How can I really motivate the Zariski topology on a scheme? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T17:33:15Z http://mathoverflow.net/feeds/question/8204 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/8204/how-can-i-really-motivate-the-zariski-topology-on-a-scheme How can I really motivate the Zariski topology on a scheme? S1 2009-12-08T16:26:04Z 2010-02-06T17:18:53Z <p>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 <a href="http://sbseminar.wordpress.com/2009/08/06/algebraic-geometry-without-prime-ideals/" rel="nofollow">Secret Blogging Seminar</a>. But I could not find an answer to a question that bugged me right from my first steps in algebraic geometry: how can I <i>really</i> motivate the Zariski topology on a scheme?</p> <p>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? </p> <p>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. </p> <p>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...</p> http://mathoverflow.net/questions/8204/how-can-i-really-motivate-the-zariski-topology-on-a-scheme/8208#8208 Answer by karlpupil for How can I really motivate the Zariski topology on a scheme? karlpupil 2009-12-08T16:45:07Z 2009-12-08T16:45:07Z <p>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 $f \in R$. </p> <p>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 $V(I)=[p \in Spec \; R: I \subset p]$.</p> http://mathoverflow.net/questions/8204/how-can-i-really-motivate-the-zariski-topology-on-a-scheme/8213#8213 Answer by Ben Webster for How can I really motivate the Zariski topology on a scheme? Ben Webster 2009-12-08T17:21:45Z 2009-12-08T19:22:55Z <p><strong>EDIT:</strong> I'm rewriting this answer.</p> <p>So, let's accept for now that considering the points of an affine scheme as a set with no topological structure makes sense.</p> <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/8204/how-can-i-really-motivate-the-zariski-topology-on-a-scheme/8218#8218 Answer by Tim Carstens for How can I really motivate the Zariski topology on a scheme? Tim Carstens 2009-12-08T17:48:43Z 2009-12-08T17:48:43Z <p>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.</p> <p>Such a map <em>can</em> 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.</p> <p>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.</p> <p>Now observe that $f^{-1}(\mathfrak m)$ need not be maximal, even if $\mathfrak m$ is, but it <em>will</em> 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.</p> http://mathoverflow.net/questions/8204/how-can-i-really-motivate-the-zariski-topology-on-a-scheme/8221#8221 Answer by r. for How can I really motivate the Zariski topology on a scheme? r. 2009-12-08T17:57:17Z 2009-12-08T17:57:17Z <p>Here is what Eisenbud and Harris ('The Geometry of Schemes') have to say on this (page 10):</p> <p>[comments by myself are in square brackets]</p> <p>"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.</p> <p>For each subset S ⊂ R, let</p> <p>V (S) = {x ∈ Spec R | f (x) = 0 for all f ∈ S} = {[p] ∈ Spec R | p ⊃ S}.</p> <p>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."</p> http://mathoverflow.net/questions/8204/how-can-i-really-motivate-the-zariski-topology-on-a-scheme/8224#8224 Answer by Andrew Critch for How can I really motivate the Zariski topology on a scheme? Andrew Critch 2009-12-08T18:17:14Z 2009-12-08T19:01:22Z <p>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:</p> <p>Given a classical variety $V$ over $\mathbb{C}$ and a "regular function" $f:V\to\mathbb{C}$, one can identify the value <br>$f(x)$ 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:</p> <p>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. </p> <p>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): $f_a(p):=image_{A/p}(a)$.</p> <p>All domains/fields have the element $0$ in common, so it makes sense to talk about the vanishing set <br>$f_a^{-1}(0)=V(a):=$ {$p\in Spec(A) | a\in p$}, and these sets form a base for the closed sets of the Zariski topology. </p> <p>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</p> <ul> <li><p>$\bigcap_{a\in E} V(a) = V(E) :=$ {$p\in Spec(A) | E \subset P$}. These are the primes where <br>"all of $E$ vanishes" in the residue domain/field.</p></li> <li><p>$V(a)\cup V(b) = V(ab)$, the primes where $a$ and $b$ "both vanish".</p></li> </ul> http://mathoverflow.net/questions/8204/how-can-i-really-motivate-the-zariski-topology-on-a-scheme/8240#8240 Answer by Qiaochu Yuan for How can I really motivate the Zariski topology on a scheme? Qiaochu Yuan 2009-12-08T19:57:47Z 2009-12-08T19:57:47Z <p>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 <strong>embedding.</strong></p> <p>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.)</p> <p>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.</p> http://mathoverflow.net/questions/8204/how-can-i-really-motivate-the-zariski-topology-on-a-scheme/8281#8281 Answer by Jonathan Wise for How can I really motivate the Zariski topology on a scheme? Jonathan Wise 2009-12-09T00:38:44Z 2009-12-09T18:02:39Z <p>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.</p> <p>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.</p> <p>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 <strike>subfunctors</strike> subobjects) in which descent works for quasi-coherent sheaves. Does anyone know?</p> http://mathoverflow.net/questions/8204/how-can-i-really-motivate-the-zariski-topology-on-a-scheme/8860#8860 Answer by Greg Kuperberg for How can I really motivate the Zariski topology on a scheme? Greg Kuperberg 2009-12-14T08:59:22Z 2009-12-14T08:59:22Z <p>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.</p> <p>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. </p> <p>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 <em>all</em> ring homomorphisms between <em>all</em> 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.</p> <p>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.</p> <p>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.</p> <p>To summarize, the framework is a ruse to study all rings and ring homomorphisms in a geometric language.</p> http://mathoverflow.net/questions/8204/how-can-i-really-motivate-the-zariski-topology-on-a-scheme/14334#14334 Answer by Peter Arndt for How can I really motivate the Zariski topology on a scheme? Peter Arndt 2010-02-06T00:36:45Z 2010-02-06T17:18:53Z <p>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.</p> <p>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).</p> <p><strong>Edit:</strong> I rephrase that less sloppily, in response to Lars' comment. So the universal property is about <em>pairs</em> (Topos, ring object in it). A map (T,R)-->(T',R') between such is a pair </p> <p>(adjunction $f_*:T \leftrightarrow T':f^*$ , morphism of ring objects $f^*R'\rightarrow R$).</p> <p>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 $f^*$.</p> <p>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). </p> <p>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.(<strong>end of edit</strong>)</p> <p>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)):</p> <p>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.</p> <p>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).</p> http://mathoverflow.net/questions/8204/how-can-i-really-motivate-the-zariski-topology-on-a-scheme/14354#14354 Answer by Allen Knutson for How can I really motivate the Zariski topology on a scheme? Allen Knutson 2010-02-06T03:11:12Z 2010-02-06T03:11:12Z <p>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.</p> <p>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. (<a href="http://at.yorku.ca/t/o/p/c/10.htm" rel="nofollow">This guy</a> has a contrary viewpoint on the field of general topology, but unsurprisingly I find Grothendieck more convincing.)</p>