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First of all, I am aware of the questions about the Zariski topology asked here and I am also aware of the discussion at the Secret Blogging Seminar. But I could not find an answer to a question that bugged me right from my first steps in algebraic geometry: how can I really motivate the Zariski topology on a scheme?

For example in classical algebraic geometry over an algebraically closed field I can define the Zariski topology as the coarsest $T_1$-topology such that all polynomial functions are continuous. I think that this is a great definition when I say that I am working with polynomials and want to make my algebraic set into a local ringed space. But what can I say in the general case of an affine scheme?

Of course I can say that I want to have a fully faithful functor from rings into local ringed spaces and this construction works, but this is not a motivation.

For example for the prime spectrum itself, all motivations I came across so far are as follows: well, over an algebraically closed field we can identify the points with maximal ideals, but in general inverse images of maximal ideals are not maximal ideals, so let's just take prime ideals and...wow, it works. But now that I know that one gets the prime spectrum from the corresponding functor (one can of course also start with a functor) by imposing an equivalence relation on geometric points (which I find very geometric!), I finally found a great motivation for this. What is left is the Zariski topology, and so far I just came across similar strange motivations as above...

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Can you explain what you mean by imposing an equivalence relation on geometric points? My understanding is that this only makes sense for algebras over a field and presumably you want to talk about commutative rings in general. –  Qiaochu Yuan Dec 8 '09 at 18:18
    
Let k be a commutative ring and A a k-algebra. Let t:A -> K, t':A -> K' be two geometric points (i.e. K,K' are fields). Say that t and t' are equivalent if and only if there exists a third geometric point s:A -> L and k-algebra morphisms f:K -> L, f':K' -> L such that s = ft = f't'. By taking kernels of the morphisms t you get a bijection between the equivalence classes of this relation and the prime ideals of A. And this is my first motivation, why I take the spectrum of A! (I think of this as compressing the corresponding zero set functor into something which still contains all I want). –  user717 Dec 8 '09 at 18:48
    
I'm not sure I understand. If t, t' are not required to be surjective then their kernels are already prime ideals and not maximal ideals. If t, t' are required to be surjective then it is not possible to recover the prime ideals of A from its maximal ideals; consider the case A = k[[x, y]], k a field. Am I misunderstanding something? –  Qiaochu Yuan Dec 8 '09 at 19:08
    
Yes, I think you misunderstand something. It's precisely my point that the kernels are prime ideals, because this is what gives you the bijection between equivalence classes (wrt the relation described above) of geometric points and prime ideals! you can read about this (probably in exactly the same words) in the 1971 edition of EGA, introduction, nr. 13.! –  user717 Dec 8 '09 at 23:33
    
I just realized that normally 'geometric point' means that the field K is algebraically closed. I used the terminology of the introduction to EGA and there 'geometric point' is what I was talking about above...Perhaps this was the source of your confusion. Sorry. –  user717 Dec 9 '09 at 14:24
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10 Answers

Here is what Eisenbud and Harris ('The Geometry of Schemes') have to say on this (page 10):

[comments by myself are in square brackets]

"By using regular functions, we make Spec R [R being a arbitrary comm. ring with 1] into a topological space; the topology is called the Zariski topology. The closed sets are defined as follows.

For each subset S ⊂ R, let

V (S) = {x ∈ Spec R | f (x) = 0 for all f ∈ S} = {[p] ∈ Spec R | p ⊃ S}.

The impulse behind this definition is to make each f ∈ R behave as much like a continuous function as possible. Of course the fields κ(x) [the residual fields at x ∈ Spec R] have no topology, and since they vary with x the usual notion of continuity makes no sense. But at least they all contain an element called zero, so one can speak of the locus of points in Spec R on which f is zero; and if f is to be like a continuous function, this locus should be closed. Since intersections of closed sets must be closed, we are led immediately to the definition above: V (S) is just the intersection of the loci where the elements of S vanish."

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In the category of sets there is no such thing as the initial local ring into which 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).

Edit: I rephrase that less sloppily, in response to Lars' comment. So the universal property is about pairs (Topos, ring object in it). A map (T,R)-->(T',R') between such is a pair

(adjunction $f_*:T \leftrightarrow T':f^* $ , morphism of ring objects $f^*R'\rightarrow R$).

Note that by convention the direction of the map is the geometric direction, the one corresponding to the direction of a map topological spaces - in my "universal local ring" picture I was stressing the algebraic direction, which is given by $f^*$.

Now for a ring R there is a map $(Sh(Spec(R)), O_{Spec(R)})\rightarrow(Set,R)$: $f^* R$ is the constant sheaf with value R on Spec(R), the map $f^* R \rightarrow O_{Spec(R)}$ is given by the inclusion of R into its localisations which occur in the definition of $O_{Spec(R)}$. This is the terminal map (T,L)-->(Set,R) from pairs with L a local ring. For a simple example you might want to work out how such a map factors, if the domain pair happens to be of the form (Set,L).

This universal property of course determines the pair up to equivalence. It thus also determines the topos half of the pair up to equivalence, and thus also the space Spec(R) up to homeomorphism.(end of edit)

An even nicer reformulation of this is the following (even more high brow, but to me it gives the true and most illuminating justification for the Zariski topology, since it singles out just the space Spec(R)):

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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This is a cool answer, but it seems completely contrary to what the OP was asking for. –  Harry Gindi Feb 6 '10 at 0:41
    
In your first construction, would the "free local ring on R" also exist in the topos of sheaves on wrt the, say, fppf or etale topology on Spec R. If so, then this argument would not "motivate" the Zariski topology, since there is still an (a priori) arbitrary choice involved. Now if the Zariski sheaves on Spec R were in some sense a minimal topos with the property that the "free local ring on R" existed... –  Lars Feb 6 '10 at 8:49
    
Ok, I was sloppy. Zariski sheaves are the "minimal" topos in the sense now explained above. Intuitively you give the ring R just enough space to spread out into a collection of local rings. –  Peter Arndt Feb 6 '10 at 14:13
    
Thanks for the edit...this is pretty neat. –  Lars Feb 6 '10 at 16:50
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And by the way, in the same fashion the etale topos gives you a universal Henselian ring. I haven't heard of any such thing for fppf, though. –  Peter Arndt Feb 6 '10 at 17:27
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If you buy into the idea that you want a topological model for your ring, then right away it becomes sensible to ask that any map Ring -> Top be functorial. Of course, m-Spec -- which is already classically motivated -- doesn't lend itself to this, simply because there isn't an obvious way to use a ring homomorphism $f : A \to B$ to move maximal ideals around.

Such a map can move around ideals, both by extension and contraction, and this is a good first thing to investigate. Your choice of whether or not you want to push ideals forward or pull them back will determine if your functor should be co- or contravariant.

To decide between these two, look at the initial and terminal objects in Ring, as well as the initial and terminal objects in Top. The ring {0,1} has a single ideal, hence its topological space has (at-most) a single point, hence should probably be sent to the final object in Top. The 0 ring has no ideals, hence its topological space has no points, hence should be sent to the initial object in Top. Your hand has thus been forced: you need a contravariant functor, hence contraction is the thing to look at.

Now observe that $f^{-1}(\mathfrak m)$ need not be maximal, even if $\mathfrak m$ is, but it will be prime. You're thus immediately led to seeing if you can put a topology on Spec the same way you did for m-Spec. It works, and you move on.

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The way I think about contravariance is that one should always associate with an ideal I the homomorphism A \to A/I. Then it's the most natural thing in the world to take a homomorphism B \to A and compose it with a homomorphism A \to A/I. The reason prime ideals pull back to prime ideals is that if A/I is an integral domain then B \to A/I is still a map with image an integral domain; however, a subring of a field need not be a field, so maximal ideals don't pull back. –  Qiaochu Yuan Dec 8 '09 at 18:10
    
ring {0,1} = k? –  user2035 Dec 8 '09 at 18:14
    
Contravariance is also sensible in light of the ideal correspondence for quotients and localization, which suggest that the topological spaces associated to A/p and A_p should both be subspaces of the space for A, a fact which runs in the opposite direction of the canonical maps A->A/p and A->A_p. –  Tim Carstens Dec 8 '09 at 18:15
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Here is my favourite way to motivate the Zariski topology: it is the coarsest topology which makes the functions defined (below) by ring elements "continuous" in the following sense:

Given a classical variety $V$ over $\mathbb{C}$ and a "regular function" $f:V\to\mathbb{C}$, one can identify the value
$f(x)$ with the the image of $f$ in $A_V/m_x$, where $A_V$ is the coordinate ring of $V$ and $m_x$ is the maximal ideal at $x$. This perpsective has the advantage of generalizing to any ring, if you allow the target field to vary from point to point:

First, motivate working with primes instead of maximal ideals because primes pull back under ring maps, and because non-maximal primes act like "generic points" in classical algebraic geometry.

Next, at each prime ideal of a ring $p\triangleleft A$, you get a domain $A/p$ (which people often like to think of as living inside a residue field, $k(p):=Frac(A/p)$). Then an element of the ring $a\in A$ defines a function $f_a$ on $Spec(A)$ taking values in various domains (or fields): $f_a(p):=image_{A/p}(a)$.

All domains/fields have the element $0$ in common, so it makes sense to talk about the vanishing set
$f_a^{-1}(0)=V(a):=$ {$p\in Spec(A) | a\in p$}, and these sets form a base for the closed sets of the Zariski topology.

Moreover, the finite unions and arbitrary intersections we need turn out to be extremely manageable because of the definition of primes, in a way that is intuitively meaningful in the context above: For any collection of basic closed sets $V(a)$ with $a$ ranging over a set $E\subset Spec(A)$, we get

  • $\bigcap_{a\in E} V(a) = V(E) := $ {$p\in Spec(A) | E \subset P $}. These are the primes where
    "all of $E$ vanishes" in the residue domain/field.

  • $V(a)\cup V(b) = V(ab)$, the primes where $a$ and $b$ "both vanish".

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This is nice. Besides, this viewpoint allows to say every element of my ring is giving rise to a continuous function from Spec(A) to the corresponding domain/field. –  Csar Lozano Huerta Dec 10 '09 at 8:43
    
This looks like the easiest and most natural way to motivate Zariski topology. –  Fei YE Feb 6 '10 at 19:11
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EDIT: I'm rewriting this answer.

So, let's accept for now that considering the points of an affine scheme as a set with no topological structure makes sense.

Then each element of your ring either vanishes at a point (i.e. lies in the ideal) or doesn't. In any topology on this set that's compatible with the idea that ring elements are sections of some bundle on it, this zero set had better be closed.

By general topology nonsense, there is a unique coarsest topology where these are closed sets, the one that uses their complements as a basis (this exists because they cover, and the intersection of the sets for a and b is the set for ab). This is the Zariski topology.

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Hm, does this make sense? Do you mean morphisms into arbitrary local ringed spaces (whether this makes sense or not)? –  user717 Dec 8 '09 at 17:40
    
So for example, in my motivation for the Zariski topology on varieties over an algebraically closed field k, you first have to equip the field k with the Zariski topology to make this definition well-defined. So, you need some "initial data". –  user717 Dec 8 '09 at 17:45
    
Sorry if I am getting on your nerves :) But I don't see why this is well-defined. Because "scheme" already carries a topology (the Zariski topology) and for "coarsest" to make sense you somehow need a (non trivial) set of topologies on your scheme. "My" case above makes sense because you just consider morphisms into one fixed space already carrying a topology and so you consider all topologies on the domain satisfying a condition and then you take the coarsest one. –  user717 Dec 8 '09 at 18:53
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Note to passersby: comments above apply to an earlier version of the answer. –  Ben Webster Dec 8 '09 at 19:24
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I don't think that you have to motivate the Zariski topology as anything other than a correct description of something that can be defined without it.

Suppose from the beginning that you are interested in commutative rings $R$, in general. Suppose that you would like to interpret the reverse category of ring homomorphisms as "geometry". After all, many other kinds of geometry are reverse categories of ring homomorphisms, for certain kinds of rings and homomorphisms. For example, any smooth map $M \to N$ between smooth manifolds is equivalent to an algebra homomorphism $C^\infty(N) \to C^\infty(M)$ with certain favorable properties. To keep things as general as possible, and in a strictly algebraic setting, let's call any ring homomorphism $R \to S$ a geometric map in the opposite direction. Let's call the map $\text{Spec}(S) \to \text{Spec}(R)$, where for the moment "Spec" doesn't mean anything other than reversing arrows. Then this is the category of affine schemes with scheme morphisms as the morphisms.

Having taken the plunge to call this an abstract "geometry", we can try to fill in a tangible geometry. Certainly maximal ideals should be called points in this "geometry", given the motivating example of polynomial rings and their ring homomorphisms. (Proposition: A homomorphism between polynomial rings is equivalent to a polynomial map between affine spaces.) Should we perhaps stop at the maximal ideals? That would be nice, if it were consistent. However, having committed ourselves to all ring homomorphisms between all rings as "geometry", it isn't consistent. For example, $\mathbb{Z} \to \mathbb{Q}$ is an important ring homomorphism. However, the inverse of maximal ideal $\{0\}$ in the target is a prime ideal which is not maximal. As this example suggests, prime ideals are the smallest viable extension of the maximal ideals in the contravariant geometry of ring homomorphisms.

They aren't the only viable extension. We could have taken all ideals instead of just the prime ideals. As far as I know, another Grothendieck and another Zariski would have defined the points of an affine scheme using all ideals instead of just the prime ideals. In any case, the prime ideals work; the maximal ideals don't.

Okay, what about topology. I think that the Zariski topology is still, as you say, the coarsest $T_1$ topology available to be able to call regular maps, by definition the maps on prime ideals induced by ring homomorphisms, continuous.

To summarize, the framework is a ruse to study all rings and ring homomorphisms in a geometric language.

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You define points to be prime ideals and not just any ideals because you want to be able to localize at points, i.e. if $R$ is supposed to be the set of functions over the space of points $SpecR$, you want to be allowed to speak of the "local functions" around each point. –  Qfwfq Mar 8 '11 at 14:19
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Fields are characterized in the category of rings by the property that an epimorphism whose source is a field must be an isomorphism. The prime ideals of a ring can be identified with the epimorphisms from that ring to fields. From a categorical perspective, the prime spectrum therefore seems a more natural object than the max. spectrum.

I don't have a good explanation for the Zariski topology. It is the coarsest topology in which the maps induced by homomorphisms of rings are continuous and the origin in $\mathbf{A}^1$ is closed. This is not very satisfying, though.

The real reason the Zariski topology is important is because descent works for lots of things. For example, a module defined Zariski locally over a ring can be glued to give a module over the original ring. I wonder if the Zariski topology is the finest topology (i.e., finest Grothendieck topology defined by covers by subfunctors subobjects) in which descent works for quasi-coherent sheaves. Does anyone know?

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Isn't there faithfully flat descent for quasi-coherent sheaves? –  Dinakar Muthiah Dec 9 '09 at 7:47
    
Maybe you should ask your last question as a new question. It is very interesting, but Grothendieck topologies may not be part of the original question here. –  Konrad Voelkel Dec 9 '09 at 15:03
    
Dinakar: yes, but the fppf topology is not defined by subobjects (I changed the wording above to emphasize this). The induced topology defined by subobjects is the Zariski topology (since an fppf embedding is an open embedding in the Zariski topology). –  Jonathan Wise Dec 9 '09 at 18:03
    
Ah, I see your distinction. –  Dinakar Muthiah Dec 10 '09 at 0:00
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think there are two questions here: (1) why study the prime spectrum, and (2) why think of it in terms of the axioms for a topology. (1) has been pretty well handled by other commenters. And a number of them point out that (2) isn't really especially useful.

Part of the problem, according to a very interesting article I read by Grothendieck (maybe where he introduces dessins d'enfants?), is that the axioms for topological spaces are "wrong". Alas, he doesn't know what the right axioms are; he's just sure that the field of general topology should never have existed. From that point of view, discovering that the prime spectrum has a topology automatically isn't that interesting. (This guy has a contrary viewpoint on the field of general topology, but unsurprisingly I find Grothendieck more convincing.)

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If you (or somebody else) can find that article, that would be awesome. –  Kevin H. Lin Feb 6 '10 at 7:50
    
to Kevin: I heard from my advisor, he said the definition of grothendieck topology(not pretopology) "is equivalent to" all the possible existing topology in various branches of mathematics –  Shizhuo Zhang Feb 6 '10 at 12:41
    
Steven Gubkin found the article; see mathoverflow.net/questions/14634/what-is-a-topological-space –  Allen Knutson Feb 8 '10 at 15:07
    
This article, and all of Grothendeick's writing, have just recently been removed from the Grothendieck Circle's webpage, apparently on Grothendieck's request (Wikipedia says the request was made in January 2010 in a letter from Grothendieck to Illusie). Too bad. –  Dan Ramras Jun 17 '10 at 5:00
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Grothendieck's article can be found here: dl.dropbox.com/u/1963234/EsquisseEng.pdf –  Gunnar Magnusson Nov 14 '10 at 1:50
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This is probably a standard question, so allow me write down (what I think) the standard answer. Viewing any ring $R$ as a ring of functions (allowing nilpotents and all that) on the prime spectrum $Spec R$, you naturally want all such functions (elements of $R$) to be continuous, thus it needs that, for any $f \in R$, $V(f) = [p \in Spec\; R: f(p) = 0 ] = [p \in Spec \; R: f \; mod \; p = 0] = [p \in Spec \; R: f \in p]$ (I used [ ] to denote a set...don't know why { } doesnt work)) being a closed set, where $p$ is a prime ideal and the 'value' of a function $f \in R$ at a point $p$ is the image of the residual of $f$ mod $p$ in the field of fractions of $R/p$ (which is an integral domain). I think there's no difficulty in showing that the field of fraction of $R/p$ is isomorphic to the residual field $k(p)$ of the local ring $O_p$, coinciding with the other definition of the 'value' of function $f \in R$.

Now any ideal $I$ of $R$ is generated by its elements, and so in order for a bunch of functions to be continuous, it needs to have the closed set $V(I)=[p \in Spec \; R: I \subset p]$.

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Hm, what do you really mean by continuity of f? I mean, f is in general not a proper function into some topological space!? –  user717 Dec 8 '09 at 16:50
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It's really a motivation for Zariski topology on an affine scheme. If you look at affine varieties over a algebraically closed field $k$, then as you say you can see why Zariski is motivated; similar here, although the only difference is that the residual fields are different (and can be of different characteristics), unlike a unique field $k$ for the "functions" to take values. These are heuristics anyway, you may not need to take the motivations too seriously - otherwise why bother a formal theory. –  user2148 Dec 8 '09 at 16:57
    
But when your "functions" aren't functions, then this motivation is just wrong, or am I wrong? Of course, I know about this intuition, but for exactly this reason I don't know what I should do with it... –  user717 Dec 8 '09 at 17:42
    
I don't think so. If you want to think of topological spaces in terms of their ring of functions, then for continuous real-valued functions into R a topological space has the initial topology if and only if it's completely regular, and this is equivalent to the zero sets of continuous functions being a basis for the closed sets. So the Zariski topology makes Spec R behave as if it were completely regular in the sense that elements of R separate points and closed sets. –  Qiaochu Yuan Dec 8 '09 at 17:49
    
You can take a look at James Milne's notes on AG. It uses the ringed space to define an affine variety over an algebraically closed field, and the structure sheaf indeed consists of regular functions (from the variety to the filed) on Zariski open subsets. The motivation from this example is clear enough, and the functions here are indeed functions. –  user2148 Dec 8 '09 at 17:52
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Here's an idea related to Tim Carstens' answer. As in Ben's answer we start from the point of view that it makes sense to think of $\text{Spec } R$ as a set. Given an ideal $I$ we have a homomorphism $R \to R/I$ and by the correspondence theorem the prime ideals of $R/I$ are precisely the prime ideals of $R$ containing $I$, so we get an injection $\text{Spec } R/I \to \text{Spec } R$. In any reasonable assignment of a topology to the spectrum this injection should be an embedding.

The Zariski topology accomplishes this by the simple requirement that the above map be both closed and continuous. Why is being closed a reasonable requirement? Well, an embedding of a compact Hausdorff spaces into another is always closed, so if we think of the relationship between $\text{Spec } R$ and $R$ as analogous to the relationship between a compact Hausdorff space $X$ and its C*-algebra then this is a natural requirement. (This is similar to the comment I made about completely regular spaces, so if you don't like that reasoning you probably won't like this argument either.)

This is perhaps a little unsatisfying until it's shown that there are no other reasonable ways to turn the map $\text{Spec } R/I \to \text{Spec } R$ into an embedding (certainly the discrete topology works, but I wouldn't call that reasonable), but hopefully somebody else has some insight here.

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