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The Nisnevich topology on $Sch$ is a Grothendieck topology strictly finer than the Zariski topology, and the etale topology is strictly finer than the Nisnevich topology.

Colin McLarty asked me for an example of a Nisnevich cover which is not a Zariski cover. The standard example I have seen in several places is rather of the distinction between etale and Zariski. Namely, a family of etale covers $\{A^1 - \{0\} \stackrel{(-)^2}{\to} A^1, A^1 - \{a\}\hookrightarrow A^1\}$ of the affine line $A^1$ over a field $k$ indexed by elements $a\in k^\times$, such that a cover is only Nisnevich if $a$ has a square root in $k$. I'm looking for something that will separate Zariski from Nisnevich.

Examples inspired by arithmetic or geometry are both good.

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What do you mean by distinction between etale and Zariski? Maybe you could mention these examples so we can see what you don't like about them. – name Jul 27 '12 at 1:43
David, your example does separate Zariski from Nisnevich: when a is a square, you have a Nisnevich cover but not a Zariski one. – Dustin Clausen Jul 27 '12 at 2:38
oh, we also need char$k \not = 2$. – David Roberts Jul 27 '12 at 2:40
Whoops. That's what comes of having $\epsilon$ experience in algebraic geometry. – David Roberts Jul 27 '12 at 2:41
I don't understand your example. The squaring map $A^1-0\to A^1$ is never Nisnevich since the extension of residue fields at the generic point is $k(x^2)\hookrightarrow k(x)$, which isn't an isomorphism. – Anton Geraschenko Jul 27 '12 at 6:07
up vote 4 down vote accepted

My standard example is an $n$-gon of $\mathbb P^1$'s covering the nodal cubic. For some reason, I especially like the case $n=2$.

Here's the affine version, which always takes me a bit to work out. It's two parabolas joined at two points covering the nodal cubic: $$ \def\spec{\mathrm{Spec\,}} \spec k[s,t]/(t^2-(s^2-1)^2) \to \spec k[x,y]/(y^2-x^2(x+1)) $$ given by $x\mapsto (s^2-1)$ and $y\mapsto st$.

This map is étale. To see it is Nisnevich, you have to check that the residue field extensions of the generic point are isomorphisms (this is where the squaring map on $\mathbb A^1-0$ fails to be Nisnevich). You can see that by restricting to the two components, giving the normalization maps $$ k[x,y]/(y^2-x^2(x+1)) \to k[s,t]/(t-(s^2-1)) \cong k[s] $$ given by $(x,y)\mapsto (t,st)$ (so $s=y/x$), and $$ k[x,y]/(y^2-x^2(x+1)) \to k[s,t]/(t+(s^2-1)) \cong k[s] $$ given by $(x,y)\mapsto (-t,st)$ (so $s=-y/x$).

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Easier to answer to your question, than your comment above. You only need to find some element of the cover such that there is an isomorphism of residue fields. For all $x\not = a$ we can use $x \in (A^1 - \{a\})$ where if it is not true I will eat my hat (not literally), but for $x=a$ we can probably take $b \in (A^1 - \{0\})$ such that $b^2 = a$. – David Roberts Jul 27 '12 at 7:17
In particular, you may need to check more than residue fields at the generic point are isomorphisms, according to the definition in Lecture Notes on Motivic Cohomology – David Roberts Jul 27 '12 at 7:20
You're absolutely right about only needing some point where the extension is trivial; I guess I've misunderstood the definition all this time. Of course you need to check more than the generic point, but the, but checking the generic point is the interesting part. The calculation of the maps on the two components shows that the residue field extensions away from the node are all isomorphisms (since the map is just the disjoint union of two isomorphisms away from the node), and the residue field extensions at the node are $k\subseteq k$. – Anton Geraschenko Jul 27 '12 at 7:48

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