Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
4  
Personally, I have found examples of Nisnevich covers un-enlightening. The most important property is: $p: U \to X$ is a Nisnevich cover if and only if it is an étale cover, and there exists a sequence $X \supset Z_1 \supset Z_2 \supset ... \supset Z_n = \varnothing$ of closed subschemes such that $p$ has a section over each $Z_i - Z_{i + 1}$. This allows induction arguments. And of course another motivation for Nisnevich is that every Nisnevich cover of a field has a section, and so the cohomological dimension of a "point" $Spec(k)$ is zero - a "failing" of the étale topology. –  name Jul 27 '12 at 1:50
    
See my question mathoverflow.net/questions/103257/… for something I would have just emailed you. –  David Roberts Jul 27 '12 at 2:42
add comment

1 Answer

up vote 9 down vote accepted

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.

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.

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.