Mixing
Singular points of a Weierstrass equation over a perfect field are rational
$begingroup$
Let $k$ be a perfect field, $E$ a curve defined by $f = y^2 + a_1 xy + a_3 y - x^3 - a_2 x^2 - a_4 x - a_6$ over $k$, and let $P = (x, y) in E(overline{k})$ be a singular point.
Then does $P in E(k)$?
Please check my proof:
$P$ satisfies the three equations $f(P) =0, a_1 y = 3 x^2 + 2a_2 x + a_4, 2y + a_1 x + a_3 = 0$. (Let us denote this condition by $(F)$.)
I'll show $x,y in k$ by case-by-case.
First, the case $operatorname{char}k = 2.$
Then $(F)$ is $ f(P) = 0, a_1 y = x^2 + a_4, a_1 x + a_3 = 0$.
It's trivial $P in E(k)$ if $a_1 neq 0$.
And even if $a_1 = 0$, since $k$ is perfect and $E$ has only one singular pint, $P in E(k)$.
Next, the case $operatorname{char}k = 3.$
Then we can assume $a_1 = a_3 = 0$, and so $(F)$ is $f(P) =0, a_2 x = a_4, y = 0$.
Thus if $a_2 neq 0$, then $x in k$.
And if $a_2 = 0$, since we have $a_4 = 0$, $x$ satisfies $x^3 + a_6 = 0$.
Now $E$ has only one singular point.
Thus this equation has only one root in any algebraically closed field of $k$.
So, since $k$ is perfect, we have $x in k$.
Finally the case $operatorname{char}k neq 2,3.$
Then we may assume $a_1 = a_2 = a_3 = 0$.
Thus $(F)$ is $y^2 = x^3 + a_4 x + a_6, 3x^2 + a_4 = 0, y = 0.$
This is equivalent to $y = 0, x^3 = a_6 / 2, a_4 x + 3a_6 /2 = 0$.
Thus if $a_4 = 0$ we have $x in k$, and if $a_4 neq 0$, $x = 3a_6 / 2a_4 in k$.
algebraic-geometry elliptic-curves
asked Jan 2 at 1:25
k.j.k.j.
35919
$endgroup$
add a comment |
$begingroup$
Let $k$ be a perfect field, $E$ a curve defined by $f = y^2 + a_1 xy + a_3 y - x^3 - a_2 x^2 - a_4 x - a_6$ over $k$, and let $P = (x, y) in E(overline{k})$ be a singular point.
Then does $P in E(k)$?
Please check my proof:
$P$ satisfies the three equations $f(P) =0, a_1 y = 3 x^2 + 2a_2 x + a_4, 2y + a_1 x + a_3 = 0$. (Let us denote this condition by $(F)$.)
I'll show $x,y in k$ by case-by-case.
First, the case $operatorname{char}k = 2.$
Then $(F)$ is $ f(P) = 0, a_1 y = x^2 + a_4, a_1 x + a_3 = 0$.
It's trivial $P in E(k)$ if $a_1 neq 0$.
And even if $a_1 = 0$, since $k$ is perfect and $E$ has only one singular pint, $P in E(k)$.
Next, the case $operatorname{char}k = 3.$
Then we can assume $a_1 = a_3 = 0$, and so $(F)$ is $f(P) =0, a_2 x = a_4, y = 0$.
Thus if $a_2 neq 0$, then $x in k$.
And if $a_2 = 0$, since we have $a_4 = 0$, $x$ satisfies $x^3 + a_6 = 0$.
Now $E$ has only one singular point.
Thus this equation has only one root in any algebraically closed field of $k$.
So, since $k$ is perfect, we have $x in k$.
Finally the case $operatorname{char}k neq 2,3.$
Then we may assume $a_1 = a_2 = a_3 = 0$.
Thus $(F)$ is $y^2 = x^3 + a_4 x + a_6, 3x^2 + a_4 = 0, y = 0.$
This is equivalent to $y = 0, x^3 = a_6 / 2, a_4 x + 3a_6 /2 = 0$.
Thus if $a_4 = 0$ we have $x in k$, and if $a_4 neq 0$, $x = 3a_6 / 2a_4 in k$.
algebraic-geometry elliptic-curves
asked Jan 2 at 1:25
k.j.k.j.
35919
$endgroup$
add a comment |
$begingroup$
Let $k$ be a perfect field, $E$ a curve defined by $f = y^2 + a_1 xy + a_3 y - x^3 - a_2 x^2 - a_4 x - a_6$ over $k$, and let $P = (x, y) in E(overline{k})$ be a singular point.
Then does $P in E(k)$?
Please check my proof:
$P$ satisfies the three equations $f(P) =0, a_1 y = 3 x^2 + 2a_2 x + a_4, 2y + a_1 x + a_3 = 0$. (Let us denote this condition by $(F)$.)
I'll show $x,y in k$ by case-by-case.
First, the case $operatorname{char}k = 2.$
Then $(F)$ is $ f(P) = 0, a_1 y = x^2 + a_4, a_1 x + a_3 = 0$.
It's trivial $P in E(k)$ if $a_1 neq 0$.
And even if $a_1 = 0$, since $k$ is perfect and $E$ has only one singular pint, $P in E(k)$.
Next, the case $operatorname{char}k = 3.$
Then we can assume $a_1 = a_3 = 0$, and so $(F)$ is $f(P) =0, a_2 x = a_4, y = 0$.
Thus if $a_2 neq 0$, then $x in k$.
And if $a_2 = 0$, since we have $a_4 = 0$, $x$ satisfies $x^3 + a_6 = 0$.
Now $E$ has only one singular point.
Thus this equation has only one root in any algebraically closed field of $k$.
So, since $k$ is perfect, we have $x in k$.
Finally the case $operatorname{char}k neq 2,3.$
Then we may assume $a_1 = a_2 = a_3 = 0$.
Thus $(F)$ is $y^2 = x^3 + a_4 x + a_6, 3x^2 + a_4 = 0, y = 0.$
This is equivalent to $y = 0, x^3 = a_6 / 2, a_4 x + 3a_6 /2 = 0$.
Thus if $a_4 = 0$ we have $x in k$, and if $a_4 neq 0$, $x = 3a_6 / 2a_4 in k$.
algebraic-geometry elliptic-curves
asked Jan 2 at 1:25
k.j.k.j.
35919
$endgroup$
Let $k$ be a perfect field, $E$ a curve defined by $f = y^2 + a_1 xy + a_3 y - x^3 - a_2 x^2 - a_4 x - a_6$ over $k$, and let $P = (x, y) in E(overline{k})$ be a singular point.
Then does $P in E(k)$?
Please check my proof:
$P$ satisfies the three equations $f(P) =0, a_1 y = 3 x^2 + 2a_2 x + a_4, 2y + a_1 x + a_3 = 0$. (Let us denote this condition by $(F)$.)
I'll show $x,y in k$ by case-by-case.
First, the case $operatorname{char}k = 2.$
Then $(F)$ is $ f(P) = 0, a_1 y = x^2 + a_4, a_1 x + a_3 = 0$.
It's trivial $P in E(k)$ if $a_1 neq 0$.
And even if $a_1 = 0$, since $k$ is perfect and $E$ has only one singular pint, $P in E(k)$.
Next, the case $operatorname{char}k = 3.$
Then we can assume $a_1 = a_3 = 0$, and so $(F)$ is $f(P) =0, a_2 x = a_4, y = 0$.
Thus if $a_2 neq 0$, then $x in k$.
And if $a_2 = 0$, since we have $a_4 = 0$, $x$ satisfies $x^3 + a_6 = 0$.
Now $E$ has only one singular point.
Thus this equation has only one root in any algebraically closed field of $k$.
So, since $k$ is perfect, we have $x in k$.
Finally the case $operatorname{char}k neq 2,3.$
Then we may assume $a_1 = a_2 = a_3 = 0$.
Thus $(F)$ is $y^2 = x^3 + a_4 x + a_6, 3x^2 + a_4 = 0, y = 0.$
This is equivalent to $y = 0, x^3 = a_6 / 2, a_4 x + 3a_6 /2 = 0$.
Thus if $a_4 = 0$ we have $x in k$, and if $a_4 neq 0$, $x = 3a_6 / 2a_4 in k$.
algebraic-geometry elliptic-curves
algebraic-geometry elliptic-curves
asked Jan 2 at 1:25
k.j.k.j.
35919
asked Jan 2 at 1:25
k.j.k.j.
35919
asked Jan 2 at 1:25
k.j.k.j.
35919
asked Jan 2 at 1:25
k.j.k.j.
35919
asked Jan 2 at 1:25
k.j.k.j.
35919
35919
add a comment |
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