Is the total ring of fractions mod the regular functions flasque?
$begingroup$
I want to show that
$$0to mathcal O_{mathbb{P}_k^1}to mathcal Kto mathcal K/mathcal O_{mathbb{P}_k^1}to 0$$
is a flasque resolution of $mathcal O_{mathbb{P}_k^1}$ with $k$ infinite, but not necessarily algebraically closed. Where $mathcal K$ is the total ring of fractions.
I was able to show that $mathcal K$ is flasque, and by extension, the presheaf $Umapsto mathcal K(U)/mathcal O_{mathbb{P}_k^1}(U)$ is 'flasque' as a presheaf. However, being a quotient, we aren't guaranteed that this is a sheaf. I tried to show that it was, but I kept running into troubles. So, instead, I tried to show flasqueness of the sheafification, but again, I couldn't lift any of the maps. Additionally, $mathcal O_{mathbb{P}_k^1}$ isn't flasque; so, I'm stuck.
Any advice is greatly appreciated! Thank you all in advance :).
algebraic-geometry sheaf-theory sheaf-cohomology
asked Dec 3 '18 at 20:20
LaarzLaarz
1098
$endgroup$
add a comment |
$begingroup$
I want to show that
$$0to mathcal O_{mathbb{P}_k^1}to mathcal Kto mathcal K/mathcal O_{mathbb{P}_k^1}to 0$$
is a flasque resolution of $mathcal O_{mathbb{P}_k^1}$ with $k$ infinite, but not necessarily algebraically closed. Where $mathcal K$ is the total ring of fractions.
I was able to show that $mathcal K$ is flasque, and by extension, the presheaf $Umapsto mathcal K(U)/mathcal O_{mathbb{P}_k^1}(U)$ is 'flasque' as a presheaf. However, being a quotient, we aren't guaranteed that this is a sheaf. I tried to show that it was, but I kept running into troubles. So, instead, I tried to show flasqueness of the sheafification, but again, I couldn't lift any of the maps. Additionally, $mathcal O_{mathbb{P}_k^1}$ isn't flasque; so, I'm stuck.
Any advice is greatly appreciated! Thank you all in advance :).
algebraic-geometry sheaf-theory sheaf-cohomology
asked Dec 3 '18 at 20:20
LaarzLaarz
1098
$endgroup$
2
$begingroup$
What can you say about $H^1(U,mathcal{O})$ if $U$ is any open subset of $mathbb{P}^1$ ?
$endgroup$
– Roland
Dec 3 '18 at 20:40
$begingroup$
@Roland I assume that you'd like me to see it's 0 somehow, which would then give me what I want, but after thinking for a few minutes, I don't see why this ought to be 0.
$endgroup$
– Laarz
Dec 3 '18 at 21:01
1
$begingroup$
Yes that is my point. There is two cases to be treated separately : $U=mathbb{P}^1$ and $Uneqmathbb{P}^1$. For the first one, it is a classical computation. For the second, use that $U$ is affine.
$endgroup$
– Roland
Dec 3 '18 at 21:04
$begingroup$
Ahhh! Well, now I feel silly. Thank you so much!!
$endgroup$
– Laarz
Dec 3 '18 at 21:06
add a comment |
$begingroup$
I want to show that
$$0to mathcal O_{mathbb{P}_k^1}to mathcal Kto mathcal K/mathcal O_{mathbb{P}_k^1}to 0$$
is a flasque resolution of $mathcal O_{mathbb{P}_k^1}$ with $k$ infinite, but not necessarily algebraically closed. Where $mathcal K$ is the total ring of fractions.
I was able to show that $mathcal K$ is flasque, and by extension, the presheaf $Umapsto mathcal K(U)/mathcal O_{mathbb{P}_k^1}(U)$ is 'flasque' as a presheaf. However, being a quotient, we aren't guaranteed that this is a sheaf. I tried to show that it was, but I kept running into troubles. So, instead, I tried to show flasqueness of the sheafification, but again, I couldn't lift any of the maps. Additionally, $mathcal O_{mathbb{P}_k^1}$ isn't flasque; so, I'm stuck.
Any advice is greatly appreciated! Thank you all in advance :).
algebraic-geometry sheaf-theory sheaf-cohomology
asked Dec 3 '18 at 20:20
LaarzLaarz
1098
$endgroup$
I want to show that
$$0to mathcal O_{mathbb{P}_k^1}to mathcal Kto mathcal K/mathcal O_{mathbb{P}_k^1}to 0$$
is a flasque resolution of $mathcal O_{mathbb{P}_k^1}$ with $k$ infinite, but not necessarily algebraically closed. Where $mathcal K$ is the total ring of fractions.
I was able to show that $mathcal K$ is flasque, and by extension, the presheaf $Umapsto mathcal K(U)/mathcal O_{mathbb{P}_k^1}(U)$ is 'flasque' as a presheaf. However, being a quotient, we aren't guaranteed that this is a sheaf. I tried to show that it was, but I kept running into troubles. So, instead, I tried to show flasqueness of the sheafification, but again, I couldn't lift any of the maps. Additionally, $mathcal O_{mathbb{P}_k^1}$ isn't flasque; so, I'm stuck.
Any advice is greatly appreciated! Thank you all in advance :).
algebraic-geometry sheaf-theory sheaf-cohomology
algebraic-geometry sheaf-theory sheaf-cohomology
asked Dec 3 '18 at 20:20
LaarzLaarz
1098
asked Dec 3 '18 at 20:20
LaarzLaarz
1098
asked Dec 3 '18 at 20:20
LaarzLaarz
1098
asked Dec 3 '18 at 20:20
LaarzLaarz
1098
asked Dec 3 '18 at 20:20
LaarzLaarz
1098
1098
2
$begingroup$
What can you say about $H^1(U,mathcal{O})$ if $U$ is any open subset of $mathbb{P}^1$ ?
$endgroup$
– Roland
Dec 3 '18 at 20:40
$begingroup$
@Roland I assume that you'd like me to see it's 0 somehow, which would then give me what I want, but after thinking for a few minutes, I don't see why this ought to be 0.
$endgroup$
– Laarz
Dec 3 '18 at 21:01
1
$begingroup$
Yes that is my point. There is two cases to be treated separately : $U=mathbb{P}^1$ and $Uneqmathbb{P}^1$. For the first one, it is a classical computation. For the second, use that $U$ is affine.
$endgroup$
– Roland
Dec 3 '18 at 21:04
$begingroup$
Ahhh! Well, now I feel silly. Thank you so much!!
$endgroup$
– Laarz
Dec 3 '18 at 21:06
add a comment |
2
$begingroup$
What can you say about $H^1(U,mathcal{O})$ if $U$ is any open subset of $mathbb{P}^1$ ?
$endgroup$
– Roland
Dec 3 '18 at 20:40
$begingroup$
@Roland I assume that you'd like me to see it's 0 somehow, which would then give me what I want, but after thinking for a few minutes, I don't see why this ought to be 0.
$endgroup$
– Laarz
Dec 3 '18 at 21:01
1
$begingroup$
Yes that is my point. There is two cases to be treated separately : $U=mathbb{P}^1$ and $Uneqmathbb{P}^1$. For the first one, it is a classical computation. For the second, use that $U$ is affine.
$endgroup$
– Roland
Dec 3 '18 at 21:04
$begingroup$
Ahhh! Well, now I feel silly. Thank you so much!!
$endgroup$
– Laarz
Dec 3 '18 at 21:06
2
2
$begingroup$
What can you say about $H^1(U,mathcal{O})$ if $U$ is any open subset of $mathbb{P}^1$ ?
$endgroup$
– Roland
Dec 3 '18 at 20:40
$begingroup$
What can you say about $H^1(U,mathcal{O})$ if $U$ is any open subset of $mathbb{P}^1$ ?
$endgroup$
– Roland
Dec 3 '18 at 20:40
$begingroup$
@Roland I assume that you'd like me to see it's 0 somehow, which would then give me what I want, but after thinking for a few minutes, I don't see why this ought to be 0.
$endgroup$
– Laarz
Dec 3 '18 at 21:01
$begingroup$
@Roland I assume that you'd like me to see it's 0 somehow, which would then give me what I want, but after thinking for a few minutes, I don't see why this ought to be 0.
$endgroup$
– Laarz
Dec 3 '18 at 21:01
1
1
$begingroup$
Yes that is my point. There is two cases to be treated separately : $U=mathbb{P}^1$ and $Uneqmathbb{P}^1$. For the first one, it is a classical computation. For the second, use that $U$ is affine.
$endgroup$
– Roland
Dec 3 '18 at 21:04
$begingroup$
Yes that is my point. There is two cases to be treated separately : $U=mathbb{P}^1$ and $Uneqmathbb{P}^1$. For the first one, it is a classical computation. For the second, use that $U$ is affine.
$endgroup$
– Roland
Dec 3 '18 at 21:04
$begingroup$
Ahhh! Well, now I feel silly. Thank you so much!!
$endgroup$
– Laarz
Dec 3 '18 at 21:06
$begingroup$
Ahhh! Well, now I feel silly. Thank you so much!!
$endgroup$
– Laarz
Dec 3 '18 at 21:06
add a comment |
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$begingroup$
What can you say about $H^1(U,mathcal{O})$ if $U$ is any open subset of $mathbb{P}^1$ ?
$endgroup$
– Roland
Dec 3 '18 at 20:40
$begingroup$
@Roland I assume that you'd like me to see it's 0 somehow, which would then give me what I want, but after thinking for a few minutes, I don't see why this ought to be 0.
$endgroup$
– Laarz
Dec 3 '18 at 21:01
1
$begingroup$
Yes that is my point. There is two cases to be treated separately : $U=mathbb{P}^1$ and $Uneqmathbb{P}^1$. For the first one, it is a classical computation. For the second, use that $U$ is affine.
$endgroup$
– Roland
Dec 3 '18 at 21:04
$begingroup$
Ahhh! Well, now I feel silly. Thank you so much!!
$endgroup$
– Laarz
Dec 3 '18 at 21:06