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I'm a bit struggling with defining the set of $\mathbb{Q}$-rational points on an elliptic curve $E:\;y^2=x^3+ax^2+bx+c$ with $a,b,c\in\mathbb{Q}$. I'm actually guessing that If we let $K$ and $L$ be fields, then I defined the set of $L$-rational points on a curve as follows.

Let $f\in K[x,y]$ be a polynomial in two variables over the field $K$. A curve $C/K$ is an equation $C:f(x,y)=0$ which describes the points of the curve $C$. Let $F\in K[x,y,z]$ be a homogenisation of $f$ such that, for $z\neq0$, $F(x,y,z)=F\left(\frac{x}{z},\frac{y}{z},1\right)=f(x,y)$. Then the set of $L$-rational points on $C$ is denoted by $$C(L):=\{[x:y:z]\in\mathbb{P}^2(L)\;:\;F(x,y,z)=0\}.$$

We will consider $\mathbb{Q}$-rational points on an elliptic curve $E:y^2=x^3+ax^2+bx+c$ where $a,\;b$ and $c$ are rational numbers. By letting $x=\frac{X}{Z}$ and $y=\frac{Y}{Z}$, where $Z\neq0$, we can write the elliptic curve in homogeneous form as \begin{align}\label{eq:homo_ell} F(X,Y,Z)=Y^2Z-X^3-aX^2Z-bXZ^2-cZ^3=0. \end{align} Now the set of $\mathbb{Q}$-rational points is, as in the definition above, defined as \begin{align*} E(\mathbb{Q}):=\{[x:y:z]\in\mathbb{P}^2(\mathbb{Q})\;:\;F(x,y,z)=0\}. \end{align*} As a remark, one should note that an element $[x:y:z]\in E(\mathbb{Q})$ is not necessarily a tuple of elements $x,\;y$ and $z$ which are rational, but there is a real number $t$ such that $tx,\;ty$ and $tz$ are all three rational.

I accidentally wrote somewhere in my thesis that $E(\mathbb{Q}):=\{[x:y:z]\in\mathbb{P}^2(\mathbb{Q})\;:\;F(x,y,z)=0\text{ and }x,y,z\in\mathbb{Q}\}$, but this doesn't seem right as now it is the case that $x,y,z\in\mathbb{Q}$.

I found a definition of the projective plane over $\mathbb{Q}$ which goes as follows

$\mathbb{P}^2(\mathbb{Q}):=\frac{\mathbb{Q}^3-0}{\mathbb{Q}^{\times}}$, so $\mathbb{Q}$-tuples without the zero element modulo the group of unity elements of $\mathbb{Q}$ which is equal to $\mathbb{Q}$.

Of course, the zero element can't be included since this does not satisfy the homogeneous equation; the point at infinity of the elliptic curve is the projective point $[0:1:0]$. But now with this definition of $\mathbb{P}^2(\mathbb{Q})$, an element $[x:y:z]\in E(\mathbb{Q})$ is just a rational tuple. This doesn't seem right. Can anyone point me at the right path here? Thanks!

EDIT I know the answers to my questions now and will add it some time soon. It has to do with this comment: The story about the embedding is really helpfull. It seems to be a bit of an abuse of notation: when there is a $t\in L-K$, but we have an embedding $\mathbb{P}^2(\mathbb{Q})\hookrightarrow\mathbb{P}^2(\mathbb{R})$, so an element $[t:0:0]=[1:0:0]\in\mathbb{P}^2(\mathbb{Q})$. Thanks for the help.

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  • $\begingroup$ Do you need to do anything else other than make the observation that as $F(x,y,z)$ is homogeneous of degree three we have $$F(tx,ty,tz)=t^3F(x,y,z).$$ So if $F(x,y,z)=0$, $F(tx,ty,tz)=0$ also (and vice versa)? $\endgroup$ – Jyrki Lahtonen Apr 8 '19 at 10:28
  • $\begingroup$ By the way, I don't think that you want to restrict to real values of $t$ only. $\endgroup$ – Jyrki Lahtonen Apr 8 '19 at 10:29
  • $\begingroup$ @JyrkiLahtonen It's an obvious property of homogeneous equations you mention here. My problem is that it seems to be that I'm unable to formally define this set in the sense that $\mathbb{P}^2(\mathbb{Q})$ is seems to give also non-rational solutions to the curve. Or put differently, if this is the right definition, then I don't understand it properly and am desperate for an explanation. The book "Rational Points on Elliptic Curves" doesn't exactly specify this. $\endgroup$ – Algebear Apr 8 '19 at 10:45
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    $\begingroup$ @Algebear I don't think we are understanding your problem. You wrote to Jurki that "$\mathbb P^2(Q)$ seems to give also non-rational solutions to the curve." Can you please give an explicit example as to what you mean? Also for your last question to Alex, there is certainly nothing special about the reals (what about the complexes, or other perhaps more exotic fields) . Assuming the fields $K\subset L $, then the set ${\mathbb P}^2 (K)$ is the subset of $P\in {\mathbb P}^2 (L)$ which can be represented by a $(x,y,z)\in K^3\setminus(0,0,0)$, as in Alex M's example. (cont) $\endgroup$ – peter a g Apr 9 '19 at 21:48
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    $\begingroup$ (cont) Namely, one can embed ${\mathbb P}^2(K)$ in ${\mathbb P}^2(L)$. But I can guarantee you're not wrong to be confused (though I am not sure what exactly your confusion is) - this certainly gets more complicated when one generalizes things. $\endgroup$ – peter a g Apr 9 '19 at 22:01
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I will just say a few words about this. I turned out to be quite simple: an elliptic curve over $\mathbb{Q}$ is defined as $$ E(\mathbb{Q})=\{(x,y)\in\mathbb{Q}|y^2=x^3+ax^2+bx+c,\;a,b,c\in\mathbb{Q}\}\cup\{\mathscr{O}\} $$ where $\mathscr{O}$ is the point $[X:Y:Z]=[0:1:0]$ on the projective curve $Y^2Z=X^3+aX^2Z+bXZ^2+cZ^3$ which a lot of people call the point at infinity which is the point in the projective plane corresponding to the vertical direction.

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