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Why is this Weil divisor not a Cartier divisor
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$begingroup$
I'm reading "Introduction to toric varieties" by "William Fulton", here is an example in page 61 illustrating the difference between Weil divisors and Cartier divisors. Here my question is only about these divisors(not about toric variety), so I change the form such that it doesn't involve the toric variety. I know only basic definition of divisors so correct me if I write something wrong.
Example: In the cone $XZ-Y^2$, the line $X=Y=0$ is a Weil divisor $D$. Then $D$ is not a Cartier divisor but $2D$ is, and $2D$ is a principal divisor.
What is the function that give the principal divisor $2D$? Is it just $X^2$? Then why function $X$ does not give a principal divisor $D$?
algebraic-geometry divisors-algebraic-geometry
asked Mar 13 at 6:22
JialeJiale
585
$endgroup$
add a comment |
$begingroup$
I'm reading "Introduction to toric varieties" by "William Fulton", here is an example in page 61 illustrating the difference between Weil divisors and Cartier divisors. Here my question is only about these divisors(not about toric variety), so I change the form such that it doesn't involve the toric variety. I know only basic definition of divisors so correct me if I write something wrong.
Example: In the cone $XZ-Y^2$, the line $X=Y=0$ is a Weil divisor $D$. Then $D$ is not a Cartier divisor but $2D$ is, and $2D$ is a principal divisor.
What is the function that give the principal divisor $2D$? Is it just $X^2$? Then why function $X$ does not give a principal divisor $D$?
algebraic-geometry divisors-algebraic-geometry
asked Mar 13 at 6:22
JialeJiale
585
$endgroup$
2
$begingroup$
I think it's the function $X$ that gives the divisor $2D$.
$endgroup$
– Nicolas Hemelsoet
Mar 13 at 9:16
add a comment |
$begingroup$
I'm reading "Introduction to toric varieties" by "William Fulton", here is an example in page 61 illustrating the difference between Weil divisors and Cartier divisors. Here my question is only about these divisors(not about toric variety), so I change the form such that it doesn't involve the toric variety. I know only basic definition of divisors so correct me if I write something wrong.
Example: In the cone $XZ-Y^2$, the line $X=Y=0$ is a Weil divisor $D$. Then $D$ is not a Cartier divisor but $2D$ is, and $2D$ is a principal divisor.
What is the function that give the principal divisor $2D$? Is it just $X^2$? Then why function $X$ does not give a principal divisor $D$?
algebraic-geometry divisors-algebraic-geometry
asked Mar 13 at 6:22
JialeJiale
585
$endgroup$
I'm reading "Introduction to toric varieties" by "William Fulton", here is an example in page 61 illustrating the difference between Weil divisors and Cartier divisors. Here my question is only about these divisors(not about toric variety), so I change the form such that it doesn't involve the toric variety. I know only basic definition of divisors so correct me if I write something wrong.
Example: In the cone $XZ-Y^2$, the line $X=Y=0$ is a Weil divisor $D$. Then $D$ is not a Cartier divisor but $2D$ is, and $2D$ is a principal divisor.
What is the function that give the principal divisor $2D$? Is it just $X^2$? Then why function $X$ does not give a principal divisor $D$?
algebraic-geometry divisors-algebraic-geometry
algebraic-geometry divisors-algebraic-geometry
asked Mar 13 at 6:22
JialeJiale
585
asked Mar 13 at 6:22
JialeJiale
585
asked Mar 13 at 6:22
JialeJiale
585
asked Mar 13 at 6:22
JialeJiale
585
asked Mar 13 at 6:22
JialeJiale
585
585
2
$begingroup$
I think it's the function $X$ that gives the divisor $2D$.
$endgroup$
– Nicolas Hemelsoet
Mar 13 at 9:16
add a comment |
2
$begingroup$
I think it's the function $X$ that gives the divisor $2D$.
$endgroup$
– Nicolas Hemelsoet
Mar 13 at 9:16
2
2
$begingroup$
I think it's the function $X$ that gives the divisor $2D$.
$endgroup$
– Nicolas Hemelsoet
Mar 13 at 9:16
$begingroup$
I think it's the function $X$ that gives the divisor $2D$.
$endgroup$
– Nicolas Hemelsoet
Mar 13 at 9:16
add a comment |
1 Answer
1
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oldest
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$begingroup$
I think I get it now. Is this correct?
The local ring $mathbb{C}[X,Y,Z]_{(X,Y)}$ has a maximal ideal $(X,Y)$, it is generated by $Y$ because $X=Y(Y/Z)$, then $X$ has valuation $2$ at $D$. Function $Y$ has valuation $1$ at $D$ but $div(Y)=D+D'$ where $D'$ is the line $Y=Z=0$.
answered Mar 13 at 15:36
JialeJiale
585
$endgroup$
add a comment |
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1 Answer
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$begingroup$
I think I get it now. Is this correct?
The local ring $mathbb{C}[X,Y,Z]_{(X,Y)}$ has a maximal ideal $(X,Y)$, it is generated by $Y$ because $X=Y(Y/Z)$, then $X$ has valuation $2$ at $D$. Function $Y$ has valuation $1$ at $D$ but $div(Y)=D+D'$ where $D'$ is the line $Y=Z=0$.
answered Mar 13 at 15:36
JialeJiale
585
$endgroup$
add a comment |
$begingroup$
I think I get it now. Is this correct?
The local ring $mathbb{C}[X,Y,Z]_{(X,Y)}$ has a maximal ideal $(X,Y)$, it is generated by $Y$ because $X=Y(Y/Z)$, then $X$ has valuation $2$ at $D$. Function $Y$ has valuation $1$ at $D$ but $div(Y)=D+D'$ where $D'$ is the line $Y=Z=0$.
answered Mar 13 at 15:36
JialeJiale
585
$endgroup$
add a comment |
$begingroup$
I think I get it now. Is this correct?
The local ring $mathbb{C}[X,Y,Z]_{(X,Y)}$ has a maximal ideal $(X,Y)$, it is generated by $Y$ because $X=Y(Y/Z)$, then $X$ has valuation $2$ at $D$. Function $Y$ has valuation $1$ at $D$ but $div(Y)=D+D'$ where $D'$ is the line $Y=Z=0$.
answered Mar 13 at 15:36
JialeJiale
585
$endgroup$
I think I get it now. Is this correct?
The local ring $mathbb{C}[X,Y,Z]_{(X,Y)}$ has a maximal ideal $(X,Y)$, it is generated by $Y$ because $X=Y(Y/Z)$, then $X$ has valuation $2$ at $D$. Function $Y$ has valuation $1$ at $D$ but $div(Y)=D+D'$ where $D'$ is the line $Y=Z=0$.
answered Mar 13 at 15:36
JialeJiale
585
answered Mar 13 at 15:36
JialeJiale
585
answered Mar 13 at 15:36
JialeJiale
585
answered Mar 13 at 15:36
JialeJiale
585
585
add a comment |
add a comment |
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I think it's the function $X$ that gives the divisor $2D$.
$endgroup$
– Nicolas Hemelsoet
Mar 13 at 9:16