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 6 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 6 hours ago
Stanley Yao Xiao
8,44142784
Are you asking about the gonality of a plane curve?
– Jason Starr
5 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
4 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
3 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 6 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 6 hours ago
Stanley Yao Xiao
8,44142784
asked 6 hours ago
Stanley Yao Xiao
8,44142784
edited 4 hours ago
asked 6 hours ago
Stanley Yao Xiao
8,44142784
asked 6 hours ago
Stanley Yao Xiao
8,44142784
asked 6 hours ago
Stanley Yao Xiao
8,44142784
8,44142784
Are you asking about the gonality of a plane curve?
– Jason Starr
5 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
4 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
3 hours ago
add a comment |
Are you asking about the gonality of a plane curve?
– Jason Starr
5 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
4 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
3 hours ago
Are you asking about the gonality of a plane curve?
– Jason Starr
5 hours ago
Are you asking about the gonality of a plane curve?
– Jason Starr
5 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
4 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
4 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
3 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
3 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 3 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
3 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
14 mins 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 3 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
3 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
14 mins 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 3 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
3 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
14 mins 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 3 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 3 hours ago
Joe Silverman
30.3k180157
answered 3 hours ago
Joe Silverman
30.3k180157
answered 3 hours ago
Joe Silverman
30.3k180157
answered 3 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
3 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
14 mins 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
3 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
14 mins 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
3 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
3 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
14 mins 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
14 mins ago
add a comment |
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Are you asking about the gonality of a plane curve?
– Jason Starr
5 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
4 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
3 hours ago