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Projective cubic curve passing through seven points in $mathbbC^2$
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Let $x_1,...,x_7$ be distinct points in $mathbbC^2$. Prove that there exists a cubic curve passing through these points which has a singularity at the point $x_1$.
My attempt so far... A related question was: show that for every five points $a_1,...,a_5inmathbbP^2$, there is a conic containing them. In the given solution it was (roughly) argued, that the space of quadratic forms is a $6$-dimensional vector space and that the condition $q(a_i)=0$ for the conic $q=0$ is a linear equation in the coefficients of $q$. Now, imposing $5$ linear conditions on a $6$-dimensional vector space leaves at least a $1$-dimensional space of solutions - which is the desired conic.
Is there a way to apply this approach to the given question? If so, I assume the space of cubic forms has dimension $10$, is this correct? I don't know how the singularity would show up here.
algebraic-geometry algebraic-curves projective-geometry
edited Jul 25 at 21:46
Armando j18eos
2,36711125
asked Jul 23 at 15:59
casimir
406
 |Â
show 6 more comments
up vote
6
down vote
favorite
Let $x_1,...,x_7$ be distinct points in $mathbbC^2$. Prove that there exists a cubic curve passing through these points which has a singularity at the point $x_1$.
My attempt so far... A related question was: show that for every five points $a_1,...,a_5inmathbbP^2$, there is a conic containing them. In the given solution it was (roughly) argued, that the space of quadratic forms is a $6$-dimensional vector space and that the condition $q(a_i)=0$ for the conic $q=0$ is a linear equation in the coefficients of $q$. Now, imposing $5$ linear conditions on a $6$-dimensional vector space leaves at least a $1$-dimensional space of solutions - which is the desired conic.
Is there a way to apply this approach to the given question? If so, I assume the space of cubic forms has dimension $10$, is this correct? I don't know how the singularity would show up here.
algebraic-geometry algebraic-curves projective-geometry
edited Jul 25 at 21:46
Armando j18eos
2,36711125
asked Jul 23 at 15:59
casimir
406
Hi @casimir could you give me the title of the books thats you are working on ?
– Bernstein
Jul 23 at 16:03
I'm not working with a book, only with lecture notes ... and they don't go into any detail with respect to the question asked.
– casimir
Jul 23 at 16:05
1
thank you very much , good luck with your question
– Bernstein
Jul 23 at 16:13
3
Sottile discusses the $12$ nodal cubics through $8$ points.
– Jan-Magnus Økland
Jul 23 at 16:35
3
This is indeed quite similar to the conic question. Can you see that there exists a cubic curve passing through $x_1,ldots,x_7$? (This corresponds to 7 equations on your coefficients). Asking for your curve to have a singularity at $x_1$ imposes some more equations (how many?).
– loch
Jul 23 at 17:17
 |Â
show 6 more comments
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Let $x_1,...,x_7$ be distinct points in $mathbbC^2$. Prove that there exists a cubic curve passing through these points which has a singularity at the point $x_1$.
My attempt so far... A related question was: show that for every five points $a_1,...,a_5inmathbbP^2$, there is a conic containing them. In the given solution it was (roughly) argued, that the space of quadratic forms is a $6$-dimensional vector space and that the condition $q(a_i)=0$ for the conic $q=0$ is a linear equation in the coefficients of $q$. Now, imposing $5$ linear conditions on a $6$-dimensional vector space leaves at least a $1$-dimensional space of solutions - which is the desired conic.
Is there a way to apply this approach to the given question? If so, I assume the space of cubic forms has dimension $10$, is this correct? I don't know how the singularity would show up here.
algebraic-geometry algebraic-curves projective-geometry
edited Jul 25 at 21:46
Armando j18eos
2,36711125
asked Jul 23 at 15:59
casimir
406
Let $x_1,...,x_7$ be distinct points in $mathbbC^2$. Prove that there exists a cubic curve passing through these points which has a singularity at the point $x_1$.
My attempt so far... A related question was: show that for every five points $a_1,...,a_5inmathbbP^2$, there is a conic containing them. In the given solution it was (roughly) argued, that the space of quadratic forms is a $6$-dimensional vector space and that the condition $q(a_i)=0$ for the conic $q=0$ is a linear equation in the coefficients of $q$. Now, imposing $5$ linear conditions on a $6$-dimensional vector space leaves at least a $1$-dimensional space of solutions - which is the desired conic.
Is there a way to apply this approach to the given question? If so, I assume the space of cubic forms has dimension $10$, is this correct? I don't know how the singularity would show up here.
algebraic-geometry algebraic-curves projective-geometry
edited Jul 25 at 21:46
Armando j18eos
2,36711125
asked Jul 23 at 15:59
casimir
406
edited Jul 25 at 21:46
Armando j18eos
2,36711125
edited Jul 25 at 21:46
Armando j18eos
2,36711125
edited Jul 25 at 21:46
Armando j18eos
2,36711125
2,36711125
asked Jul 23 at 15:59
casimir
406
asked Jul 23 at 15:59
casimir
406
asked Jul 23 at 15:59
casimir
406
406
Hi @casimir could you give me the title of the books thats you are working on ?
– Bernstein
Jul 23 at 16:03
I'm not working with a book, only with lecture notes ... and they don't go into any detail with respect to the question asked.
– casimir
Jul 23 at 16:05
1
thank you very much , good luck with your question
– Bernstein
Jul 23 at 16:13
3
Sottile discusses the $12$ nodal cubics through $8$ points.
– Jan-Magnus Økland
Jul 23 at 16:35
3
This is indeed quite similar to the conic question. Can you see that there exists a cubic curve passing through $x_1,ldots,x_7$? (This corresponds to 7 equations on your coefficients). Asking for your curve to have a singularity at $x_1$ imposes some more equations (how many?).
– loch
Jul 23 at 17:17
 |Â
show 6 more comments
Hi @casimir could you give me the title of the books thats you are working on ?
– Bernstein
Jul 23 at 16:03
I'm not working with a book, only with lecture notes ... and they don't go into any detail with respect to the question asked.
– casimir
Jul 23 at 16:05
1
thank you very much , good luck with your question
– Bernstein
Jul 23 at 16:13
3
Sottile discusses the $12$ nodal cubics through $8$ points.
– Jan-Magnus Økland
Jul 23 at 16:35
3
This is indeed quite similar to the conic question. Can you see that there exists a cubic curve passing through $x_1,ldots,x_7$? (This corresponds to 7 equations on your coefficients). Asking for your curve to have a singularity at $x_1$ imposes some more equations (how many?).
– loch
Jul 23 at 17:17
Hi @casimir could you give me the title of the books thats you are working on ?
– Bernstein
Jul 23 at 16:03
Hi @casimir could you give me the title of the books thats you are working on ?
– Bernstein
Jul 23 at 16:03
I'm not working with a book, only with lecture notes ... and they don't go into any detail with respect to the question asked.
– casimir
Jul 23 at 16:05
I'm not working with a book, only with lecture notes ... and they don't go into any detail with respect to the question asked.
– casimir
Jul 23 at 16:05
1
1
thank you very much , good luck with your question
– Bernstein
Jul 23 at 16:13
thank you very much , good luck with your question
– Bernstein
Jul 23 at 16:13
3
3
Sottile discusses the $12$ nodal cubics through $8$ points.
– Jan-Magnus Økland
Jul 23 at 16:35
Sottile discusses the $12$ nodal cubics through $8$ points.
– Jan-Magnus Økland
Jul 23 at 16:35
3
3
This is indeed quite similar to the conic question. Can you see that there exists a cubic curve passing through $x_1,ldots,x_7$? (This corresponds to 7 equations on your coefficients). Asking for your curve to have a singularity at $x_1$ imposes some more equations (how many?).
– loch
Jul 23 at 17:17
This is indeed quite similar to the conic question. Can you see that there exists a cubic curve passing through $x_1,ldots,x_7$? (This corresponds to 7 equations on your coefficients). Asking for your curve to have a singularity at $x_1$ imposes some more equations (how many?).
– loch
Jul 23 at 17:17
 |Â
show 6 more comments
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Let $mathbbP^NequivmathbbP(mathbbC[z_0,z_1,z_2]_3)$ be the space of plane cubic (projective) curves; its dimension $N$ is
beginequation
binom3-1+33-1=10-1=9.
endequation
Let $displaystyle F=sum_i_0+i_1+i_2=3a_i_0i_1i_2z_0^i_0z_1^i_1z_2^i^2$ be the generic homogeneous polynomial of degree $3$ in three variables; let
beginequation
nu_2,3:mathbbP^2tomathbbP^9
endequation
be the Veronese embedding of the (complex) projective plane in $mathbbP^9$; one knows that the image of $F$ via $nu_2,3$ is a hyperplane $H$.
The conditions $F([1:x_k])=0$, where $kin1,dots,7$ and $[1:x_k]$ are the homogeneous cordinates of $x_k$'s in $mathbbP^2$, constraint $H$ passing through $7$ distinct points of $mathbbP^9$: this is possible, so there exist cubic plane curves passing through the points $x_k$.
Remark 1. If one has $9$ distinct points in $mathbbP^2$, then the hyperplane $H$ is uniquely determined; in other words, on $9$ distinct points in $mathbbP^2$ gets through a unique cubic plane curve.
Let $F_z_1equiv F_1$ and $F_z_2equiv F_2$ be the partial derivaties of $F$; the curve $Gamma=F(1:z_1:z_2)=0$ has a (unique) singular point if the system
beginequation
begincases
F(1:z_1:z_2)=0\
F_1(1:z_1:z_2)=0\
F_2(1:z_1:z_2)=0
endcases
endequation
admits a (unique) solution; by the previous reasoning, one fixes $7$ of $10$ coefficients of $F$, so the previous system can have at most a unique solution after fixed the other $3$ coefficients of $F$.
Finded this solution, one has determined a cubic plane curve $Gamma$ passing through the points $x_k$ with a singular point.
Remark 2. The projective closure of $Gamma$ is an elliptic curve.
answered Jul 26 at 12:44
Armando j18eos
2,36711125
Very helpful, thank you!
– casimir
Jul 27 at 9:57
1
You are welcome. P.S.: The trick of Veronese embedding works also for the generic (algebraic) hypersurface of degree $dgeq1$. ;)
– Armando j18eos
Jul 27 at 10:01
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $mathbbP^NequivmathbbP(mathbbC[z_0,z_1,z_2]_3)$ be the space of plane cubic (projective) curves; its dimension $N$ is
beginequation
binom3-1+33-1=10-1=9.
endequation
Let $displaystyle F=sum_i_0+i_1+i_2=3a_i_0i_1i_2z_0^i_0z_1^i_1z_2^i^2$ be the generic homogeneous polynomial of degree $3$ in three variables; let
beginequation
nu_2,3:mathbbP^2tomathbbP^9
endequation
be the Veronese embedding of the (complex) projective plane in $mathbbP^9$; one knows that the image of $F$ via $nu_2,3$ is a hyperplane $H$.
The conditions $F([1:x_k])=0$, where $kin1,dots,7$ and $[1:x_k]$ are the homogeneous cordinates of $x_k$'s in $mathbbP^2$, constraint $H$ passing through $7$ distinct points of $mathbbP^9$: this is possible, so there exist cubic plane curves passing through the points $x_k$.
Remark 1. If one has $9$ distinct points in $mathbbP^2$, then the hyperplane $H$ is uniquely determined; in other words, on $9$ distinct points in $mathbbP^2$ gets through a unique cubic plane curve.
Let $F_z_1equiv F_1$ and $F_z_2equiv F_2$ be the partial derivaties of $F$; the curve $Gamma=F(1:z_1:z_2)=0$ has a (unique) singular point if the system
beginequation
begincases
F(1:z_1:z_2)=0\
F_1(1:z_1:z_2)=0\
F_2(1:z_1:z_2)=0
endcases
endequation
admits a (unique) solution; by the previous reasoning, one fixes $7$ of $10$ coefficients of $F$, so the previous system can have at most a unique solution after fixed the other $3$ coefficients of $F$.
Finded this solution, one has determined a cubic plane curve $Gamma$ passing through the points $x_k$ with a singular point.
Remark 2. The projective closure of $Gamma$ is an elliptic curve.
answered Jul 26 at 12:44
Armando j18eos
2,36711125
Very helpful, thank you!
– casimir
Jul 27 at 9:57
1
You are welcome. P.S.: The trick of Veronese embedding works also for the generic (algebraic) hypersurface of degree $dgeq1$. ;)
– Armando j18eos
Jul 27 at 10:01
add a comment |Â
up vote
1
down vote
accepted
Let $mathbbP^NequivmathbbP(mathbbC[z_0,z_1,z_2]_3)$ be the space of plane cubic (projective) curves; its dimension $N$ is
beginequation
binom3-1+33-1=10-1=9.
endequation
Let $displaystyle F=sum_i_0+i_1+i_2=3a_i_0i_1i_2z_0^i_0z_1^i_1z_2^i^2$ be the generic homogeneous polynomial of degree $3$ in three variables; let
beginequation
nu_2,3:mathbbP^2tomathbbP^9
endequation
be the Veronese embedding of the (complex) projective plane in $mathbbP^9$; one knows that the image of $F$ via $nu_2,3$ is a hyperplane $H$.
The conditions $F([1:x_k])=0$, where $kin1,dots,7$ and $[1:x_k]$ are the homogeneous cordinates of $x_k$'s in $mathbbP^2$, constraint $H$ passing through $7$ distinct points of $mathbbP^9$: this is possible, so there exist cubic plane curves passing through the points $x_k$.
Remark 1. If one has $9$ distinct points in $mathbbP^2$, then the hyperplane $H$ is uniquely determined; in other words, on $9$ distinct points in $mathbbP^2$ gets through a unique cubic plane curve.
Let $F_z_1equiv F_1$ and $F_z_2equiv F_2$ be the partial derivaties of $F$; the curve $Gamma=F(1:z_1:z_2)=0$ has a (unique) singular point if the system
beginequation
begincases
F(1:z_1:z_2)=0\
F_1(1:z_1:z_2)=0\
F_2(1:z_1:z_2)=0
endcases
endequation
admits a (unique) solution; by the previous reasoning, one fixes $7$ of $10$ coefficients of $F$, so the previous system can have at most a unique solution after fixed the other $3$ coefficients of $F$.
Finded this solution, one has determined a cubic plane curve $Gamma$ passing through the points $x_k$ with a singular point.
Remark 2. The projective closure of $Gamma$ is an elliptic curve.
answered Jul 26 at 12:44
Armando j18eos
2,36711125
Very helpful, thank you!
– casimir
Jul 27 at 9:57
1
You are welcome. P.S.: The trick of Veronese embedding works also for the generic (algebraic) hypersurface of degree $dgeq1$. ;)
– Armando j18eos
Jul 27 at 10:01
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $mathbbP^NequivmathbbP(mathbbC[z_0,z_1,z_2]_3)$ be the space of plane cubic (projective) curves; its dimension $N$ is
beginequation
binom3-1+33-1=10-1=9.
endequation
Let $displaystyle F=sum_i_0+i_1+i_2=3a_i_0i_1i_2z_0^i_0z_1^i_1z_2^i^2$ be the generic homogeneous polynomial of degree $3$ in three variables; let
beginequation
nu_2,3:mathbbP^2tomathbbP^9
endequation
be the Veronese embedding of the (complex) projective plane in $mathbbP^9$; one knows that the image of $F$ via $nu_2,3$ is a hyperplane $H$.
The conditions $F([1:x_k])=0$, where $kin1,dots,7$ and $[1:x_k]$ are the homogeneous cordinates of $x_k$'s in $mathbbP^2$, constraint $H$ passing through $7$ distinct points of $mathbbP^9$: this is possible, so there exist cubic plane curves passing through the points $x_k$.
Remark 1. If one has $9$ distinct points in $mathbbP^2$, then the hyperplane $H$ is uniquely determined; in other words, on $9$ distinct points in $mathbbP^2$ gets through a unique cubic plane curve.
Let $F_z_1equiv F_1$ and $F_z_2equiv F_2$ be the partial derivaties of $F$; the curve $Gamma=F(1:z_1:z_2)=0$ has a (unique) singular point if the system
beginequation
begincases
F(1:z_1:z_2)=0\
F_1(1:z_1:z_2)=0\
F_2(1:z_1:z_2)=0
endcases
endequation
admits a (unique) solution; by the previous reasoning, one fixes $7$ of $10$ coefficients of $F$, so the previous system can have at most a unique solution after fixed the other $3$ coefficients of $F$.
Finded this solution, one has determined a cubic plane curve $Gamma$ passing through the points $x_k$ with a singular point.
Remark 2. The projective closure of $Gamma$ is an elliptic curve.
answered Jul 26 at 12:44
Armando j18eos
2,36711125
Let $mathbbP^NequivmathbbP(mathbbC[z_0,z_1,z_2]_3)$ be the space of plane cubic (projective) curves; its dimension $N$ is
beginequation
binom3-1+33-1=10-1=9.
endequation
Let $displaystyle F=sum_i_0+i_1+i_2=3a_i_0i_1i_2z_0^i_0z_1^i_1z_2^i^2$ be the generic homogeneous polynomial of degree $3$ in three variables; let
beginequation
nu_2,3:mathbbP^2tomathbbP^9
endequation
be the Veronese embedding of the (complex) projective plane in $mathbbP^9$; one knows that the image of $F$ via $nu_2,3$ is a hyperplane $H$.
The conditions $F([1:x_k])=0$, where $kin1,dots,7$ and $[1:x_k]$ are the homogeneous cordinates of $x_k$'s in $mathbbP^2$, constraint $H$ passing through $7$ distinct points of $mathbbP^9$: this is possible, so there exist cubic plane curves passing through the points $x_k$.
Remark 1. If one has $9$ distinct points in $mathbbP^2$, then the hyperplane $H$ is uniquely determined; in other words, on $9$ distinct points in $mathbbP^2$ gets through a unique cubic plane curve.
Let $F_z_1equiv F_1$ and $F_z_2equiv F_2$ be the partial derivaties of $F$; the curve $Gamma=F(1:z_1:z_2)=0$ has a (unique) singular point if the system
beginequation
begincases
F(1:z_1:z_2)=0\
F_1(1:z_1:z_2)=0\
F_2(1:z_1:z_2)=0
endcases
endequation
admits a (unique) solution; by the previous reasoning, one fixes $7$ of $10$ coefficients of $F$, so the previous system can have at most a unique solution after fixed the other $3$ coefficients of $F$.
Finded this solution, one has determined a cubic plane curve $Gamma$ passing through the points $x_k$ with a singular point.
Remark 2. The projective closure of $Gamma$ is an elliptic curve.
answered Jul 26 at 12:44
Armando j18eos
2,36711125
edited Jul 26 at 13:21
answered Jul 26 at 12:44
Armando j18eos
2,36711125
answered Jul 26 at 12:44
Armando j18eos
2,36711125
answered Jul 26 at 12:44
Armando j18eos
2,36711125
2,36711125
Very helpful, thank you!
– casimir
Jul 27 at 9:57
1
You are welcome. P.S.: The trick of Veronese embedding works also for the generic (algebraic) hypersurface of degree $dgeq1$. ;)
– Armando j18eos
Jul 27 at 10:01
add a comment |Â
Very helpful, thank you!
– casimir
Jul 27 at 9:57
1
You are welcome. P.S.: The trick of Veronese embedding works also for the generic (algebraic) hypersurface of degree $dgeq1$. ;)
– Armando j18eos
Jul 27 at 10:01
Very helpful, thank you!
– casimir
Jul 27 at 9:57
Very helpful, thank you!
– casimir
Jul 27 at 9:57
1
1
You are welcome. P.S.: The trick of Veronese embedding works also for the generic (algebraic) hypersurface of degree $dgeq1$. ;)
– Armando j18eos
Jul 27 at 10:01
You are welcome. P.S.: The trick of Veronese embedding works also for the generic (algebraic) hypersurface of degree $dgeq1$. ;)
– Armando j18eos
Jul 27 at 10:01
add a comment |Â
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Hi @casimir could you give me the title of the books thats you are working on ?
– Bernstein
Jul 23 at 16:03
I'm not working with a book, only with lecture notes ... and they don't go into any detail with respect to the question asked.
– casimir
Jul 23 at 16:05
1
thank you very much , good luck with your question
– Bernstein
Jul 23 at 16:13
3
Sottile discusses the $12$ nodal cubics through $8$ points.
– Jan-Magnus Økland
Jul 23 at 16:35
3
This is indeed quite similar to the conic question. Can you see that there exists a cubic curve passing through $x_1,ldots,x_7$? (This corresponds to 7 equations on your coefficients). Asking for your curve to have a singularity at $x_1$ imposes some more equations (how many?).
– loch
Jul 23 at 17:17