Degree of point $\infty$ for $f(x,y) = x$ on an elliptic curve I am studying divisors of rational functions over elliptic curves and am bothered by the definition of a degree through a uniformizer function. In the book Elliptic curves number theory and cryptography (Lawrence C. Washington), page 341, there is an example of calculating the degree of a rational function $f(x,y) = x$ at $\infty$ and it's stated that $u_\infty(x,y) = \frac{x}{y}$ is a uniformizer function on an elliptic curve $y^2 = x^3 + Ax + B$ (with $A, B$ giving a nonzero discriminant) and that
$$x = \left(\frac{x}{y}\right)^{-2} \left(1 + \frac{A}{x^2} + \frac{B}{x^3}\right)^{-1},$$
For $x$ to have a degree $-2$ at $\infty$, the right multiplier on the right side should (by definition of a degree) have neither zero nor a pole at the point $\infty$. But treating this projectively and thus introducing the coordinate $x \to x /z$ gives me
$$\left(1 + \frac{A}{(x / z)^2} + \frac{B}{(x/z)^3}\right)^{-1} = \cdots 
= \frac{x^3}{y^2z},$$
which gives $0 / 0$ at the point $\infty = [0,1,0]$.
How do I see that this function is non-zero and has no pole at $\infty$? This is not explained in the book (at least not explicitly).
 A: A quick way to see the order of vanishing of $x$ at $\infty$ is -2 is to compute $div(x)$ in the affine xy-plane, and then use the fact globally degree of $div(x)$ has to be zero (as is the case for the divisor of any rational function on a projective variety). So $x$ vanishes where $y^2=B$, there are two points in xy-plane so degree of $div(x)$ in xy-plane is 2, so degree at infinity is -2.
A more concrete computational argument: let $X, Y, Z$ be projective coordinates. Then our function we are computing the degree of vanishing of is $x=X/Z$. Let $x'=X/Y$ (the prime to distinguish it from $x=X/Z$) and let $z=Z/Y$. Then $$x=(x')/z$$
and so we need to compute the order of $(x')/z$ at $\infty =(x',z)=(0,0)$ in the $Y=1$ affine patch. The homogeneous equation of the cubic $Y^2Z=X^3+AXZ^2 + BZ^3$ becomes $z=(x')^3 + Ax'z^2 + Bz^3$ in this patch. Since the partial wrt z is non-vanishing, we get that $x'$ is uniformizer. Then if $d:=ord_{x'}(z)$ then the equation $z=(x')^3 + Ax'z^2 + Bz^3$ shows that $ord_{x'}(z)=3$ (since $d \geq 2$ because if $d=1$, then $z$ would also be a uniformizer, but it can't because the partial with respect to x vanishes). So the order of vanishing of $x=(x')/z$ is $1-3=-2$.
To answer your question at the end, what you write $x/z$ I'm writing $x'/z$, and we see it has order vanishing $-2$, so your function
$$\left(1 + \frac{A}{(x / z)^2} + \frac{B}{(x/z)^3}\right)$$ is non-vanishing at $\infty$ (since e.g. $\frac{A}{(x/z)^2}$ has order of vanishing $+2$)
