Skip to main content
7 events
when toggle format what by license comment
Sep 5, 2020 at 18:57 comment added Francesco Polizzi This is double and not bi-double, for instance because $[z_0:z_1]=[-z_0: -z_1]$ and $[-z_0:z_1]=[z_0:-z_1]$.
Sep 5, 2020 at 18:50 comment added Francesco Polizzi If $k=3$ then $k-1=2$. The covering in the affine chart is clearly a bi-double cover, non tri-double. So the extension to the projective plane must be bi-double, too. Recall that $z_0$, $z_1$, $z_2$ are homogeneous coordinates. Think of the double cover given in affine coordinates by $x \mapsto x^2$: in homogeneous coordinates it becomes $[z_0 \, : \,z_1] \mapsto[z_0^2 \, : \,z_1^2]$.
Sep 5, 2020 at 15:06 comment added Federico Fallucca But in this case $\pi$ would be a $(Z/2Z)^3$ covering and not a $(Z/2Z)^2$-covering, right?
Aug 14, 2020 at 10:09 history edited Francesco Polizzi CC BY-SA 4.0
edited body
Aug 13, 2020 at 17:54 history edited Francesco Polizzi CC BY-SA 4.0
edited body
Aug 13, 2020 at 16:04 history edited Francesco Polizzi CC BY-SA 4.0
added 353 characters in body
Aug 13, 2020 at 15:58 history answered Francesco Polizzi CC BY-SA 4.0