying topos of local rings is what corresponds to the structure sheaf.

Isn't that nice? See Monique Hakim's "Schemas relatifs et Topos anelles" for all this (the original reference, free of logic), or alternatively Moerdijk/MacLane's "Sheaves in Geometry and Logic" (with logic and formal languages).

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That is indeed nice! – Tom Leinster Feb 9 '10 at 7:22

Let me elaborate a little on Pete's answer from a perspective which I briefly outlined in this blog post. The point of looking at affine schemes as opposed to commutative rings is that the category of affine schemes behaves more like $\text{Set}$ than $\text{CRing}$ does: the product distributes over the coproduct instead of the other way around, for example. That means that $\text{CRing}$ should be thought of as a generalization of $\text{Set}^{op}$, one incarnation of which is the category of Boolean rings a certain subcategory of the category of Boolean rings.

Now, there is a general construction in category theory called "subobject" which I describe in the above post. Applied to $\text{Set}$, it gives the lattice of subsets. Applied to $\text{CRing}^{op}$, it gives the lattice of ideals. So, the ultimate goal of a "geometric" approach to commutative rings should be to give a category equivalent to $\text{CRing}^{op}$ whose objects have "points" with the property that the lattice of ideals behaves like the lattice of subsets of these points (so we can apply intuition from $\text{Set}$). This is quite wrong. But one still gets the lattice of ideals as a certain sublattice. More precisely, given a ring $R$ its subobjects in $\text{CRing}^{op}$ are epimorphisms $R \to S$. I wanted these to correspond to quotients $R \to R/I$, but this is wrong. But it seems that replacing "epimorphism" with a stronger definition like extremal epimorphism is supposed to work.

This approach makes the most sense in the subcategory of reduced rings, where the subobject construction gives the lattice of radical ideals, which is as close to the lattice of subsets of the prime ideals as you're going to get. (It's not 100% there because primes can contain each other.) And in the further subcategory of finitely-generated reduced rings over an algebraically closed field the subobject construction gives the lattice of algebraic subsets of a given algebraic set. The point I'm trying to make here is that you shouldn't think of radical ideals as points, but rather as a natural candidate for "subsets of points," which should then lead you to a definition of "point." From this perspective, for reduced rings it makes good sense to think of the prime ideals as the correct candidate for "points," and for reduced Jacobson rings it makes good sense to think of the maximal ideals as the correct candidate for "points." Whether this is the best approach for all commutative rings is a matter of taste, I think.

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Greg Kuperberg's answer to this question is also, I think, relevant:… – Qiaochu Yuan Feb 5 '10 at 22:03
Thanks for an informative answer - I like how the subobject concept unifies these things. I've posted a separate community wiki question about lattice theory books, but do you have a good reference for learning about this stuff? – Zev Chonoles Feb 6 '10 at 1:43
Your answer is very interesting, but I found it difficult to understand. One immediate suggestion: italicize the "The point I'm trying to make" sentence. But also, I'd appreciate an expansion of these thoughts. – Ilya Grigoriev Feb 9 '10 at 3:52
I've italicized two sentences. Which parts would you like expanded? I have to admit I have only really thought about the most basic aspects of the subject like the definitions. – Qiaochu Yuan Feb 9 '10 at 4:02
"you shouldn't think of radical ideals as points, but rather as a natural candidate for 'subsets of points'" I'd like to stress this sentiment. For any commutative ring (not necessarily reduced or finitely generated), the frame of open subsets of $\mathrm{Spec}A$ is canonically isomorphic to the frame of radical ideals of $A$. There is a notion of "point of a frame" (rather "point of the induced locale"). The points of this frame are in canonical bijection with the prime filters of $A$. If classical logic is available, then these in turn are in canonical bijection with the prime ideals of $A$. – Ingo Blechschmidt Nov 30 '15 at 18:09

Pete's answer gave a very good reason, and indeed contained what I planned to say at the beginning of this post.

However, let me provide some mildly positive news, namely some cases when we are forced to work with only radical ideals: In many situations involving determinantal rings or ideals defining the commuting matrices, one would like to know if certain ideals are radical. For example, this is useful in proving that ideals of minors are perfect, which implies that their resolution is as short as it can be.

One of the useful methods (the other uses algebra with straightening law) has been Hochster's principal radical system, which proves an ideal is radical by inductively proving a whole set of ideals are radical! The precise statement, and some applications to algebraic statistic, was provided in this paper, Section 7 by Kirkup. I think one of Hochster's lecture notes contains a more general and formal treatment, but I forgot which one (sorry!). ADDED: I found it, it is Math 711, Fall 2005, it was discussed starting on the Lecture of Sept 9.

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Hochster assigned the Jacobian conjecture as a bonus problem this week =\. – Harry Gindi Feb 5 '10 at 22:36
Ah! Good to see he is going strong! – Hailong Dao Feb 5 '10 at 23:13

The category of reduced $k$-algebras over a field $k$ is an example of a so-called Zariski category; see the book "Categories of commutative algebras" by Yves Diers. In every Zariski category, basic notions of commutative algebra can be developped and over every Zariski category $A$ we can consider the category of $A$-schemes. The theory shows that, in fact, most of the basic algebraic geometry takes over to this general setting. And if $A$ is a so-called rational reduced Zariski category, it turns out that the basic notions of varieties take over (such as birational equivalence is detected by the function field object, and Hilbert's Nullstellensatz). If $A$ denotes the category of reduced $k$-algebras, we get the usual category of reduced $k$-schemes. So in some sense, this is an algebraic geometry in its own right which you can study. The functor of points approach also works with functors $A \to \text{Sets}$. Note, however, that here the spectrum of a reduced $k$-algebra also consists of the usual prime ideals, not all radical ideals. Nevertheless, you may read Section 2.4 in Dier's book. There radical congruences in Zariski categories are studied in general.

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