Is there a typical example of Nisnevich covers? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T19:10:49Z http://mathoverflow.net/feeds/question/103258 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103258/is-there-a-typical-example-of-nisnevich-covers Is there a typical example of Nisnevich covers? Colin McLarty 2012-07-27T01:37:32Z 2012-07-27T18:26:45Z <p>There is a popular (and I think helpful) example of etale covers, namely covers of Riemann surfaces with ramification points removed. Is there a similarly accessible example to motivate Nisnevich covers? </p> http://mathoverflow.net/questions/103258/is-there-a-typical-example-of-nisnevich-covers/103260#103260 Answer by Dustin Clausen for Is there a typical example of Nisnevich covers? Dustin Clausen 2012-07-27T02:10:01Z 2012-07-27T18:26:45Z <p>Well, a representative example is where you take some arbitrary etale cover Y of X which splits over a closed subvariety Z of X, then form the Nisnevich cover of X consisting of the open complement X - Z together with the open subscheme Y' of Y where you remove all but one of the copies of Z lying above Z. For instance Z could be a point, and our field could be algebraically closed.</p> <p>The intuition I find helpful is that descent for the Nisnevich topology is meant to be an easier-to-precisely-phrase consequence of the principle "X is gotten by gluing X-Z to a tubular neighborhood of Z, along the punctured tubular neighborhood of Z". The idea being that, in the situation of the previous paragraph, the tubular neighborhoods of Z in X and Z in Y should be the same, Y --> X being etale.</p> <p>As an example of this intuition, note that if X is a smooth variety of dimension n (over an algebraically closed field for simplicity) and x is a point of X, then by choosing n independent parameters at x you can find a Zariski neighborhood X' of x and an etale map X' --> A^n which, together with A^n - 0 ---> A^n, makes a Nisnevich cover of A^n with intersection equal to X'-x. This corresponds (ish) to the fact that the tubular neighborhood of x in X should be the same as the tubular neighborhood of 0 in A^n.</p>