Algebraic points on a curve with small degree
Let $d geq 2$ be a positive integer, and let $K_d$ denote the compositum of all fields of degree $d$ over $mathbb{Q}$.
Let $Y$ be an algebraic curve defined over the rationals and has genus $g geq 2$. It is a well-known theorem of Faltings that asserts that for any number field $K$, the set of $K$-points of $Y$ is finite. However, if $Y$ is given by a plane model of the form $F(x,y,z) = 0$ for some homogeneous polynomial $F$ with rational coefficients and degree $d$, then certainly $Y$ has infinitely many points over $K_d$. Indeed, one can choose $x,y$ to be any two rational numbers, and the corresponding equation in $z$ will surely have a solution over $K_d$.
Sometimes $Y$ has infinitely many points over $K_d$ for a much smaller $d$. For instance if $Y$ is hyperelliptic, then $Y$ always has infinitely many points over $K_2$.
The gonality of $Y$, given as a plane curve, is the smallest positive integer $k$ for which $Y$ admits a degree $k$, non-constant map to the projective line $mathbb{P}^1$. Hyperelliptic curves have gonality equal to two, which explains the behaviour above. In general, if $m$ is the gonality of $Y$, then $Y(K_m)$ will be infinite.
In general, if $Y$ is given as a plane curve by a homogeneous polynomial $F(x,y,z) = 0$ of gonality $m$, do we expect $Y(K_s)$ to be finite or infinite for $s < m$?
Edit: thanks to Jason Starr for pointing out the notion of gonality.
nt.number-theory arithmetic-geometry
asked 12 hours ago
Stanley Yao Xiao
8,44142784
add a comment |
Let $d geq 2$ be a positive integer, and let $K_d$ denote the compositum of all fields of degree $d$ over $mathbb{Q}$.
Let $Y$ be an algebraic curve defined over the rationals and has genus $g geq 2$. It is a well-known theorem of Faltings that asserts that for any number field $K$, the set of $K$-points of $Y$ is finite. However, if $Y$ is given by a plane model of the form $F(x,y,z) = 0$ for some homogeneous polynomial $F$ with rational coefficients and degree $d$, then certainly $Y$ has infinitely many points over $K_d$. Indeed, one can choose $x,y$ to be any two rational numbers, and the corresponding equation in $z$ will surely have a solution over $K_d$.
Sometimes $Y$ has infinitely many points over $K_d$ for a much smaller $d$. For instance if $Y$ is hyperelliptic, then $Y$ always has infinitely many points over $K_2$.
The gonality of $Y$, given as a plane curve, is the smallest positive integer $k$ for which $Y$ admits a degree $k$, non-constant map to the projective line $mathbb{P}^1$. Hyperelliptic curves have gonality equal to two, which explains the behaviour above. In general, if $m$ is the gonality of $Y$, then $Y(K_m)$ will be infinite.
In general, if $Y$ is given as a plane curve by a homogeneous polynomial $F(x,y,z) = 0$ of gonality $m$, do we expect $Y(K_s)$ to be finite or infinite for $s < m$?
Edit: thanks to Jason Starr for pointing out the notion of gonality.
nt.number-theory arithmetic-geometry
asked 12 hours ago
Stanley Yao Xiao
8,44142784
Are you asking about the gonality of a plane curve?
– Jason Starr
11 hours ago
I was not aware of the notion of gonality before; I will edit the question to reflect this new notion.
– Stanley Yao Xiao
10 hours ago
It depends also if it has a map of degree say $r$ to an elliptic curve of rank >0. For example bielliptic curves (that have a degree 2 map to an elliptic curve) have infinitely many degree 2 points over any field where some elliptic quotient has rank >0.
– Xarles
9 hours ago
add a comment |
Let $d geq 2$ be a positive integer, and let $K_d$ denote the compositum of all fields of degree $d$ over $mathbb{Q}$.
Let $Y$ be an algebraic curve defined over the rationals and has genus $g geq 2$. It is a well-known theorem of Faltings that asserts that for any number field $K$, the set of $K$-points of $Y$ is finite. However, if $Y$ is given by a plane model of the form $F(x,y,z) = 0$ for some homogeneous polynomial $F$ with rational coefficients and degree $d$, then certainly $Y$ has infinitely many points over $K_d$. Indeed, one can choose $x,y$ to be any two rational numbers, and the corresponding equation in $z$ will surely have a solution over $K_d$.
Sometimes $Y$ has infinitely many points over $K_d$ for a much smaller $d$. For instance if $Y$ is hyperelliptic, then $Y$ always has infinitely many points over $K_2$.
The gonality of $Y$, given as a plane curve, is the smallest positive integer $k$ for which $Y$ admits a degree $k$, non-constant map to the projective line $mathbb{P}^1$. Hyperelliptic curves have gonality equal to two, which explains the behaviour above. In general, if $m$ is the gonality of $Y$, then $Y(K_m)$ will be infinite.
In general, if $Y$ is given as a plane curve by a homogeneous polynomial $F(x,y,z) = 0$ of gonality $m$, do we expect $Y(K_s)$ to be finite or infinite for $s < m$?
Edit: thanks to Jason Starr for pointing out the notion of gonality.
nt.number-theory arithmetic-geometry
asked 12 hours ago
Stanley Yao Xiao
8,44142784
Let $d geq 2$ be a positive integer, and let $K_d$ denote the compositum of all fields of degree $d$ over $mathbb{Q}$.
Let $Y$ be an algebraic curve defined over the rationals and has genus $g geq 2$. It is a well-known theorem of Faltings that asserts that for any number field $K$, the set of $K$-points of $Y$ is finite. However, if $Y$ is given by a plane model of the form $F(x,y,z) = 0$ for some homogeneous polynomial $F$ with rational coefficients and degree $d$, then certainly $Y$ has infinitely many points over $K_d$. Indeed, one can choose $x,y$ to be any two rational numbers, and the corresponding equation in $z$ will surely have a solution over $K_d$.
Sometimes $Y$ has infinitely many points over $K_d$ for a much smaller $d$. For instance if $Y$ is hyperelliptic, then $Y$ always has infinitely many points over $K_2$.
The gonality of $Y$, given as a plane curve, is the smallest positive integer $k$ for which $Y$ admits a degree $k$, non-constant map to the projective line $mathbb{P}^1$. Hyperelliptic curves have gonality equal to two, which explains the behaviour above. In general, if $m$ is the gonality of $Y$, then $Y(K_m)$ will be infinite.
In general, if $Y$ is given as a plane curve by a homogeneous polynomial $F(x,y,z) = 0$ of gonality $m$, do we expect $Y(K_s)$ to be finite or infinite for $s < m$?
Edit: thanks to Jason Starr for pointing out the notion of gonality.
nt.number-theory arithmetic-geometry
nt.number-theory arithmetic-geometry
asked 12 hours ago
Stanley Yao Xiao
8,44142784
asked 12 hours ago
Stanley Yao Xiao
8,44142784
edited 10 hours ago
asked 12 hours ago
Stanley Yao Xiao
8,44142784
asked 12 hours ago
Stanley Yao Xiao
8,44142784
asked 12 hours ago
Stanley Yao Xiao
8,44142784
8,44142784
Are you asking about the gonality of a plane curve?
– Jason Starr
11 hours ago
I was not aware of the notion of gonality before; I will edit the question to reflect this new notion.
– Stanley Yao Xiao
10 hours ago
It depends also if it has a map of degree say $r$ to an elliptic curve of rank >0. For example bielliptic curves (that have a degree 2 map to an elliptic curve) have infinitely many degree 2 points over any field where some elliptic quotient has rank >0.
– Xarles
9 hours ago
add a comment |
Are you asking about the gonality of a plane curve?
– Jason Starr
11 hours ago
I was not aware of the notion of gonality before; I will edit the question to reflect this new notion.
– Stanley Yao Xiao
10 hours ago
It depends also if it has a map of degree say $r$ to an elliptic curve of rank >0. For example bielliptic curves (that have a degree 2 map to an elliptic curve) have infinitely many degree 2 points over any field where some elliptic quotient has rank >0.
– Xarles
9 hours ago
Are you asking about the gonality of a plane curve?
– Jason Starr
11 hours ago
Are you asking about the gonality of a plane curve?
– Jason Starr
11 hours ago
I was not aware of the notion of gonality before; I will edit the question to reflect this new notion.
– Stanley Yao Xiao
10 hours ago
I was not aware of the notion of gonality before; I will edit the question to reflect this new notion.
– Stanley Yao Xiao
10 hours ago
It depends also if it has a map of degree say $r$ to an elliptic curve of rank >0. For example bielliptic curves (that have a degree 2 map to an elliptic curve) have infinitely many degree 2 points over any field where some elliptic quotient has rank >0.
– Xarles
9 hours ago
It depends also if it has a map of degree say $r$ to an elliptic curve of rank >0. For example bielliptic curves (that have a degree 2 map to an elliptic curve) have infinitely many degree 2 points over any field where some elliptic quotient has rank >0.
– Xarles
9 hours ago
add a comment |
1 Answer
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In the first place, there's also the possibility that there is a map of degree $m$ from $Y$ to an elliptic curve $E$ such that $E(mathbb Q)$ is infinite, in which case $Y(K_m)$ will clearly be infinite. There's actually a bunch of literature on this subject, including for example:
MR1055774 Harris, Joe; Silverman, Joe; Bielliptic curves and symmetric products. Proc. Amer. Math. Soc. 112 (1991), no. 2, 347–356. (This deals with quadratic points, so in your notation, with $K_2$. The answer is that in order for $Y(K_2)$ to be infinite, the curve $Y$ must be hyperelliptic or admit a degree 2 map to an elliptic curve.)
MR11047 Abramovich, Dan; Harris, Joe; Abelian varieties and curves in $W_d(C)$. Compositio Math. 78 (1991), no. 2, 227–238. (This deals with the general case. The answer is a bit more complicated than just looking at gonality.)
In all cases, the results rely on Faltings' theorem describing subvarieties of $A(K)$ having having a Zariski dense set of $K$-rational points.
answered 9 hours ago
Joe Silverman
30.3k180157
I think the OP asks about the points in what he call $K_2$, which is the composite of all the degree 2 extensions of $K$, not just the degree 2 points (in the usual sense of your paper with Harris). I guess there are (modular?) curves with infinitely many points in $K_2$ but finitely many quadratic points for $K=mathbb{Q}$.
– Xarles
9 hours ago
@Xarles You're right, the OP said that $K_d$ is the compositum of all fields of degree $d$, but from the rest of his post, it really sounds as if he meant to define it to be the union. In any case, as you say, the papers that I cited deal with points defined over a field of degree $d$, not the compositum of all such fields.
– Joe Silverman
6 hours ago
add a comment |
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In the first place, there's also the possibility that there is a map of degree $m$ from $Y$ to an elliptic curve $E$ such that $E(mathbb Q)$ is infinite, in which case $Y(K_m)$ will clearly be infinite. There's actually a bunch of literature on this subject, including for example:
MR1055774 Harris, Joe; Silverman, Joe; Bielliptic curves and symmetric products. Proc. Amer. Math. Soc. 112 (1991), no. 2, 347–356. (This deals with quadratic points, so in your notation, with $K_2$. The answer is that in order for $Y(K_2)$ to be infinite, the curve $Y$ must be hyperelliptic or admit a degree 2 map to an elliptic curve.)
MR11047 Abramovich, Dan; Harris, Joe; Abelian varieties and curves in $W_d(C)$. Compositio Math. 78 (1991), no. 2, 227–238. (This deals with the general case. The answer is a bit more complicated than just looking at gonality.)
In all cases, the results rely on Faltings' theorem describing subvarieties of $A(K)$ having having a Zariski dense set of $K$-rational points.
answered 9 hours ago
Joe Silverman
30.3k180157
I think the OP asks about the points in what he call $K_2$, which is the composite of all the degree 2 extensions of $K$, not just the degree 2 points (in the usual sense of your paper with Harris). I guess there are (modular?) curves with infinitely many points in $K_2$ but finitely many quadratic points for $K=mathbb{Q}$.
– Xarles
9 hours ago
@Xarles You're right, the OP said that $K_d$ is the compositum of all fields of degree $d$, but from the rest of his post, it really sounds as if he meant to define it to be the union. In any case, as you say, the papers that I cited deal with points defined over a field of degree $d$, not the compositum of all such fields.
– Joe Silverman
6 hours ago
add a comment |
In the first place, there's also the possibility that there is a map of degree $m$ from $Y$ to an elliptic curve $E$ such that $E(mathbb Q)$ is infinite, in which case $Y(K_m)$ will clearly be infinite. There's actually a bunch of literature on this subject, including for example:
MR1055774 Harris, Joe; Silverman, Joe; Bielliptic curves and symmetric products. Proc. Amer. Math. Soc. 112 (1991), no. 2, 347–356. (This deals with quadratic points, so in your notation, with $K_2$. The answer is that in order for $Y(K_2)$ to be infinite, the curve $Y$ must be hyperelliptic or admit a degree 2 map to an elliptic curve.)
MR11047 Abramovich, Dan; Harris, Joe; Abelian varieties and curves in $W_d(C)$. Compositio Math. 78 (1991), no. 2, 227–238. (This deals with the general case. The answer is a bit more complicated than just looking at gonality.)
In all cases, the results rely on Faltings' theorem describing subvarieties of $A(K)$ having having a Zariski dense set of $K$-rational points.
answered 9 hours ago
Joe Silverman
30.3k180157
I think the OP asks about the points in what he call $K_2$, which is the composite of all the degree 2 extensions of $K$, not just the degree 2 points (in the usual sense of your paper with Harris). I guess there are (modular?) curves with infinitely many points in $K_2$ but finitely many quadratic points for $K=mathbb{Q}$.
– Xarles
9 hours ago
@Xarles You're right, the OP said that $K_d$ is the compositum of all fields of degree $d$, but from the rest of his post, it really sounds as if he meant to define it to be the union. In any case, as you say, the papers that I cited deal with points defined over a field of degree $d$, not the compositum of all such fields.
– Joe Silverman
6 hours ago
add a comment |
In the first place, there's also the possibility that there is a map of degree $m$ from $Y$ to an elliptic curve $E$ such that $E(mathbb Q)$ is infinite, in which case $Y(K_m)$ will clearly be infinite. There's actually a bunch of literature on this subject, including for example:
MR1055774 Harris, Joe; Silverman, Joe; Bielliptic curves and symmetric products. Proc. Amer. Math. Soc. 112 (1991), no. 2, 347–356. (This deals with quadratic points, so in your notation, with $K_2$. The answer is that in order for $Y(K_2)$ to be infinite, the curve $Y$ must be hyperelliptic or admit a degree 2 map to an elliptic curve.)
MR11047 Abramovich, Dan; Harris, Joe; Abelian varieties and curves in $W_d(C)$. Compositio Math. 78 (1991), no. 2, 227–238. (This deals with the general case. The answer is a bit more complicated than just looking at gonality.)
In all cases, the results rely on Faltings' theorem describing subvarieties of $A(K)$ having having a Zariski dense set of $K$-rational points.
answered 9 hours ago
Joe Silverman
30.3k180157
In the first place, there's also the possibility that there is a map of degree $m$ from $Y$ to an elliptic curve $E$ such that $E(mathbb Q)$ is infinite, in which case $Y(K_m)$ will clearly be infinite. There's actually a bunch of literature on this subject, including for example:
MR1055774 Harris, Joe; Silverman, Joe; Bielliptic curves and symmetric products. Proc. Amer. Math. Soc. 112 (1991), no. 2, 347–356. (This deals with quadratic points, so in your notation, with $K_2$. The answer is that in order for $Y(K_2)$ to be infinite, the curve $Y$ must be hyperelliptic or admit a degree 2 map to an elliptic curve.)
MR11047 Abramovich, Dan; Harris, Joe; Abelian varieties and curves in $W_d(C)$. Compositio Math. 78 (1991), no. 2, 227–238. (This deals with the general case. The answer is a bit more complicated than just looking at gonality.)
In all cases, the results rely on Faltings' theorem describing subvarieties of $A(K)$ having having a Zariski dense set of $K$-rational points.
answered 9 hours ago
Joe Silverman
30.3k180157
answered 9 hours ago
Joe Silverman
30.3k180157
answered 9 hours ago
Joe Silverman
30.3k180157
answered 9 hours ago
Joe Silverman
30.3k180157
30.3k180157
I think the OP asks about the points in what he call $K_2$, which is the composite of all the degree 2 extensions of $K$, not just the degree 2 points (in the usual sense of your paper with Harris). I guess there are (modular?) curves with infinitely many points in $K_2$ but finitely many quadratic points for $K=mathbb{Q}$.
– Xarles
9 hours ago
@Xarles You're right, the OP said that $K_d$ is the compositum of all fields of degree $d$, but from the rest of his post, it really sounds as if he meant to define it to be the union. In any case, as you say, the papers that I cited deal with points defined over a field of degree $d$, not the compositum of all such fields.
– Joe Silverman
6 hours ago
add a comment |
I think the OP asks about the points in what he call $K_2$, which is the composite of all the degree 2 extensions of $K$, not just the degree 2 points (in the usual sense of your paper with Harris). I guess there are (modular?) curves with infinitely many points in $K_2$ but finitely many quadratic points for $K=mathbb{Q}$.
– Xarles
9 hours ago
@Xarles You're right, the OP said that $K_d$ is the compositum of all fields of degree $d$, but from the rest of his post, it really sounds as if he meant to define it to be the union. In any case, as you say, the papers that I cited deal with points defined over a field of degree $d$, not the compositum of all such fields.
– Joe Silverman
6 hours ago
I think the OP asks about the points in what he call $K_2$, which is the composite of all the degree 2 extensions of $K$, not just the degree 2 points (in the usual sense of your paper with Harris). I guess there are (modular?) curves with infinitely many points in $K_2$ but finitely many quadratic points for $K=mathbb{Q}$.
– Xarles
9 hours ago
I think the OP asks about the points in what he call $K_2$, which is the composite of all the degree 2 extensions of $K$, not just the degree 2 points (in the usual sense of your paper with Harris). I guess there are (modular?) curves with infinitely many points in $K_2$ but finitely many quadratic points for $K=mathbb{Q}$.
– Xarles
9 hours ago
@Xarles You're right, the OP said that $K_d$ is the compositum of all fields of degree $d$, but from the rest of his post, it really sounds as if he meant to define it to be the union. In any case, as you say, the papers that I cited deal with points defined over a field of degree $d$, not the compositum of all such fields.
– Joe Silverman
6 hours ago
@Xarles You're right, the OP said that $K_d$ is the compositum of all fields of degree $d$, but from the rest of his post, it really sounds as if he meant to define it to be the union. In any case, as you say, the papers that I cited deal with points defined over a field of degree $d$, not the compositum of all such fields.
– Joe Silverman
6 hours ago
add a comment |
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Are you asking about the gonality of a plane curve?
– Jason Starr
11 hours ago
I was not aware of the notion of gonality before; I will edit the question to reflect this new notion.
– Stanley Yao Xiao
10 hours ago
It depends also if it has a map of degree say $r$ to an elliptic curve of rank >0. For example bielliptic curves (that have a degree 2 map to an elliptic curve) have infinitely many degree 2 points over any field where some elliptic quotient has rank >0.
– Xarles
9 hours ago