$[2]Q=P$ with the rational coordinate of $Q$ in an elliptic curve

Let $$\tag{1} y^2=x^3+Ax+B$$ be an elliptic curve, $$A, B$$ are integers and $$P$$ is a point on the curve with rational coordinates. Is there always a point $$Q$$ such that $$[2]Q=P$$ such that the $$x$$-coordinate of $$Q$$ is rational?

The formula we get form doubling points, tells us coordinates of $$Q$$ depends on $$a$$ and the coordinates of $$P$$, if we put those values in the formula, we get the value of x coordinate of $$Q$$, and thus get $$y$$ coordinate of $$Q$$ from equation $$(1)$$, , we see from the forumula given below in image, x coordinate of $$Q$$ is always rational (Given that the x-coordinate of $$P$$ and $$a$$ are rational), is it correct? am I missing something ?

Here, $$f(x)=y^2.$$

• There is always a point $Q$ such that $[2] Q = P$ as you say, but it is not necessarily true that $Q$ has rational coordinates. Commented Mar 10, 2020 at 17:21
• @user45878 the x coordinate of $Q$ is rational, the y coordinate might not be, right? Commented Mar 10, 2020 at 17:29
• I don't see why that would be true. In general, the $x$ coordinate of $Q$ can be obtained by solving a degree $4$ equation in the $x$ coordinate of $P$ with coefficients depending on $A$ and $B$, (See page 54 of Silverman, for example). Commented Mar 10, 2020 at 17:34
• @user45878 I have attached an image, note that if $X, a$ are rational then the formula of doubling point gives us a rational x-coordinate of $Q$, because the slope $\lambda$, even if it is not a rational becomes rational by squaring it, note $\lambda$ becomes irrational because of $Y$, if $\lambda$ is irrational then there is a chance of getting irrational x-coordinate of $Q$, please check. For the formula, see page 10 of :math.uchicago.edu/~may/REU2013/REUPapers/Galperin.pdf Commented Mar 13, 2020 at 0:23

2 Answers

For an elliptic curve over $$\mathbb{Q}$$, then $$E_{\text{tor}}(\mathbb{Q})$$ can be $$\mathbb{Z}/4\mathbb{Z}$$(see this), so $$E(\mathbb{Q})=\mathbb{Z}/4\mathbb{Z}\bigoplus\mathbb{Z}^N$$. Let $$P$$ be $$(\bar{1},0,...0)$$, then there doesn't exist such a $$Q\in E(\mathbb{Q})$$ such that $$2Q=P$$.

But as you see, by solving these equations, we get these roots in $$\bar{\mathbb{Q}}$$, so there does exist a point $$Q\in E(\bar{\mathbb{Q}})$$ such that $$2Q=P$$.

Let $$E$$ be the elliptic curve $$y^2=x^3-2x$$. Then, one can easily show that $$E(\mathbb{Q})\cong \mathbb{Z}/2\mathbb{Z}$$, so that the only points on $$E$$ with rational coordinates are the point $$\mathcal{O}$$ at infinity and $$P=(0,0)$$. If there was a point $$Q$$ with rational coordinates such that $$2Q=P$$, then $$4Q = 2(2Q)=2P=\mathcal{O}$$, so the order of $$Q$$ would be $$4$$. However, that would imply a few contradictions (with the size of $$E(\mathbb{Q})$$, with the isomorphism type of the torsion subgroup, etc), so it is impossible, and no such point $$Q$$ can exist.

Edit to add: the $$x$$-coordinates of the points of order $$4$$ on $$E$$ are given by the quotient of the $$4$$-division polynomial, by the $$2$$-division polynomial, which is $$(x^2 - 4x + 2)(x^2 + 2)(x^2 + 4x + 2),$$ and all three irreducible polynomial divisors over $$\mathbb{Q}$$ are of degree $$2$$, so no point $$Q$$ of order $$4$$ has an $$x$$-coordinate that is rational.

• In this post only the $x$ coordinate has to be rational, $y$ might be irrational, I am not sure Mazur's theorem (where both $x, y$ are rational) agrees with this condition, also plz see this post,according to which $x$ is rational point on elliptic curve: math.stackexchange.com/questions/3580905/… Commented Mar 15, 2020 at 14:28
• @Andrew I've added an explanation that no point of order $4$ in this case has a rational $x$-coordinate. Commented Mar 16, 2020 at 1:20
• Hello, I am thinking about the above problem again, if I try to check in an algorithmic way, whether a point $P$ with a rational $x$ coordinate exists or not, when $x$ coordinate of $Q,$ and coefficient $A, B$ are given, it should be easy, but I couldn't find the algorithm, could you please hint me how should I try? Thank you. Commented Nov 21, 2022 at 13:38
• @ConsiderNon-TrivialCases you can simply check if $x^3+Ax+B$ is a square in the rationals. Commented Nov 30, 2022 at 21:29
• Hi! Is there a way to know the percentage of such $P$ that will have an elliptic curve and a point $Q$ on the curve as described in the post? So, we can say, for example, at-least $20 \%$ out of all pairs of rational number have an elliptic curve such that $P=[n]Q, n\geq2$ (note it could be part of infinite order, I don't mean torsion-group points only). Commented May 9, 2023 at 19:19