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I'm interested in the arithmetic/diophantine equation applications of arithmetic/algebraic geometry. From what I understand, many of the difficult/technical aspects of the latter theories (sheaves, cohomologies, schemes, ...) are due to the desire to access continuity. The benefits of this continuity must be great due to the substantial difficulties in setting it up, but what are they? For example, I have read that Grothendieck's cohomology was crucial in the solution of the Riemann hypothesis for varieties over finite fields, but why is continuity so important for this result?

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I don't really understand the question. For instance, the Zariski topology on a variety over a finite field is the discrete topology, so indeed (Zariski-)continuity is not important at all: all functions are automatically continuous. Are you perhaps trying to ask about etale cohomology? Do you realize that this is not a cohomology theory on topological spaces but rather a cohomology theory with respect to sheaves on a Grothendieck topology? (Here the fact that algebraic maps induce "continuous" morphisms on the etale sites is quite straightforward...) –  Pete L. Clark May 26 '10 at 3:47
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Pete, varieties of positive dimension over any field (finite or otherwise) never have the discrete topology; think of the generic points or even the affine line. Even if abusing terminology and ignoring non-closed points (which I assume you wouldn't do), it's still non-discrete (since inclusion of set of closed points with subspace topology is a "quasi-homeomorphism", as for any Jacobson scheme). Maybe you're referring to just the finite set of rational points (which has no useful geometric structure at all)? Not sure. –  BCnrd May 26 '10 at 4:46
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Pete, what is something useful done with Zariski topology on set of rational points over a field that is not separably closed? (Statements that rational points constitute a Zariski-dense set in the underlying $K$-scheme are important, but that's an entirely separate matter.) It sounds dangerous, since there can be plenty of smooth closed subvarieties of the $K$-scheme which have no $K$-points. (Springer's book makes some errors related to that.) What's an example where it's better to think in such terms instead of "rational points in a $K$-scheme"? –  BCnrd May 26 '10 at 6:10
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P.S.: I am not saying that there is any situation in which this relatively naive language covers more ground than the scheme-theoretic language: it is an easy exercise in the "soberification" of a topological space to see that the two points of view are consistent. But there is some merit in not speaking of scheme-theoretic closure unless one really needs to: there are an order of magnitude more mathematicians who can understand "coarsest topology on $K^n$ making the polynomials continuous" than "the scheme-theoretic closure".... –  Pete L. Clark May 26 '10 at 7:45
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@OP, it's not exactly clear to me what you're asking, and forgive me, but it seems like it's not really clear to you either. –  Harry Gindi May 26 '10 at 8:34
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closed as not a real question by S. Carnahan, Harry Gindi, Andrew Stacey, Anton Geraschenko Jun 1 '10 at 1:34

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