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.



A Nisnevich cover which is not Zariski
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 $y\mapsto st$.





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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 $s=-y/x$).



