Mixing
Constructing an invertible sheaf from a Cartier divisor?
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Let $X$ be a scheme. By definition, a Cartier divisor $D$ is a global section of the sheaf $mathcal{M}_X^{times}/mathcal{O}_X^{times}$, where $mathcal{M}_X^{times}$ is the sheaf of multiplicative units of the sheaf of rational functions. This in turn is equivalent to the data of a cover ${ U_i}$ of $X$ together with sections $f_i in mathcal{M}_X^{times}$, that is, $f_i = g_i/h_i$ for nonzero divisors $g_i, h_i in mathcal{O}_X(U_i)$.
I want to use this data to construct an invertible sheaf $mathcal{L}$ from a Cartier divisor $D$.
This is part of an exercise that is proving to be too hard, so hints rather than answers would be appreciated. Thank you in advance.
abstract-algebra algebraic-geometry sheaf-theory schemes divisors-algebraic-geometry
asked Jan 15 at 19:51
user313212user313212
358520
$endgroup$
add a comment |
$begingroup$
Let $X$ be a scheme. By definition, a Cartier divisor $D$ is a global section of the sheaf $mathcal{M}_X^{times}/mathcal{O}_X^{times}$, where $mathcal{M}_X^{times}$ is the sheaf of multiplicative units of the sheaf of rational functions. This in turn is equivalent to the data of a cover ${ U_i}$ of $X$ together with sections $f_i in mathcal{M}_X^{times}$, that is, $f_i = g_i/h_i$ for nonzero divisors $g_i, h_i in mathcal{O}_X(U_i)$.
I want to use this data to construct an invertible sheaf $mathcal{L}$ from a Cartier divisor $D$.
This is part of an exercise that is proving to be too hard, so hints rather than answers would be appreciated. Thank you in advance.
abstract-algebra algebraic-geometry sheaf-theory schemes divisors-algebraic-geometry
asked Jan 15 at 19:51
user313212user313212
358520
$endgroup$
2
$begingroup$
You're missing two conditions. One is that $g_iinmathcal{O}_X^times(U_i)$ and the other is that $(f_i|_{ij})(f_j|_{ij})^{-1}inmathcal{O}_X(U_{ij})^{times}$. To construct a locally free sheaf from this simply define $mathcal{L}|_{U_i}=frac{1}{f_i}cdotmathcal{O}_X|_{U_i}subseteqmathcal{M}_X$. Then you need to check that there are isomorphisms $phi_{ij}:(mathcal{L}|_{i})|_{ij}to(mathcal{L}_j)|_{ij}$ such that $phi_{ii}=id$ and $phi_{jk}circphi_{ij}=phi_{ik}$.
$endgroup$
– user347489
Jan 15 at 20:10
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@user347489 I'm not seeing why $g_i$ should be in $mathcal{O}_X^times(U_i)$. I do agree with the rest of your comment though.
$endgroup$
– jgon
Jan 16 at 1:09
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@jgon because otherwise $f_i=g_i/h_inotinmathcal{M}_X^times$. But now I see that the OP wrote "nonzero", making my remark redundant.
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– user347489
Jan 16 at 1:45
$begingroup$
@user347489 Ah yes, though I should point out that being in $mathcal{O}_X^times(U_i)$ is a much stronger condition than being nonzero (which is all that we need here). Hence my question, no worries.
$endgroup$
– jgon
Jan 16 at 1:53
$begingroup$
@jgon you're right.
$endgroup$
– user347489
Jan 16 at 7:32
add a comment |
$begingroup$
Let $X$ be a scheme. By definition, a Cartier divisor $D$ is a global section of the sheaf $mathcal{M}_X^{times}/mathcal{O}_X^{times}$, where $mathcal{M}_X^{times}$ is the sheaf of multiplicative units of the sheaf of rational functions. This in turn is equivalent to the data of a cover ${ U_i}$ of $X$ together with sections $f_i in mathcal{M}_X^{times}$, that is, $f_i = g_i/h_i$ for nonzero divisors $g_i, h_i in mathcal{O}_X(U_i)$.
I want to use this data to construct an invertible sheaf $mathcal{L}$ from a Cartier divisor $D$.
This is part of an exercise that is proving to be too hard, so hints rather than answers would be appreciated. Thank you in advance.
abstract-algebra algebraic-geometry sheaf-theory schemes divisors-algebraic-geometry
asked Jan 15 at 19:51
user313212user313212
358520
$endgroup$
Let $X$ be a scheme. By definition, a Cartier divisor $D$ is a global section of the sheaf $mathcal{M}_X^{times}/mathcal{O}_X^{times}$, where $mathcal{M}_X^{times}$ is the sheaf of multiplicative units of the sheaf of rational functions. This in turn is equivalent to the data of a cover ${ U_i}$ of $X$ together with sections $f_i in mathcal{M}_X^{times}$, that is, $f_i = g_i/h_i$ for nonzero divisors $g_i, h_i in mathcal{O}_X(U_i)$.
I want to use this data to construct an invertible sheaf $mathcal{L}$ from a Cartier divisor $D$.
This is part of an exercise that is proving to be too hard, so hints rather than answers would be appreciated. Thank you in advance.
abstract-algebra algebraic-geometry sheaf-theory schemes divisors-algebraic-geometry
abstract-algebra algebraic-geometry sheaf-theory schemes divisors-algebraic-geometry
asked Jan 15 at 19:51
user313212user313212
358520
asked Jan 15 at 19:51
user313212user313212
358520
asked Jan 15 at 19:51
user313212user313212
358520
asked Jan 15 at 19:51
user313212user313212
358520
asked Jan 15 at 19:51
user313212user313212
358520
358520
2
$begingroup$
You're missing two conditions. One is that $g_iinmathcal{O}_X^times(U_i)$ and the other is that $(f_i|_{ij})(f_j|_{ij})^{-1}inmathcal{O}_X(U_{ij})^{times}$. To construct a locally free sheaf from this simply define $mathcal{L}|_{U_i}=frac{1}{f_i}cdotmathcal{O}_X|_{U_i}subseteqmathcal{M}_X$. Then you need to check that there are isomorphisms $phi_{ij}:(mathcal{L}|_{i})|_{ij}to(mathcal{L}_j)|_{ij}$ such that $phi_{ii}=id$ and $phi_{jk}circphi_{ij}=phi_{ik}$.
$endgroup$
– user347489
Jan 15 at 20:10
$begingroup$
@user347489 I'm not seeing why $g_i$ should be in $mathcal{O}_X^times(U_i)$. I do agree with the rest of your comment though.
$endgroup$
– jgon
Jan 16 at 1:09
$begingroup$
@jgon because otherwise $f_i=g_i/h_inotinmathcal{M}_X^times$. But now I see that the OP wrote "nonzero", making my remark redundant.
$endgroup$
– user347489
Jan 16 at 1:45
$begingroup$
@user347489 Ah yes, though I should point out that being in $mathcal{O}_X^times(U_i)$ is a much stronger condition than being nonzero (which is all that we need here). Hence my question, no worries.
$endgroup$
– jgon
Jan 16 at 1:53
$begingroup$
@jgon you're right.
$endgroup$
– user347489
Jan 16 at 7:32
add a comment |
2
$begingroup$
You're missing two conditions. One is that $g_iinmathcal{O}_X^times(U_i)$ and the other is that $(f_i|_{ij})(f_j|_{ij})^{-1}inmathcal{O}_X(U_{ij})^{times}$. To construct a locally free sheaf from this simply define $mathcal{L}|_{U_i}=frac{1}{f_i}cdotmathcal{O}_X|_{U_i}subseteqmathcal{M}_X$. Then you need to check that there are isomorphisms $phi_{ij}:(mathcal{L}|_{i})|_{ij}to(mathcal{L}_j)|_{ij}$ such that $phi_{ii}=id$ and $phi_{jk}circphi_{ij}=phi_{ik}$.
$endgroup$
– user347489
Jan 15 at 20:10
$begingroup$
@user347489 I'm not seeing why $g_i$ should be in $mathcal{O}_X^times(U_i)$. I do agree with the rest of your comment though.
$endgroup$
– jgon
Jan 16 at 1:09
$begingroup$
@jgon because otherwise $f_i=g_i/h_inotinmathcal{M}_X^times$. But now I see that the OP wrote "nonzero", making my remark redundant.
$endgroup$
– user347489
Jan 16 at 1:45
$begingroup$
@user347489 Ah yes, though I should point out that being in $mathcal{O}_X^times(U_i)$ is a much stronger condition than being nonzero (which is all that we need here). Hence my question, no worries.
$endgroup$
– jgon
Jan 16 at 1:53
$begingroup$
@jgon you're right.
$endgroup$
– user347489
Jan 16 at 7:32
2
2
$begingroup$
You're missing two conditions. One is that $g_iinmathcal{O}_X^times(U_i)$ and the other is that $(f_i|_{ij})(f_j|_{ij})^{-1}inmathcal{O}_X(U_{ij})^{times}$. To construct a locally free sheaf from this simply define $mathcal{L}|_{U_i}=frac{1}{f_i}cdotmathcal{O}_X|_{U_i}subseteqmathcal{M}_X$. Then you need to check that there are isomorphisms $phi_{ij}:(mathcal{L}|_{i})|_{ij}to(mathcal{L}_j)|_{ij}$ such that $phi_{ii}=id$ and $phi_{jk}circphi_{ij}=phi_{ik}$.
$endgroup$
– user347489
Jan 15 at 20:10
$begingroup$
You're missing two conditions. One is that $g_iinmathcal{O}_X^times(U_i)$ and the other is that $(f_i|_{ij})(f_j|_{ij})^{-1}inmathcal{O}_X(U_{ij})^{times}$. To construct a locally free sheaf from this simply define $mathcal{L}|_{U_i}=frac{1}{f_i}cdotmathcal{O}_X|_{U_i}subseteqmathcal{M}_X$. Then you need to check that there are isomorphisms $phi_{ij}:(mathcal{L}|_{i})|_{ij}to(mathcal{L}_j)|_{ij}$ such that $phi_{ii}=id$ and $phi_{jk}circphi_{ij}=phi_{ik}$.
$endgroup$
– user347489
Jan 15 at 20:10
$begingroup$
@user347489 I'm not seeing why $g_i$ should be in $mathcal{O}_X^times(U_i)$. I do agree with the rest of your comment though.
$endgroup$
– jgon
Jan 16 at 1:09
$begingroup$
@user347489 I'm not seeing why $g_i$ should be in $mathcal{O}_X^times(U_i)$. I do agree with the rest of your comment though.
$endgroup$
– jgon
Jan 16 at 1:09
$begingroup$
@jgon because otherwise $f_i=g_i/h_inotinmathcal{M}_X^times$. But now I see that the OP wrote "nonzero", making my remark redundant.
$endgroup$
– user347489
Jan 16 at 1:45
$begingroup$
@jgon because otherwise $f_i=g_i/h_inotinmathcal{M}_X^times$. But now I see that the OP wrote "nonzero", making my remark redundant.
$endgroup$
– user347489
Jan 16 at 1:45
$begingroup$
@user347489 Ah yes, though I should point out that being in $mathcal{O}_X^times(U_i)$ is a much stronger condition than being nonzero (which is all that we need here). Hence my question, no worries.
$endgroup$
– jgon
Jan 16 at 1:53
$begingroup$
@user347489 Ah yes, though I should point out that being in $mathcal{O}_X^times(U_i)$ is a much stronger condition than being nonzero (which is all that we need here). Hence my question, no worries.
$endgroup$
– jgon
Jan 16 at 1:53
$begingroup$
@jgon you're right.
$endgroup$
– user347489
Jan 16 at 7:32
$begingroup$
@jgon you're right.
$endgroup$
– user347489
Jan 16 at 7:32
add a comment |
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$begingroup$
You're missing two conditions. One is that $g_iinmathcal{O}_X^times(U_i)$ and the other is that $(f_i|_{ij})(f_j|_{ij})^{-1}inmathcal{O}_X(U_{ij})^{times}$. To construct a locally free sheaf from this simply define $mathcal{L}|_{U_i}=frac{1}{f_i}cdotmathcal{O}_X|_{U_i}subseteqmathcal{M}_X$. Then you need to check that there are isomorphisms $phi_{ij}:(mathcal{L}|_{i})|_{ij}to(mathcal{L}_j)|_{ij}$ such that $phi_{ii}=id$ and $phi_{jk}circphi_{ij}=phi_{ik}$.
$endgroup$
– user347489
Jan 15 at 20:10
$begingroup$
@user347489 I'm not seeing why $g_i$ should be in $mathcal{O}_X^times(U_i)$. I do agree with the rest of your comment though.
$endgroup$
– jgon
Jan 16 at 1:09
$begingroup$
@jgon because otherwise $f_i=g_i/h_inotinmathcal{M}_X^times$. But now I see that the OP wrote "nonzero", making my remark redundant.
$endgroup$
– user347489
Jan 16 at 1:45
$begingroup$
@user347489 Ah yes, though I should point out that being in $mathcal{O}_X^times(U_i)$ is a much stronger condition than being nonzero (which is all that we need here). Hence my question, no worries.
$endgroup$
– jgon
Jan 16 at 1:53
$begingroup$
@jgon you're right.
$endgroup$
– user347489
Jan 16 at 7:32