Existence of rational parametrization of elliptic curves
1
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I read somewhere that it is not possible to have rational parametrization for elliptic curves. So there is possibility of the existence of rational parametrization for a 'part' of elliptic curve or which may just miss finite number of points on the curve?
elliptic-curves
asked Dec 6 '18 at 22:26
ershersh
357113
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add a comment |
1
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I read somewhere that it is not possible to have rational parametrization for elliptic curves. So there is possibility of the existence of rational parametrization for a 'part' of elliptic curve or which may just miss finite number of points on the curve?
elliptic-curves
asked Dec 6 '18 at 22:26
ershersh
357113
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2
$begingroup$
In that case the function field of the curve would be $cong overline{K}(x)$ and its genus would be $0$.
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– reuns
Dec 6 '18 at 22:55
1
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To expand on @reuns, you can’t get a rational parametrization of even the slightest part of an elliptic curve, so long as that part isn’t finite, of course.
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– Lubin
Dec 7 '18 at 1:04
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Got it. Thank you
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– ersh
Dec 7 '18 at 15:19
add a comment |
1
1
1
$begingroup$
I read somewhere that it is not possible to have rational parametrization for elliptic curves. So there is possibility of the existence of rational parametrization for a 'part' of elliptic curve or which may just miss finite number of points on the curve?
elliptic-curves
asked Dec 6 '18 at 22:26
ershersh
357113
$endgroup$
I read somewhere that it is not possible to have rational parametrization for elliptic curves. So there is possibility of the existence of rational parametrization for a 'part' of elliptic curve or which may just miss finite number of points on the curve?
elliptic-curves
elliptic-curves
asked Dec 6 '18 at 22:26
ershersh
357113
asked Dec 6 '18 at 22:26
ershersh
357113
asked Dec 6 '18 at 22:26
ershersh
357113
asked Dec 6 '18 at 22:26
ershersh
357113
asked Dec 6 '18 at 22:26
ershersh
357113
357113
2
$begingroup$
In that case the function field of the curve would be $cong overline{K}(x)$ and its genus would be $0$.
$endgroup$
– reuns
Dec 6 '18 at 22:55
1
$begingroup$
To expand on @reuns, you can’t get a rational parametrization of even the slightest part of an elliptic curve, so long as that part isn’t finite, of course.
$endgroup$
– Lubin
Dec 7 '18 at 1:04
$begingroup$
Got it. Thank you
$endgroup$
– ersh
Dec 7 '18 at 15:19
add a comment |
2
$begingroup$
In that case the function field of the curve would be $cong overline{K}(x)$ and its genus would be $0$.
$endgroup$
– reuns
Dec 6 '18 at 22:55
1
$begingroup$
To expand on @reuns, you can’t get a rational parametrization of even the slightest part of an elliptic curve, so long as that part isn’t finite, of course.
$endgroup$
– Lubin
Dec 7 '18 at 1:04
$begingroup$
Got it. Thank you
$endgroup$
– ersh
Dec 7 '18 at 15:19
2
2
$begingroup$
In that case the function field of the curve would be $cong overline{K}(x)$ and its genus would be $0$.
$endgroup$
– reuns
Dec 6 '18 at 22:55
$begingroup$
In that case the function field of the curve would be $cong overline{K}(x)$ and its genus would be $0$.
$endgroup$
– reuns
Dec 6 '18 at 22:55
1
1
$begingroup$
To expand on @reuns, you can’t get a rational parametrization of even the slightest part of an elliptic curve, so long as that part isn’t finite, of course.
$endgroup$
– Lubin
Dec 7 '18 at 1:04
$begingroup$
To expand on @reuns, you can’t get a rational parametrization of even the slightest part of an elliptic curve, so long as that part isn’t finite, of course.
$endgroup$
– Lubin
Dec 7 '18 at 1:04
$begingroup$
Got it. Thank you
$endgroup$
– ersh
Dec 7 '18 at 15:19
$begingroup$
Got it. Thank you
$endgroup$
– ersh
Dec 7 '18 at 15:19
add a comment |
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2
$begingroup$
In that case the function field of the curve would be $cong overline{K}(x)$ and its genus would be $0$.
$endgroup$
– reuns
Dec 6 '18 at 22:55
1
$begingroup$
To expand on @reuns, you can’t get a rational parametrization of even the slightest part of an elliptic curve, so long as that part isn’t finite, of course.
$endgroup$
– Lubin
Dec 7 '18 at 1:04
$begingroup$
Got it. Thank you
$endgroup$
– ersh
Dec 7 '18 at 15:19