Cech cohomology does not compute étale cohomology - Explanation of an example












0












$begingroup$


In the first answer to this MO post the author says that the $H^2$ of $X$ can be compute using the Cech-to-derived functor spectral sequence, i.e. in that case the Mayer-Vietoris sequence.



I'm having trouble in understanding how it works: as for $H^1(Ucap V)$, this should be equal to $$H^1_{et}(operatorname{Spec} k((t)), A)=Cont(Gal(k((t))^s,k((t))),mathbb Z/n)=Cont((1),mathbb Z/n)congmathbb Z/n.$$
(See the answer to this post, second part.)
Is this correct?



But when I come to compute $H^1_{et}(U,A)$ and $H^1_{et}(V,A)$, which I would like to be $0$ in order to conclude that $H^2_{et}(X,A)cong H^1_{et}(Ucap V,A)cong mathbb Z/n$, I am stuck. Can someone give me a clue?



Thank you in advance.






algebraic-geometry schemes derived-functors etale-cohomology







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $k[[t]]$ is a strictly henselian ring, so there is no non-trivial étale covering. In other word, this is a point for the étale topology. Hence, it has vanishing higher cohomology.
    $endgroup$
    – Roland
    Jan 7 at 18:36










  • $begingroup$
    Ok, thanks. There is just onething I'm missing: why is the same false for $k((t))$?
    $endgroup$
    – W. Rether
    Jan 7 at 18:57






  • 1




    $begingroup$
    Because $k((t))$ is not algebraically closed. As you wrote, its $H^1$ is the group of continuous homomorphism from $operatorname{Gal}(k((t))^s/k((t)))$ to $A$ (not $mathbb{Z}/nmathbb{Z}$). A classical picture : $k[[t]]$ is a small disc and $k((t))$ is the punctured disc. So $k[[t]]$ is contractible and it has no cohomology, $k((t))$ is homotopy equivalent to a circle so it has a $H^1simeq A$.
    $endgroup$
    – Roland
    Jan 7 at 19:30






  • 1




    $begingroup$
    Sure it is possible, but you didn't specify $A$. This is just for the formula $Cont(Gal,A)simeqmathbb{Z}/nmathbb{Z}$ (which is false if $Aneqmathbb{Z}/nmathbb{Z}$). I just have a doubt : I believe, one need that $k$ is of characteristic $0$ or that $A$ has torsion prime to the characteristic.
    $endgroup$
    – Roland
    Jan 7 at 20:01








  • 1




    $begingroup$
    This is because the Galois group of $k((t))$ as a huge $p$-torsion part. If $A$ has no $p$-torsion, then $Cont(G,A)$ factor through the "prime to $p$" quotient which is $mathbb{Z}'(1)=prod_{lneq p}mathbb{Z}_l(1)$. But if $A$ has $p$ torsion, this may be very different.
    $endgroup$
    – Roland
    Jan 7 at 20:30
















0












$begingroup$


In the first answer to this MO post the author says that the $H^2$ of $X$ can be compute using the Cech-to-derived functor spectral sequence, i.e. in that case the Mayer-Vietoris sequence.



I'm having trouble in understanding how it works: as for $H^1(Ucap V)$, this should be equal to $$H^1_{et}(operatorname{Spec} k((t)), A)=Cont(Gal(k((t))^s,k((t))),mathbb Z/n)=Cont((1),mathbb Z/n)congmathbb Z/n.$$
(See the answer to this post, second part.)
Is this correct?



But when I come to compute $H^1_{et}(U,A)$ and $H^1_{et}(V,A)$, which I would like to be $0$ in order to conclude that $H^2_{et}(X,A)cong H^1_{et}(Ucap V,A)cong mathbb Z/n$, I am stuck. Can someone give me a clue?



Thank you in advance.






algebraic-geometry schemes derived-functors etale-cohomology







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $k[[t]]$ is a strictly henselian ring, so there is no non-trivial étale covering. In other word, this is a point for the étale topology. Hence, it has vanishing higher cohomology.
    $endgroup$
    – Roland
    Jan 7 at 18:36










  • $begingroup$
    Ok, thanks. There is just onething I'm missing: why is the same false for $k((t))$?
    $endgroup$
    – W. Rether
    Jan 7 at 18:57






  • 1




    $begingroup$
    Because $k((t))$ is not algebraically closed. As you wrote, its $H^1$ is the group of continuous homomorphism from $operatorname{Gal}(k((t))^s/k((t)))$ to $A$ (not $mathbb{Z}/nmathbb{Z}$). A classical picture : $k[[t]]$ is a small disc and $k((t))$ is the punctured disc. So $k[[t]]$ is contractible and it has no cohomology, $k((t))$ is homotopy equivalent to a circle so it has a $H^1simeq A$.
    $endgroup$
    – Roland
    Jan 7 at 19:30






  • 1




    $begingroup$
    Sure it is possible, but you didn't specify $A$. This is just for the formula $Cont(Gal,A)simeqmathbb{Z}/nmathbb{Z}$ (which is false if $Aneqmathbb{Z}/nmathbb{Z}$). I just have a doubt : I believe, one need that $k$ is of characteristic $0$ or that $A$ has torsion prime to the characteristic.
    $endgroup$
    – Roland
    Jan 7 at 20:01








  • 1




    $begingroup$
    This is because the Galois group of $k((t))$ as a huge $p$-torsion part. If $A$ has no $p$-torsion, then $Cont(G,A)$ factor through the "prime to $p$" quotient which is $mathbb{Z}'(1)=prod_{lneq p}mathbb{Z}_l(1)$. But if $A$ has $p$ torsion, this may be very different.
    $endgroup$
    – Roland
    Jan 7 at 20:30














0












0








0


1



$begingroup$


In the first answer to this MO post the author says that the $H^2$ of $X$ can be compute using the Cech-to-derived functor spectral sequence, i.e. in that case the Mayer-Vietoris sequence.



I'm having trouble in understanding how it works: as for $H^1(Ucap V)$, this should be equal to $$H^1_{et}(operatorname{Spec} k((t)), A)=Cont(Gal(k((t))^s,k((t))),mathbb Z/n)=Cont((1),mathbb Z/n)congmathbb Z/n.$$
(See the answer to this post, second part.)
Is this correct?



But when I come to compute $H^1_{et}(U,A)$ and $H^1_{et}(V,A)$, which I would like to be $0$ in order to conclude that $H^2_{et}(X,A)cong H^1_{et}(Ucap V,A)cong mathbb Z/n$, I am stuck. Can someone give me a clue?



Thank you in advance.






algebraic-geometry schemes derived-functors etale-cohomology







share|cite|improve this question











$endgroup$




In the first answer to this MO post the author says that the $H^2$ of $X$ can be compute using the Cech-to-derived functor spectral sequence, i.e. in that case the Mayer-Vietoris sequence.



I'm having trouble in understanding how it works: as for $H^1(Ucap V)$, this should be equal to $$H^1_{et}(operatorname{Spec} k((t)), A)=Cont(Gal(k((t))^s,k((t))),mathbb Z/n)=Cont((1),mathbb Z/n)congmathbb Z/n.$$
(See the answer to this post, second part.)
Is this correct?



But when I come to compute $H^1_{et}(U,A)$ and $H^1_{et}(V,A)$, which I would like to be $0$ in order to conclude that $H^2_{et}(X,A)cong H^1_{et}(Ucap V,A)cong mathbb Z/n$, I am stuck. Can someone give me a clue?



Thank you in advance.






algebraic-geometry schemes derived-functors etale-cohomology




algebraic-geometry schemes derived-functors etale-cohomology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 20:23







W. Rether

















asked Jan 7 at 16:22









W. RetherW. Rether

728417




728417








  • 1




    $begingroup$
    $k[[t]]$ is a strictly henselian ring, so there is no non-trivial étale covering. In other word, this is a point for the étale topology. Hence, it has vanishing higher cohomology.
    $endgroup$
    – Roland
    Jan 7 at 18:36










  • $begingroup$
    Ok, thanks. There is just onething I'm missing: why is the same false for $k((t))$?
    $endgroup$
    – W. Rether
    Jan 7 at 18:57






  • 1




    $begingroup$
    Because $k((t))$ is not algebraically closed. As you wrote, its $H^1$ is the group of continuous homomorphism from $operatorname{Gal}(k((t))^s/k((t)))$ to $A$ (not $mathbb{Z}/nmathbb{Z}$). A classical picture : $k[[t]]$ is a small disc and $k((t))$ is the punctured disc. So $k[[t]]$ is contractible and it has no cohomology, $k((t))$ is homotopy equivalent to a circle so it has a $H^1simeq A$.
    $endgroup$
    – Roland
    Jan 7 at 19:30






  • 1




    $begingroup$
    Sure it is possible, but you didn't specify $A$. This is just for the formula $Cont(Gal,A)simeqmathbb{Z}/nmathbb{Z}$ (which is false if $Aneqmathbb{Z}/nmathbb{Z}$). I just have a doubt : I believe, one need that $k$ is of characteristic $0$ or that $A$ has torsion prime to the characteristic.
    $endgroup$
    – Roland
    Jan 7 at 20:01








  • 1




    $begingroup$
    This is because the Galois group of $k((t))$ as a huge $p$-torsion part. If $A$ has no $p$-torsion, then $Cont(G,A)$ factor through the "prime to $p$" quotient which is $mathbb{Z}'(1)=prod_{lneq p}mathbb{Z}_l(1)$. But if $A$ has $p$ torsion, this may be very different.
    $endgroup$
    – Roland
    Jan 7 at 20:30














  • 1




    $begingroup$
    $k[[t]]$ is a strictly henselian ring, so there is no non-trivial étale covering. In other word, this is a point for the étale topology. Hence, it has vanishing higher cohomology.
    $endgroup$
    – Roland
    Jan 7 at 18:36










  • $begingroup$
    Ok, thanks. There is just onething I'm missing: why is the same false for $k((t))$?
    $endgroup$
    – W. Rether
    Jan 7 at 18:57






  • 1




    $begingroup$
    Because $k((t))$ is not algebraically closed. As you wrote, its $H^1$ is the group of continuous homomorphism from $operatorname{Gal}(k((t))^s/k((t)))$ to $A$ (not $mathbb{Z}/nmathbb{Z}$). A classical picture : $k[[t]]$ is a small disc and $k((t))$ is the punctured disc. So $k[[t]]$ is contractible and it has no cohomology, $k((t))$ is homotopy equivalent to a circle so it has a $H^1simeq A$.
    $endgroup$
    – Roland
    Jan 7 at 19:30






  • 1




    $begingroup$
    Sure it is possible, but you didn't specify $A$. This is just for the formula $Cont(Gal,A)simeqmathbb{Z}/nmathbb{Z}$ (which is false if $Aneqmathbb{Z}/nmathbb{Z}$). I just have a doubt : I believe, one need that $k$ is of characteristic $0$ or that $A$ has torsion prime to the characteristic.
    $endgroup$
    – Roland
    Jan 7 at 20:01








  • 1




    $begingroup$
    This is because the Galois group of $k((t))$ as a huge $p$-torsion part. If $A$ has no $p$-torsion, then $Cont(G,A)$ factor through the "prime to $p$" quotient which is $mathbb{Z}'(1)=prod_{lneq p}mathbb{Z}_l(1)$. But if $A$ has $p$ torsion, this may be very different.
    $endgroup$
    – Roland
    Jan 7 at 20:30








1




1




$begingroup$
$k[[t]]$ is a strictly henselian ring, so there is no non-trivial étale covering. In other word, this is a point for the étale topology. Hence, it has vanishing higher cohomology.
$endgroup$
– Roland
Jan 7 at 18:36




$begingroup$
$k[[t]]$ is a strictly henselian ring, so there is no non-trivial étale covering. In other word, this is a point for the étale topology. Hence, it has vanishing higher cohomology.
$endgroup$
– Roland
Jan 7 at 18:36












$begingroup$
Ok, thanks. There is just onething I'm missing: why is the same false for $k((t))$?
$endgroup$
– W. Rether
Jan 7 at 18:57




$begingroup$
Ok, thanks. There is just onething I'm missing: why is the same false for $k((t))$?
$endgroup$
– W. Rether
Jan 7 at 18:57




1




1




$begingroup$
Because $k((t))$ is not algebraically closed. As you wrote, its $H^1$ is the group of continuous homomorphism from $operatorname{Gal}(k((t))^s/k((t)))$ to $A$ (not $mathbb{Z}/nmathbb{Z}$). A classical picture : $k[[t]]$ is a small disc and $k((t))$ is the punctured disc. So $k[[t]]$ is contractible and it has no cohomology, $k((t))$ is homotopy equivalent to a circle so it has a $H^1simeq A$.
$endgroup$
– Roland
Jan 7 at 19:30




$begingroup$
Because $k((t))$ is not algebraically closed. As you wrote, its $H^1$ is the group of continuous homomorphism from $operatorname{Gal}(k((t))^s/k((t)))$ to $A$ (not $mathbb{Z}/nmathbb{Z}$). A classical picture : $k[[t]]$ is a small disc and $k((t))$ is the punctured disc. So $k[[t]]$ is contractible and it has no cohomology, $k((t))$ is homotopy equivalent to a circle so it has a $H^1simeq A$.
$endgroup$
– Roland
Jan 7 at 19:30




1




1




$begingroup$
Sure it is possible, but you didn't specify $A$. This is just for the formula $Cont(Gal,A)simeqmathbb{Z}/nmathbb{Z}$ (which is false if $Aneqmathbb{Z}/nmathbb{Z}$). I just have a doubt : I believe, one need that $k$ is of characteristic $0$ or that $A$ has torsion prime to the characteristic.
$endgroup$
– Roland
Jan 7 at 20:01






$begingroup$
Sure it is possible, but you didn't specify $A$. This is just for the formula $Cont(Gal,A)simeqmathbb{Z}/nmathbb{Z}$ (which is false if $Aneqmathbb{Z}/nmathbb{Z}$). I just have a doubt : I believe, one need that $k$ is of characteristic $0$ or that $A$ has torsion prime to the characteristic.
$endgroup$
– Roland
Jan 7 at 20:01






1




1




$begingroup$
This is because the Galois group of $k((t))$ as a huge $p$-torsion part. If $A$ has no $p$-torsion, then $Cont(G,A)$ factor through the "prime to $p$" quotient which is $mathbb{Z}'(1)=prod_{lneq p}mathbb{Z}_l(1)$. But if $A$ has $p$ torsion, this may be very different.
$endgroup$
– Roland
Jan 7 at 20:30




$begingroup$
This is because the Galois group of $k((t))$ as a huge $p$-torsion part. If $A$ has no $p$-torsion, then $Cont(G,A)$ factor through the "prime to $p$" quotient which is $mathbb{Z}'(1)=prod_{lneq p}mathbb{Z}_l(1)$. But if $A$ has $p$ torsion, this may be very different.
$endgroup$
– Roland
Jan 7 at 20:30










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