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In a comment to this question the quotation "The Zariski Topology is the 'Wrong' topology for Algebraic Geometry" appears.

Well, so some spontaneous questions arise:

1) What is Zariski topology good for in algebraic geometry? In other words, what can you do with it, without referring directly to some finer Grothendieck topologies?

2) On the other hand, which concepts which are analogs of concepts in -say- analytic geometry or topology, really need finer Grothendieck topologies to be generalized to the algebro geometric setting?

I think one could mention vector bundles for the 1), and principal bundles and projective bundles for 2). And also cohomology for 2), since it seems that the étale topology is best suited for reproducing a cohomology that resembles the singular one in the analytic setting. And also sheaf cohomology for 1).

Edit: I think it would also be nice if in the answers there were some brief comments about the sufficiency (or not) of Zariski topology to capture the essential geometric picture borrowed from analytic geometry and/or topology about (some of) the following contexts:

  • Patching of morphisms
  • Inverse function theorem & implicit function theorem
  • Existence of tubular neighbourhoods (?)
  • Local reducubility vs. global reducibility (analytic branches of a variety at a point..)
  • "Infinitesimal" properties given by the local ring at a point (Zariski vs. Hensel)
  • Vector bundles
  • Coherent sheaves (and patching thereof...)
  • Principal bundles (local triviality...)
  • Projective bundles (as above...)
  • In general, "locally trivial" fiber bundles (isotriviality vs. triviality...)
  • Sheaf cohomology with constant coefficients vs. of coherent sheaves
  • Cech cohomology (as above...)
  • Covering spaces and fundamental group
  • Extracting topological/homotopical information via coverings
  • (any other topic you find interesting)
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  • $\begingroup$ I heard somewhere (although I don't remember if it was a reputable source) that the étale topology is a wholesale replacement for the Zariski topology for pretty much every use. Could someone elaborate on whether or not this is even true? $\endgroup$
    – Harry Gindi
    Commented Apr 15, 2010 at 23:16
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    $\begingroup$ No way, not true. Zariski topology is useful for doing actual computations, constructing analytifications, formal schemes (for formal GAGA), open Bruhat cell, applications to commutative algebra, and so on. Proofs in etale cohomology use Zariski-stratifications. Useful to have actual space in which can make affine opens to do computations. Just look at the proofs of most serious theorems in etale sheaf theory. If one never reads proofs of hard theorems it may seem otherwise, but ignoring Zariski topology sounds bad. Each has its own uses. It's algebraic geometry. $\endgroup$
    – BCnrd
    Commented Apr 15, 2010 at 23:48
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    $\begingroup$ Thanks for the clarification. I knew it sounded suspect. I think I read it on Wikipedia, which is never ever ever wrong ;). $\endgroup$
    – Harry Gindi
    Commented Apr 15, 2010 at 23:51
  • $\begingroup$ Let me tack on to BCnrd's answer the trivial statement that one cannot even talk about etale morphisms without having already developed a theory of algebraic geometry based on the Zariski topology. For example, the definition of etale mentions the Zariski local rings. $\endgroup$
    – Ryan Reich
    Commented Jun 29, 2010 at 15:06
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    $\begingroup$ @Ryan: That's definitely not true. A morphism of rings is etale if it is formally etale and finitely presented. It is formally etale if it has the unique LLP with respect to all quotients by a square-zero ideal. $\endgroup$
    – Harry Gindi
    Commented Jul 31, 2010 at 18:56

6 Answers 6

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The Zariski topology is part of the basic structure of varieties and schemes. Unlike other, fancier Grothendieck topologies, it is actually a topology, defined by subsets of the variety/scheme, and so gives rise to the notion of closed, as well as of open, subset. The closed subsets are the algebraic subsets (in the variety case) or (the spaces underlying) the closed subschemes (in the scheme case), which are what the study of algebraic geometry is to a large extent about.

If you look at Hartshorne, Chapters IV and V, you will find a lot of geometry of curves and surfaces (as well as some geometry of higher dimensional varieties too), all of it treated without recourse to any topology other than the Zariski topology.

Added: A good example, in my opinion, is the (direct) proof that the addition law on a smooth plane cubic curve is associative. (By direct, I mean the proof working from the definition of the group law in terms of collinear points, not the more involved proof that proceeds by identifying the curve with its Picard variety.) It is not hard to see that associativity holds when the three points being added are in suitably general position. To conclude, one can either make complicated special cases arguments in the various non-generic situations, or, one can make a continuity argument in the Zariski topology. The latter argument is simple (in that it uses the most basic kinds of arguments about continuous maps in general topology) and decisive. It illustrates perfectly the role of the Zariski topology in geometric arguments.

One could generalize from this example as follows: two of the most fundamental notions of algebraic geometry are the complementary concepts of general and special position, and these notions are precisely what the Zariski topology captures (just as the topology on a metric space captures the notion of closeness in the metric).

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    $\begingroup$ Harry, here's a "definition" of non-negativity in $\mathbf{Z}$: being a sum of 4 perfect squares. Do you think it is reasonable to develop number theory based on that definition? Suggesting one can develop the notion of closedness via etale topology by avoiding the Zariski topology makes about as much sense as that. $\endgroup$
    – BCnrd
    Commented Apr 16, 2010 at 3:22
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    $\begingroup$ @Brian: I think your example, if anything, makes too much sense. (It is known to have some model-theoretic consequences.) How about this? We don't need to define the real (or complex, or p-adic numbers) explicitly, because they can be defined as quotient rings of the adele ring of a global field. This is presumably possible (non-circularly, I mean) but is clearly looking through the wrong end of the telescope. Just because something is logically possible doesn't make it a good approach to thinking about the subject, at any level. $\endgroup$
    – Pete L. Clark
    Commented Apr 16, 2010 at 4:11
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    $\begingroup$ Pete, even though that characterization of non-negativity has consequences, I think it's a perfectly good illustration of the last sentence of your comment. To be honest, I doubt it is even logically feasible in the sense that one couldn't really develop elementary number theory on the basis of such an absurd definition. $\endgroup$
    – BCnrd
    Commented Apr 16, 2010 at 4:15
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    $\begingroup$ Dear Harry, I'm not sure that "just as easily" has the seem meaning for me as for you. $\endgroup$
    – Emerton
    Commented Apr 16, 2010 at 4:51
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    $\begingroup$ Dear Harry: We know how these defns go for etale top. But a defn not saying "Zariski topology" doesn't allow do much interesting without many results proved in Zar. top. Many results for alg. spaces are proved via reduction to schemes with Zar. topology. Real test is not making defns, but proving nontrivial thms. For lots in geometry of varieties, need Zariski-local rings, not henselian local rings (e.g., function field is local ring at generic point for Zar. top., and irreducibility is not etale local.) You would be well-served studying deeper theorems instead of more formalism. $\endgroup$
    – BCnrd
    Commented Apr 16, 2010 at 12:37
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While the Zariski topology has its limitations, it amazes me how well it does work. A few brief points in its defense:

  1. It's easy to define. In the classical case of affine spaces over a field, it's the weakest topology for which points are closed and polynomials are continuous.

  2. It can be used to give precise meaning to the word "generic" or "general", as in "a general matrix is diagonalizable, therefore to prove Cayley-Hamilton it suffices..."

  3. For coherent sheaves, it's the right topology; cohomology works as expected. At a more sophisticated level, cohomology is upper semicontinuous in the Zariski topology, and this is very important for many arguments.

So for the younger generation out there who are thinking of doing away with it: please don't!

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  • $\begingroup$ Is sheaf cohomology of coherent sheaves always the same in the Zariski topology and in finer topologies? $\endgroup$
    – Qfwfq
    Commented Apr 16, 2010 at 12:58
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    $\begingroup$ Yes for etale etc. [cf. Milne p 114]. Morally yes for analytic topology of a proper variety over C by GAGA. $\endgroup$
    – Donu Arapura
    Commented Apr 16, 2010 at 13:08
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    $\begingroup$ As I have pointed out several times on MO already, to completely prove Cayley-Hamilton directly it suffices to write one line, $$p_A(\lambda)=(A-\lambda)\text{adj}(A-\lambda), \lambda\to A \implies p_A(A)=0,$$ whereas to prove that a general matrix is diagonalizable requires a lot more. I am just wondering about the origin of this misapprehension, apparently common among a certain generation of algebraic geometers (or AG students) at MO, that linear algebra results are to be proved in a roundabout way with AG tools: is there a canonical textbook that advocates this philosophy? $\endgroup$
    – Victor Protsak
    Commented Jul 31, 2010 at 23:08
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    $\begingroup$ Victor, while I don't disagree with you that Cayley-Hamilton can and should be approached directly when teaching linear algebra, I still maintain that it is an instructive use of the Zariski topology. I don't know where other people learned this, I found it for myself while teaching a beginning algebraic class back in the mid 90's. I might have given as a homework problem. $\endgroup$
    – Donu Arapura
    Commented Aug 3, 2010 at 14:46
  • $\begingroup$ @Victor: my apologies, I don't know if you're even checking MO anymore, but I don't see how this proof works. $\lambda$ is a scalar and $A - \lambda$ is $A$ minus a scalar matrix, it doesn't make any sense to substitute $A$ for $\lambda$ on the RHS. Doesn't this run into exactly the same problem as the naive proof where we substitute $\lambda = A$ directly into $\det(A - \lambda)$? $\endgroup$
    – Qiaochu Yuan
    Commented Jun 16 at 21:27
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As Kevin said, the higher cohomology groups of constant sheaves on irreducible varieties are zero when working with the Zariski topology. Also, "fibre bundles aren't locally trivial" and "the inverse mapping theorem fails". Instead of writing about the last two myself, let me refer you to the wonderful notes of fellow MOer JS Milne on étale cohomology (page 10, The inadequacy of the Zariski topology)

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(I guess this is an elaboration on one of your comments in your last paragraph.)

One reason why the Zariski topology is "bad" is that the higher cohomologies ($H^i$ for $i > 0$) of a constant sheaf on an irreducible space (in particular irreducible varieties with the Zariski topology) are zero, because such a sheaf will be flasque (exercise!).

I guess this is one of the motivations for the etale topology and for etale cohomology. The etale cohomology of (certain) constant sheaves will be nontrivial, indeed, it will coincide with the singular cohomology of the associated analytic variety. This is the "Comparison Theorem", see for example Milne's book on etale cohomology for the details.

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algebraic geometry is not my field at all, but it seems like this is a bit of an answer, and i don't really know the details but maybe someone can fill them in. The Zariski topology is not good for doing homotopy theory. I have heard this at many seminars, specifically from someone who does motivic homotopy theory. So from that perspective it is not the right topology, but i can only say that i have heard this not that i understand why the Zariski topology is bad. I guess just naively it seems like it would be pretty hard doing homotopy theory on "any" line where your only open sets are finite complements.

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In light of your edit, Remark 2.1.2 of these notes of Christoph Sorger might be useful.

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