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

Question

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

Answer 1

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.

Answer 2

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 $2Q=P$.