Castelnuovo's basepoint free pencil trick

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I am having trouble solving Exercise 4.13 in Eisenbud's "Geometry of Syzygies":

Suppose that $mathscrL$ is a line bundle on a (smooth) curve $Xsubset mathbbP^r$ over an infinite field, and suppose that $mathscrL$ is basepoint-free. Show that we may choose 2 sections $sigma_1, sigma_2$ of $mathscrL$ which together form a base-point free pencil--that is, $V=langle sigma_1,sigma_2 rangle$ is a two-dimensional subspace of $H^0(mathscrL)$ which generates $mathscrL$ locally everywhere. Show that the Koszul complex on $sigma_1, sigma_2$
$$
mathbbK: 0to mathscrL^-2to mathscrL^-1oplus mathscrL^-1to mathscrLto 0
$$
is exact, and remains exact when tensored with any sheaf.

We know that $mathscrL$ is generated by a vector space $Wsubseteq H^0(mathscrL)$ if there is a surjection $Wotimes mathcalO_Xto mathscrL$.

• How do we show that $W=V$ is two-dimensional?


• Where do the bundles $mathscrL^-1oplus mathscrL^-1$ and $mathscrL^-2$ come from?

My ultimate interest is the final assertion of the exercise:

Suppose that $X$ is embedded in $mathbbP^r$ as a curve of degree $dgeq 2g+1$, where $g$ is the genus of $X$. Use the argument above to show that
$$
H^0(mathcalO_X(1))otimes H^0(mathcalO_X(n))to H^0(mathcalO_X(n+1))
$$
is surjective...

I am hoping that someone can work out this exercise for some so I may understand the "pencil trick". I'm happy to assume that the field is $mathbbC$, as opposed to "infinite field" in the statement of the exercise. This trick is used frequently in the study of syzygies of curves and I cannot find a proof anywhere. I've tried to do this a few times in the past, and there must be something silly that I'm missing. Thanks.

algebraic-geometry complex-geometry

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edited Jul 19 at 18:19

asked Jul 19 at 18:08

Justine

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I am having trouble solving Exercise 4.13 in Eisenbud's "Geometry of Syzygies":

Suppose that $mathscrL$ is a line bundle on a (smooth) curve $Xsubset mathbbP^r$ over an infinite field, and suppose that $mathscrL$ is basepoint-free. Show that we may choose 2 sections $sigma_1, sigma_2$ of $mathscrL$ which together form a base-point free pencil--that is, $V=langle sigma_1,sigma_2 rangle$ is a two-dimensional subspace of $H^0(mathscrL)$ which generates $mathscrL$ locally everywhere. Show that the Koszul complex on $sigma_1, sigma_2$
$$
mathbbK: 0to mathscrL^-2to mathscrL^-1oplus mathscrL^-1to mathscrLto 0
$$
is exact, and remains exact when tensored with any sheaf.

We know that $mathscrL$ is generated by a vector space $Wsubseteq H^0(mathscrL)$ if there is a surjection $Wotimes mathcalO_Xto mathscrL$.

• How do we show that $W=V$ is two-dimensional?


• Where do the bundles $mathscrL^-1oplus mathscrL^-1$ and $mathscrL^-2$ come from?

My ultimate interest is the final assertion of the exercise:

Suppose that $X$ is embedded in $mathbbP^r$ as a curve of degree $dgeq 2g+1$, where $g$ is the genus of $X$. Use the argument above to show that
$$
H^0(mathcalO_X(1))otimes H^0(mathcalO_X(n))to H^0(mathcalO_X(n+1))
$$
is surjective...

I am hoping that someone can work out this exercise for some so I may understand the "pencil trick". I'm happy to assume that the field is $mathbbC$, as opposed to "infinite field" in the statement of the exercise. This trick is used frequently in the study of syzygies of curves and I cannot find a proof anywhere. I've tried to do this a few times in the past, and there must be something silly that I'm missing. Thanks.

algebraic-geometry complex-geometry

share|cite|improve this question

edited Jul 19 at 18:19

asked Jul 19 at 18:08

Justine

508211

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

I am having trouble solving Exercise 4.13 in Eisenbud's "Geometry of Syzygies":

Suppose that $mathscrL$ is a line bundle on a (smooth) curve $Xsubset mathbbP^r$ over an infinite field, and suppose that $mathscrL$ is basepoint-free. Show that we may choose 2 sections $sigma_1, sigma_2$ of $mathscrL$ which together form a base-point free pencil--that is, $V=langle sigma_1,sigma_2 rangle$ is a two-dimensional subspace of $H^0(mathscrL)$ which generates $mathscrL$ locally everywhere. Show that the Koszul complex on $sigma_1, sigma_2$
$$
mathbbK: 0to mathscrL^-2to mathscrL^-1oplus mathscrL^-1to mathscrLto 0
$$
is exact, and remains exact when tensored with any sheaf.

We know that $mathscrL$ is generated by a vector space $Wsubseteq H^0(mathscrL)$ if there is a surjection $Wotimes mathcalO_Xto mathscrL$.

• How do we show that $W=V$ is two-dimensional?


• Where do the bundles $mathscrL^-1oplus mathscrL^-1$ and $mathscrL^-2$ come from?

My ultimate interest is the final assertion of the exercise:

Suppose that $X$ is embedded in $mathbbP^r$ as a curve of degree $dgeq 2g+1$, where $g$ is the genus of $X$. Use the argument above to show that
$$
H^0(mathcalO_X(1))otimes H^0(mathcalO_X(n))to H^0(mathcalO_X(n+1))
$$
is surjective...

I am hoping that someone can work out this exercise for some so I may understand the "pencil trick". I'm happy to assume that the field is $mathbbC$, as opposed to "infinite field" in the statement of the exercise. This trick is used frequently in the study of syzygies of curves and I cannot find a proof anywhere. I've tried to do this a few times in the past, and there must be something silly that I'm missing. Thanks.

algebraic-geometry complex-geometry

share|cite|improve this question

edited Jul 19 at 18:19

asked Jul 19 at 18:08

Justine

508211

I am having trouble solving Exercise 4.13 in Eisenbud's "Geometry of Syzygies":

Suppose that $mathscrL$ is a line bundle on a (smooth) curve $Xsubset mathbbP^r$ over an infinite field, and suppose that $mathscrL$ is basepoint-free. Show that we may choose 2 sections $sigma_1, sigma_2$ of $mathscrL$ which together form a base-point free pencil--that is, $V=langle sigma_1,sigma_2 rangle$ is a two-dimensional subspace of $H^0(mathscrL)$ which generates $mathscrL$ locally everywhere. Show that the Koszul complex on $sigma_1, sigma_2$
$$
mathbbK: 0to mathscrL^-2to mathscrL^-1oplus mathscrL^-1to mathscrLto 0
$$
is exact, and remains exact when tensored with any sheaf.

We know that $mathscrL$ is generated by a vector space $Wsubseteq H^0(mathscrL)$ if there is a surjection $Wotimes mathcalO_Xto mathscrL$.

• How do we show that $W=V$ is two-dimensional?


• Where do the bundles $mathscrL^-1oplus mathscrL^-1$ and $mathscrL^-2$ come from?

My ultimate interest is the final assertion of the exercise:

Suppose that $X$ is embedded in $mathbbP^r$ as a curve of degree $dgeq 2g+1$, where $g$ is the genus of $X$. Use the argument above to show that
$$
H^0(mathcalO_X(1))otimes H^0(mathcalO_X(n))to H^0(mathcalO_X(n+1))
$$
is surjective...

I am hoping that someone can work out this exercise for some so I may understand the "pencil trick". I'm happy to assume that the field is $mathbbC$, as opposed to "infinite field" in the statement of the exercise. This trick is used frequently in the study of syzygies of curves and I cannot find a proof anywhere. I've tried to do this a few times in the past, and there must be something silly that I'm missing. Thanks.

algebraic-geometry complex-geometry

share|cite|improve this question

edited Jul 19 at 18:19

asked Jul 19 at 18:08

Justine

508211

share|cite|improve this question

share|cite|improve this question

edited Jul 19 at 18:19

edited Jul 19 at 18:19

edited Jul 19 at 18:19

asked Jul 19 at 18:08

Justine

508211

asked Jul 19 at 18:08

Justine

508211

asked Jul 19 at 18:08

Justine

508211

508211

add a comment | 

add a comment | 

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Let $0neq s$ be a section of $L$ (calligraphy is too much to type). Then, we have an exact sequence $0to mathcalO_Xstackrelsto Lto Tto 0$. Since $T$ is a coherent sheaf and zero at the generic point, we see that it is a skyscraper sheaf and since it is a quotient of a line bundle, easy to see that $TcongmathcalO_Z$ for some proper closed subset $Zsubset X$. Since $L$ is globally generated, we have images global sections of $L$ generate $T$ and then (using infinite field), there is a single section $t$ of $L$ such that its image generate $T$. Then, the subspace generated by $s,t$ will do what you want.

Your Koszul complex is not correct. It should be, $0to L^-2to L^-1oplus L^-1to mathcalO_Xto 0$, and comes from the surjection $mathcalO_X^2to L$ given by $s,t$. The kernel is $L^-1$ by determinant considerations and twist the sequence by $L^-1$ to get what you want.

Finally, for your last part, most of what you want can be proved by the vanishing of $H^1$ for all large degree line bundles, inducting on $n$. The case of $n=1$ requires some care.

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answered Jul 19 at 21:19

Mohan

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Let $0neq s$ be a section of $L$ (calligraphy is too much to type). Then, we have an exact sequence $0to mathcalO_Xstackrelsto Lto Tto 0$. Since $T$ is a coherent sheaf and zero at the generic point, we see that it is a skyscraper sheaf and since it is a quotient of a line bundle, easy to see that $TcongmathcalO_Z$ for some proper closed subset $Zsubset X$. Since $L$ is globally generated, we have images global sections of $L$ generate $T$ and then (using infinite field), there is a single section $t$ of $L$ such that its image generate $T$. Then, the subspace generated by $s,t$ will do what you want.

Your Koszul complex is not correct. It should be, $0to L^-2to L^-1oplus L^-1to mathcalO_Xto 0$, and comes from the surjection $mathcalO_X^2to L$ given by $s,t$. The kernel is $L^-1$ by determinant considerations and twist the sequence by $L^-1$ to get what you want.

Finally, for your last part, most of what you want can be proved by the vanishing of $H^1$ for all large degree line bundles, inducting on $n$. The case of $n=1$ requires some care.

share|cite|improve this answer

answered Jul 19 at 21:19

Mohan

11k1816

add a comment | 

up vote
1
down vote

Let $0neq s$ be a section of $L$ (calligraphy is too much to type). Then, we have an exact sequence $0to mathcalO_Xstackrelsto Lto Tto 0$. Since $T$ is a coherent sheaf and zero at the generic point, we see that it is a skyscraper sheaf and since it is a quotient of a line bundle, easy to see that $TcongmathcalO_Z$ for some proper closed subset $Zsubset X$. Since $L$ is globally generated, we have images global sections of $L$ generate $T$ and then (using infinite field), there is a single section $t$ of $L$ such that its image generate $T$. Then, the subspace generated by $s,t$ will do what you want.

Your Koszul complex is not correct. It should be, $0to L^-2to L^-1oplus L^-1to mathcalO_Xto 0$, and comes from the surjection $mathcalO_X^2to L$ given by $s,t$. The kernel is $L^-1$ by determinant considerations and twist the sequence by $L^-1$ to get what you want.

Finally, for your last part, most of what you want can be proved by the vanishing of $H^1$ for all large degree line bundles, inducting on $n$. The case of $n=1$ requires some care.

share|cite|improve this answer

answered Jul 19 at 21:19

Mohan

11k1816

add a comment | 

up vote
1
down vote

up vote
1
down vote

Let $0neq s$ be a section of $L$ (calligraphy is too much to type). Then, we have an exact sequence $0to mathcalO_Xstackrelsto Lto Tto 0$. Since $T$ is a coherent sheaf and zero at the generic point, we see that it is a skyscraper sheaf and since it is a quotient of a line bundle, easy to see that $TcongmathcalO_Z$ for some proper closed subset $Zsubset X$. Since $L$ is globally generated, we have images global sections of $L$ generate $T$ and then (using infinite field), there is a single section $t$ of $L$ such that its image generate $T$. Then, the subspace generated by $s,t$ will do what you want.

Your Koszul complex is not correct. It should be, $0to L^-2to L^-1oplus L^-1to mathcalO_Xto 0$, and comes from the surjection $mathcalO_X^2to L$ given by $s,t$. The kernel is $L^-1$ by determinant considerations and twist the sequence by $L^-1$ to get what you want.

Finally, for your last part, most of what you want can be proved by the vanishing of $H^1$ for all large degree line bundles, inducting on $n$. The case of $n=1$ requires some care.

share|cite|improve this answer

answered Jul 19 at 21:19

Mohan

11k1816

Let $0neq s$ be a section of $L$ (calligraphy is too much to type). Then, we have an exact sequence $0to mathcalO_Xstackrelsto Lto Tto 0$. Since $T$ is a coherent sheaf and zero at the generic point, we see that it is a skyscraper sheaf and since it is a quotient of a line bundle, easy to see that $TcongmathcalO_Z$ for some proper closed subset $Zsubset X$. Since $L$ is globally generated, we have images global sections of $L$ generate $T$ and then (using infinite field), there is a single section $t$ of $L$ such that its image generate $T$. Then, the subspace generated by $s,t$ will do what you want.

Your Koszul complex is not correct. It should be, $0to L^-2to L^-1oplus L^-1to mathcalO_Xto 0$, and comes from the surjection $mathcalO_X^2to L$ given by $s,t$. The kernel is $L^-1$ by determinant considerations and twist the sequence by $L^-1$ to get what you want.

Finally, for your last part, most of what you want can be proved by the vanishing of $H^1$ for all large degree line bundles, inducting on $n$. The case of $n=1$ requires some care.

share|cite|improve this answer

answered Jul 19 at 21:19

Mohan

11k1816

share|cite|improve this answer

share|cite|improve this answer

answered Jul 19 at 21:19

Mohan

11k1816

answered Jul 19 at 21:19

Mohan

11k1816

answered Jul 19 at 21:19

Mohan

11k1816

11k1816

add a comment | 

add a comment | 

 

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