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Cartier divisor example in Harthsorne

Questions about the definition of a Cartier Divisor from Liu pg 256Question about cusp cubic example in HartshorneQuestion about Hartshorne Example 6.11.4 in Chapter IIRepresentation of an effective Cartier divisor: is there an imprecision in Liu's book?Multiplicity of Cartier divisor on locally noetherian scheme is only non-zero at generic pointMapping Cartier Divisors to Picard GroupPreimage of Cartier divisor under finite morphismExtending a Cartier divisor on a curve to singular points?Constructing an invertible sheaf from a Cartier divisor?Why is this Weil divisor not a Cartier divisor

1

$begingroup$

This question is about example 6.11.4 in Chapter II of Hartshorne. The example is about computing the Cartier divisor class group of the cuspidal cubic curve $y^2z = x^3$ in $mathbbP_k^2$. He begins by saying "note that any Cartier divisor is linearly equivalent to one whose local function is invertible in some neighbourhood of the singular point $Z = (0,0,1)$".

I am not sure what this statement means at all. A Cartier divisor is, by definition, represented by a family $(U_i, f_i) $ where the $U_i$ cover $X$ and the $f_i in mathcalK^*(U_i)$. But the elements of $mathcalK^*(U_i)$ are invertible by definition. So every local function is invertible, right? So what does Hartshorne mean by hsi statement?

algebraic-geometry proof-explanation schemes divisors-algebraic-geometry

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asked Mar 21 at 8:07

LukeLuke

91937

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1

$begingroup$

This question is about example 6.11.4 in Chapter II of Hartshorne. The example is about computing the Cartier divisor class group of the cuspidal cubic curve $y^2z = x^3$ in $mathbbP_k^2$. He begins by saying "note that any Cartier divisor is linearly equivalent to one whose local function is invertible in some neighbourhood of the singular point $Z = (0,0,1)$".

I am not sure what this statement means at all. A Cartier divisor is, by definition, represented by a family $(U_i, f_i) $ where the $U_i$ cover $X$ and the $f_i in mathcalK^*(U_i)$. But the elements of $mathcalK^*(U_i)$ are invertible by definition. So every local function is invertible, right? So what does Hartshorne mean by hsi statement?

algebraic-geometry proof-explanation schemes divisors-algebraic-geometry

share|cite|improve this question

asked Mar 21 at 8:07

LukeLuke

91937

$endgroup$

add a comment | 

$begingroup$

This question is about example 6.11.4 in Chapter II of Hartshorne. The example is about computing the Cartier divisor class group of the cuspidal cubic curve $y^2z = x^3$ in $mathbbP_k^2$. He begins by saying "note that any Cartier divisor is linearly equivalent to one whose local function is invertible in some neighbourhood of the singular point $Z = (0,0,1)$".

I am not sure what this statement means at all. A Cartier divisor is, by definition, represented by a family $(U_i, f_i) $ where the $U_i$ cover $X$ and the $f_i in mathcalK^*(U_i)$. But the elements of $mathcalK^*(U_i)$ are invertible by definition. So every local function is invertible, right? So what does Hartshorne mean by hsi statement?

algebraic-geometry proof-explanation schemes divisors-algebraic-geometry

share|cite|improve this question

asked Mar 21 at 8:07

LukeLuke

91937

$endgroup$

share|cite|improve this question

asked Mar 21 at 8:07

LukeLuke

91937

share|cite|improve this question

asked Mar 21 at 8:07

LukeLuke

91937

asked Mar 21 at 8:07

LukeLuke

91937

asked Mar 21 at 8:07

LukeLuke

91937

1 Answer
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0

$begingroup$

I agree he did not phrase it very well. Let $U$ be an open chart containing the cusp. By removing the cuspidal point from the other charts, we may assume it is the only one containing the cuspidal point. Let $f$ be the rational function on $U$. What he means, is that we may assume that $f$ is invertible in $mathcalO$ (and not in $mathcalK$, which would be obvious as $mathcalK$ is a field). So, he is assuming that $f in mathcalO^*(V)$, where $V subset U$ is an open set containing the cusp.

The way to achieve it is the following. Notice that $f^-1$ is an element of $mathcalK^*$, as so is $f$. Thus, $f^-1$ gives a principal Cartier divisor. So, the divisor $lbrace (U_i,f_i) rbrace$ is the same as $lbrace (U_i,f_i cdot f^-1) rbrace$. Since $U = U_i_0$ for some index $i_0$, this shows that we may assume that on that chart the divisor is represented by $(U,1)$.

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answered Mar 21 at 14:49

StefanoStefano

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0

$begingroup$

I agree he did not phrase it very well. Let $U$ be an open chart containing the cusp. By removing the cuspidal point from the other charts, we may assume it is the only one containing the cuspidal point. Let $f$ be the rational function on $U$. What he means, is that we may assume that $f$ is invertible in $mathcalO$ (and not in $mathcalK$, which would be obvious as $mathcalK$ is a field). So, he is assuming that $f in mathcalO^*(V)$, where $V subset U$ is an open set containing the cusp.

The way to achieve it is the following. Notice that $f^-1$ is an element of $mathcalK^*$, as so is $f$. Thus, $f^-1$ gives a principal Cartier divisor. So, the divisor $lbrace (U_i,f_i) rbrace$ is the same as $lbrace (U_i,f_i cdot f^-1) rbrace$. Since $U = U_i_0$ for some index $i_0$, this shows that we may assume that on that chart the divisor is represented by $(U,1)$.

share|cite|improve this answer

answered Mar 21 at 14:49

StefanoStefano

2,223931

$endgroup$

add a comment | 

0

$begingroup$

I agree he did not phrase it very well. Let $U$ be an open chart containing the cusp. By removing the cuspidal point from the other charts, we may assume it is the only one containing the cuspidal point. Let $f$ be the rational function on $U$. What he means, is that we may assume that $f$ is invertible in $mathcalO$ (and not in $mathcalK$, which would be obvious as $mathcalK$ is a field). So, he is assuming that $f in mathcalO^*(V)$, where $V subset U$ is an open set containing the cusp.

The way to achieve it is the following. Notice that $f^-1$ is an element of $mathcalK^*$, as so is $f$. Thus, $f^-1$ gives a principal Cartier divisor. So, the divisor $lbrace (U_i,f_i) rbrace$ is the same as $lbrace (U_i,f_i cdot f^-1) rbrace$. Since $U = U_i_0$ for some index $i_0$, this shows that we may assume that on that chart the divisor is represented by $(U,1)$.

share|cite|improve this answer

answered Mar 21 at 14:49

StefanoStefano

2,223931

$endgroup$

add a comment | 

$begingroup$

I agree he did not phrase it very well. Let $U$ be an open chart containing the cusp. By removing the cuspidal point from the other charts, we may assume it is the only one containing the cuspidal point. Let $f$ be the rational function on $U$. What he means, is that we may assume that $f$ is invertible in $mathcalO$ (and not in $mathcalK$, which would be obvious as $mathcalK$ is a field). So, he is assuming that $f in mathcalO^*(V)$, where $V subset U$ is an open set containing the cusp.

The way to achieve it is the following. Notice that $f^-1$ is an element of $mathcalK^*$, as so is $f$. Thus, $f^-1$ gives a principal Cartier divisor. So, the divisor $lbrace (U_i,f_i) rbrace$ is the same as $lbrace (U_i,f_i cdot f^-1) rbrace$. Since $U = U_i_0$ for some index $i_0$, this shows that we may assume that on that chart the divisor is represented by $(U,1)$.

share|cite|improve this answer

answered Mar 21 at 14:49

StefanoStefano

2,223931

$endgroup$

share|cite|improve this answer

answered Mar 21 at 14:49

StefanoStefano

2,223931

answered Mar 21 at 14:49

StefanoStefano

2,223931

answered Mar 21 at 14:49

StefanoStefano

2,223931