Different definitions of Cartier divisor and when they agreeAmple Cartier divisors and coherent sheavesDefinition of Cartier divisorsCartier Divisor corresponds on $X$ to $(U_i,f_i)$Showing that a power of an ample sheaf is equivalent to an effective Cartier divisorWhy do we have two definitions of Cartier divisor?Sheaf associated to a Cartier divisorEach divisor arises from a rational sectionTwisted sheaf and restriction to a divisorPreimage of Cartier divisor under finite morphismConstructing an invertible sheaf from a Cartier divisor?
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Different definitions of Cartier divisor and when they agree
Ample Cartier divisors and coherent sheavesDefinition of Cartier divisorsCartier Divisor corresponds on $X$ to $(U_i,f_i)$Showing that a power of an ample sheaf is equivalent to an effective Cartier divisorWhy do we have two definitions of Cartier divisor?Sheaf associated to a Cartier divisorEach divisor arises from a rational sectionTwisted sheaf and restriction to a divisorPreimage of Cartier divisor under finite morphismConstructing an invertible sheaf from a Cartier divisor?
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$begingroup$
On a scheme $X$, the most general definition of Cartier divisor is a global section in $Gamma(X, mathcalK^*/mathcalO^*)$, where $mathcalK^*$ is the sheaf of invertible elements of the sheaf of total quotient rings.
Alternatively, some texts refer to a Cartier divisor as an equivalence class of pairs $(mathcalL, s)$, where $mathcalL$ is an invertible sheaf on $X$ and $s$ is a non-zero rational section of $mathcalL$.
I have been able to show that these definitions are equivalent in the case that $X$ is a noetherian integral scheme. But how general does this equivalence go? Are they always the same? I expect the problem may occur when you drop reducedness.
algebraic-geometry sheaf-theory schemes divisors-algebraic-geometry
asked Jul 28 at 6:22
LukeLuke
9373 silver badges8 bronze badges
$endgroup$
add a comment |
$begingroup$
On a scheme $X$, the most general definition of Cartier divisor is a global section in $Gamma(X, mathcalK^*/mathcalO^*)$, where $mathcalK^*$ is the sheaf of invertible elements of the sheaf of total quotient rings.
Alternatively, some texts refer to a Cartier divisor as an equivalence class of pairs $(mathcalL, s)$, where $mathcalL$ is an invertible sheaf on $X$ and $s$ is a non-zero rational section of $mathcalL$.
I have been able to show that these definitions are equivalent in the case that $X$ is a noetherian integral scheme. But how general does this equivalence go? Are they always the same? I expect the problem may occur when you drop reducedness.
algebraic-geometry sheaf-theory schemes divisors-algebraic-geometry
asked Jul 28 at 6:22
LukeLuke
9373 silver badges8 bronze badges
$endgroup$
$begingroup$
how do you define a rational section of a line bundle on a general scheme?
$endgroup$
– user690882
Jul 28 at 6:29
$begingroup$
Actually good question. I guess for an irreducible scheme it is fine, but I don't know how it would be defined more generally than that.
$endgroup$
– Luke
Jul 28 at 6:36
add a comment |
$begingroup$
On a scheme $X$, the most general definition of Cartier divisor is a global section in $Gamma(X, mathcalK^*/mathcalO^*)$, where $mathcalK^*$ is the sheaf of invertible elements of the sheaf of total quotient rings.
Alternatively, some texts refer to a Cartier divisor as an equivalence class of pairs $(mathcalL, s)$, where $mathcalL$ is an invertible sheaf on $X$ and $s$ is a non-zero rational section of $mathcalL$.
I have been able to show that these definitions are equivalent in the case that $X$ is a noetherian integral scheme. But how general does this equivalence go? Are they always the same? I expect the problem may occur when you drop reducedness.
algebraic-geometry sheaf-theory schemes divisors-algebraic-geometry
asked Jul 28 at 6:22
LukeLuke
9373 silver badges8 bronze badges
$endgroup$
On a scheme $X$, the most general definition of Cartier divisor is a global section in $Gamma(X, mathcalK^*/mathcalO^*)$, where $mathcalK^*$ is the sheaf of invertible elements of the sheaf of total quotient rings.
Alternatively, some texts refer to a Cartier divisor as an equivalence class of pairs $(mathcalL, s)$, where $mathcalL$ is an invertible sheaf on $X$ and $s$ is a non-zero rational section of $mathcalL$.
I have been able to show that these definitions are equivalent in the case that $X$ is a noetherian integral scheme. But how general does this equivalence go? Are they always the same? I expect the problem may occur when you drop reducedness.
algebraic-geometry sheaf-theory schemes divisors-algebraic-geometry
algebraic-geometry sheaf-theory schemes divisors-algebraic-geometry
asked Jul 28 at 6:22
LukeLuke
9373 silver badges8 bronze badges
asked Jul 28 at 6:22
LukeLuke
9373 silver badges8 bronze badges
asked Jul 28 at 6:22
LukeLuke
9373 silver badges8 bronze badges
asked Jul 28 at 6:22
LukeLuke
9373 silver badges8 bronze badges
asked Jul 28 at 6:22
LukeLuke
9373 silver badges8 bronze badges
9373 silver badges8 bronze badges
$begingroup$
how do you define a rational section of a line bundle on a general scheme?
$endgroup$
– user690882
Jul 28 at 6:29
$begingroup$
Actually good question. I guess for an irreducible scheme it is fine, but I don't know how it would be defined more generally than that.
$endgroup$
– Luke
Jul 28 at 6:36
add a comment |
$begingroup$
how do you define a rational section of a line bundle on a general scheme?
$endgroup$
– user690882
Jul 28 at 6:29
$begingroup$
Actually good question. I guess for an irreducible scheme it is fine, but I don't know how it would be defined more generally than that.
$endgroup$
– Luke
Jul 28 at 6:36
$begingroup$
how do you define a rational section of a line bundle on a general scheme?
$endgroup$
– user690882
Jul 28 at 6:29
$begingroup$
how do you define a rational section of a line bundle on a general scheme?
$endgroup$
– user690882
Jul 28 at 6:29
$begingroup$
Actually good question. I guess for an irreducible scheme it is fine, but I don't know how it would be defined more generally than that.
$endgroup$
– Luke
Jul 28 at 6:36
$begingroup$
Actually good question. I guess for an irreducible scheme it is fine, but I don't know how it would be defined more generally than that.
$endgroup$
– Luke
Jul 28 at 6:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Cartier divisors can be defined on any scheme. And to any Cartier divisor $D$ on any scheme $X$ you can associate a line bundle (invertible $mathcalO_X$-module) which is usually denoted by $mathcalO_X(D)$. So there is always a map $$ beginalign mathrmDiv(X) &to mathrmPic(X) \ D &to mathcalO_X(D), endalign $$ from the group of Cartier divisors to the group of line bundles. This further induces an injective map $mathrmDivCl(X) to mathrmPic(X)$, where $mathrmDivCl(X)$ denotes the group $mathrmDiv(X)$ modulo linear equivalence. This map is surjective if $X$ is an integral scheme, but for general schemes not all line bundles come from Cartier divisors.
Now to answer your question, for any scheme $X$ and a line bundle $mathcalL$ on it, you have the following natural one to one correspondence
$$ texteffective cartier divisors D text such that mathcalO_X(D) cong mathcalL leftrightarrow textnon-zero divisors of Gamma(X, mathcalL) big/sim, $$ where $ s sim s'$ if and only if $s' = us$ for some $u in Gamma(X, mathcalO_X^times).$ Note that we are only considering effective Cartier divisors, because the associated line bundle of any divisor has a nonzero global section if and only if it is linearly equivalent to an effective one.
answered Jul 28 at 7:28
Parthiv BasuParthiv Basu
1,03010 bronze badges
$endgroup$
$begingroup$
If you want explicit details, see pages $302-305$ of Algebraic Geometry $1$ by Görtz, Wedhorn.
$endgroup$
– Parthiv Basu
Jul 28 at 8:00
add a comment |
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$begingroup$
Cartier divisors can be defined on any scheme. And to any Cartier divisor $D$ on any scheme $X$ you can associate a line bundle (invertible $mathcalO_X$-module) which is usually denoted by $mathcalO_X(D)$. So there is always a map $$ beginalign mathrmDiv(X) &to mathrmPic(X) \ D &to mathcalO_X(D), endalign $$ from the group of Cartier divisors to the group of line bundles. This further induces an injective map $mathrmDivCl(X) to mathrmPic(X)$, where $mathrmDivCl(X)$ denotes the group $mathrmDiv(X)$ modulo linear equivalence. This map is surjective if $X$ is an integral scheme, but for general schemes not all line bundles come from Cartier divisors.
Now to answer your question, for any scheme $X$ and a line bundle $mathcalL$ on it, you have the following natural one to one correspondence
$$ texteffective cartier divisors D text such that mathcalO_X(D) cong mathcalL leftrightarrow textnon-zero divisors of Gamma(X, mathcalL) big/sim, $$ where $ s sim s'$ if and only if $s' = us$ for some $u in Gamma(X, mathcalO_X^times).$ Note that we are only considering effective Cartier divisors, because the associated line bundle of any divisor has a nonzero global section if and only if it is linearly equivalent to an effective one.
answered Jul 28 at 7:28
Parthiv BasuParthiv Basu
1,03010 bronze badges
$endgroup$
$begingroup$
If you want explicit details, see pages $302-305$ of Algebraic Geometry $1$ by Görtz, Wedhorn.
$endgroup$
– Parthiv Basu
Jul 28 at 8:00
add a comment |
$begingroup$
Cartier divisors can be defined on any scheme. And to any Cartier divisor $D$ on any scheme $X$ you can associate a line bundle (invertible $mathcalO_X$-module) which is usually denoted by $mathcalO_X(D)$. So there is always a map $$ beginalign mathrmDiv(X) &to mathrmPic(X) \ D &to mathcalO_X(D), endalign $$ from the group of Cartier divisors to the group of line bundles. This further induces an injective map $mathrmDivCl(X) to mathrmPic(X)$, where $mathrmDivCl(X)$ denotes the group $mathrmDiv(X)$ modulo linear equivalence. This map is surjective if $X$ is an integral scheme, but for general schemes not all line bundles come from Cartier divisors.
Now to answer your question, for any scheme $X$ and a line bundle $mathcalL$ on it, you have the following natural one to one correspondence
$$ texteffective cartier divisors D text such that mathcalO_X(D) cong mathcalL leftrightarrow textnon-zero divisors of Gamma(X, mathcalL) big/sim, $$ where $ s sim s'$ if and only if $s' = us$ for some $u in Gamma(X, mathcalO_X^times).$ Note that we are only considering effective Cartier divisors, because the associated line bundle of any divisor has a nonzero global section if and only if it is linearly equivalent to an effective one.
answered Jul 28 at 7:28
Parthiv BasuParthiv Basu
1,03010 bronze badges
$endgroup$
$begingroup$
If you want explicit details, see pages $302-305$ of Algebraic Geometry $1$ by Görtz, Wedhorn.
$endgroup$
– Parthiv Basu
Jul 28 at 8:00
add a comment |
$begingroup$
Cartier divisors can be defined on any scheme. And to any Cartier divisor $D$ on any scheme $X$ you can associate a line bundle (invertible $mathcalO_X$-module) which is usually denoted by $mathcalO_X(D)$. So there is always a map $$ beginalign mathrmDiv(X) &to mathrmPic(X) \ D &to mathcalO_X(D), endalign $$ from the group of Cartier divisors to the group of line bundles. This further induces an injective map $mathrmDivCl(X) to mathrmPic(X)$, where $mathrmDivCl(X)$ denotes the group $mathrmDiv(X)$ modulo linear equivalence. This map is surjective if $X$ is an integral scheme, but for general schemes not all line bundles come from Cartier divisors.
Now to answer your question, for any scheme $X$ and a line bundle $mathcalL$ on it, you have the following natural one to one correspondence
$$ texteffective cartier divisors D text such that mathcalO_X(D) cong mathcalL leftrightarrow textnon-zero divisors of Gamma(X, mathcalL) big/sim, $$ where $ s sim s'$ if and only if $s' = us$ for some $u in Gamma(X, mathcalO_X^times).$ Note that we are only considering effective Cartier divisors, because the associated line bundle of any divisor has a nonzero global section if and only if it is linearly equivalent to an effective one.
answered Jul 28 at 7:28
Parthiv BasuParthiv Basu
1,03010 bronze badges
$endgroup$
Cartier divisors can be defined on any scheme. And to any Cartier divisor $D$ on any scheme $X$ you can associate a line bundle (invertible $mathcalO_X$-module) which is usually denoted by $mathcalO_X(D)$. So there is always a map $$ beginalign mathrmDiv(X) &to mathrmPic(X) \ D &to mathcalO_X(D), endalign $$ from the group of Cartier divisors to the group of line bundles. This further induces an injective map $mathrmDivCl(X) to mathrmPic(X)$, where $mathrmDivCl(X)$ denotes the group $mathrmDiv(X)$ modulo linear equivalence. This map is surjective if $X$ is an integral scheme, but for general schemes not all line bundles come from Cartier divisors.
Now to answer your question, for any scheme $X$ and a line bundle $mathcalL$ on it, you have the following natural one to one correspondence
$$ texteffective cartier divisors D text such that mathcalO_X(D) cong mathcalL leftrightarrow textnon-zero divisors of Gamma(X, mathcalL) big/sim, $$ where $ s sim s'$ if and only if $s' = us$ for some $u in Gamma(X, mathcalO_X^times).$ Note that we are only considering effective Cartier divisors, because the associated line bundle of any divisor has a nonzero global section if and only if it is linearly equivalent to an effective one.
answered Jul 28 at 7:28
Parthiv BasuParthiv Basu
1,03010 bronze badges
edited Jul 28 at 17:35
answered Jul 28 at 7:28
Parthiv BasuParthiv Basu
1,03010 bronze badges
answered Jul 28 at 7:28
Parthiv BasuParthiv Basu
1,03010 bronze badges
answered Jul 28 at 7:28
Parthiv BasuParthiv Basu
1,03010 bronze badges
1,03010 bronze badges
$begingroup$
If you want explicit details, see pages $302-305$ of Algebraic Geometry $1$ by Görtz, Wedhorn.
$endgroup$
– Parthiv Basu
Jul 28 at 8:00
add a comment |
$begingroup$
If you want explicit details, see pages $302-305$ of Algebraic Geometry $1$ by Görtz, Wedhorn.
$endgroup$
– Parthiv Basu
Jul 28 at 8:00
$begingroup$
If you want explicit details, see pages $302-305$ of Algebraic Geometry $1$ by Görtz, Wedhorn.
$endgroup$
– Parthiv Basu
Jul 28 at 8:00
$begingroup$
If you want explicit details, see pages $302-305$ of Algebraic Geometry $1$ by Görtz, Wedhorn.
$endgroup$
– Parthiv Basu
Jul 28 at 8:00
add a comment |
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$begingroup$
how do you define a rational section of a line bundle on a general scheme?
$endgroup$
– user690882
Jul 28 at 6:29
$begingroup$
Actually good question. I guess for an irreducible scheme it is fine, but I don't know how it would be defined more generally than that.
$endgroup$
– Luke
Jul 28 at 6:36