This is from J.S. Milne's book 'Elliptic Curves'-

With $E(\mathbb{Q}) : Y^2Z = X^3 +aX Z^2+bZ^3$ and given a point on $P =(x,y)$ on it's Weierstrass equation dehomogenized $E: y^2= x^3+ax+b$ where $a,b \in \mathbb{Z}$, the duplication formula gives x-coordinate and y-coordinate of the point $2P$ as -

$$ x(2P). = \frac{x^4-2ax^2-8bx+a^2}{4y^2}$$ and, $y(2P) = \frac{a polynomial in x }{8y^3}$

Proposition $4.4$- says that there exists a constant $A$ such that $|h(2P)-4h(P)| \leq A$

The proof runs as- let $P= (x:y:z)$ and $2P = (x' : y' : z')$ be points on $E(\mathbb{Q})$. According to the duplication formula,

$$ (x' : z') = (F(x) : G(x))$$

where $F(X,Z)$ and $G(X,Z)$ are homogeneous polynomials of degree 4 such that $$ F(X,1) = (3X^2+a)^2 - 8X (X^3+aX+b)$$

$$G(X,1) = 4(X^3 +aX +b)$$

And so on....

My problem- I'm not sure how to interpret the point $(x':z')$ and this is my guess:

We start with a point $P= (x:y:z) \in E(\mathbb{Q}), z\neq 0$ which is as good as taking the point $(x:y:1)$ on it and it corresponds to the point $(x,y)$ on $y^2= x^3+ax+b$ then we find $x(2P)$ and $y(2P)$ using the duplication formula say, it is $(\frac{F_1(x)}{4y^2}, \frac{H_1(x)}{8y^3})$ where $F_1, H_1,$ are polynomials in $\mathbb{Z[x]}$ and it'll correspond to the point $(\frac{F_1(x)}{4y^2} : \frac{H_1(x)}{8y^3} : 1)$ on $E(\mathbb{Q})$.

Then multiplying by the LCM of $G_1, H_1$we get $2P = ( F(x) 2y: H(x) : 8y^3)$ and so, $$ (x' : z') = (F(x) :4y^2)$$

It'd be great if someone here could check my work and comment on where it is wrong or right. It's causijg so much confusion because accordijg to what I've read, duplication formula does not give z-coordinate of the point $2P$ so it must be how we find it.

Thank you!


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