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Let $L$ be a subfield of the algebraically closed field $K$, let $D$ ⊂ $\mathbb{P}_K^2$ be an elliptic curve defined by a polynomial equation with coefficients in $L$, and let $E$ $\in D_L$, where $D_L$ denote the set of points of $D$ with coordinates in $L$.

Prove that $D_L$ is a subgroup in the group defined on $D$, with $E$ the zero element.

I know the geometric meaning of the low group defined on $D$. I don't know what is the link between low group and coordinates.

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    If you consider the "geometric construction" of the group law, then you can compute explicitly formulas for $P+Q$ and for $-P$. Since $E$ is defined over $L$, these will be rational functions with coefficients in $L$.2017-01-13
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    @Watson How can I compute explicitly formulas for $P + Q$ and $-P$? My definition for an elliptic curve is: a projective, plane, cubic, smooth curve.2017-01-13
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    What is your definition of $P+Q$ ? Through the Picard group?2017-01-13
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    @Watson For $P$ and $Q$ on the elliptic curve, first I consider the intersection of the line $PQ$ with curve. I denote this point with $P*Q$. Now I consider the intersection of the line $E(P*Q)$ with curve. This point is $P+Q$.2017-01-14
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    Well, you probably know how to write the equation of the line $PQ$, and you can find the coordinates of its intersection with $E$. Then you'll find the coordinates of $P*Q$. Do you see how to proceed?2017-01-14
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    @Watson I don't work with Weirstrass form.2017-01-14
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    At least you are given a cubic curve. Maybe with Riemann-Roch theorem you can get the Weierstrass equations (which simplifies in characteristic $p \neq 2, 3$).2017-01-14
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    You are being unreasonably timid in the face of pencil-and-paper computation. I assure you that with the (nonsingular) cubic plane curve in hand, it is easy to compute the sum of $P_1=(x_1,y_1)$ and $P_2=(x_2,y_2)$, two points on the curve.2017-01-20
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    @Lubin I haven't an equation of curve.2017-01-20
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    Well, I just don’t understand, then. If you have a “a projective, plane, cubic, smooth curve”, what are you in the position of knowing about it beyond those words?2017-01-21

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