Calculating divisor of function on elliptic curve I read Pairings for Beginners by Craig Costello.
In the example 3.1.1 at 37-th page we consider $ E/F_{103} : y^2 = x^3 + 20x + 20$, with
points $ P = (26, 20), Q = (63, 78), R = (59, 95), T = (77, 84)$ all on $E$. The author defines function $f = \frac{6y + 71x^2 + 91x + 91}{x^2 + 70x + 11} $ and states that it's divisor is $(f) = (P) + (Q) - (R) - (T)$. To check that there is no zero/pole at $\mathcal{O} = (0 : 1 : 0)$ we look at $f$ in projective space $f = \frac{6YZ + 71X^2 + 91XZ + 91Z^2}{X^2 + 70XZ + 11Z^2}$. At infinity both the numerator and the denominator are zero and the book says that they cancels out. But I see that if we substitute $Y=1$ and $X=Z \to 0$ the numerator will tend to zero linearly while the denominator quadraticly. And that should give us first order pole at $\mathcal{O}$.
My version of $f$ is $f' =\frac{y+4x+82}{y+75x+12}$ and it clearly evaluates to $1$ at $\mathcal{O}$.
Do I calculating divisors wrong?
 A: The example problem concerns an elliptic curve
$$ y^2 = x^3 + 20x + 20. \tag{1} $$
If $\,x\,$ goes to infinity, then let $\,x=1/t^2\,$ where
$\,t\,$ goes to zero and then
$$ y = \sqrt{x^3 + 20x + 20} = 1/t^3 + 10t + 10t^3 -50t^5
+O(t^7). \tag{2} $$
Thus, $\,x\,$ has a pole of order $2$ and $\,y\,$ has a
pole of order $3$ at infinity. More easily seen by eliminating
all but the leading terms of equation $(1)$ to get
$\,y^2=x^3\,$ and then $\,y=1/t^3.$
The standard parametrization
$$ x = X/Z,\quad y = Y/Z \tag{3} $$ obscures this situation.
A better one for this purpose is
$$ x = X/T^2,\quad y = Y/T^3. \tag{4} $$
Now, given that
$$ f = \frac{6y + 71x^2 + 91x + 91}{x^2 + 70x + 11}, \tag{5} $$
the new projectivization in equation $(4)$ gives
$$ f = \frac{71X^2 + 6TY + 91T^2 + 91T^4}{X^2+70XT^2+11T^4} . \tag{6} $$
The elliptic curve equation $(1)$ in the new parametrization is
$$ Y^2 = X^3+20XT^4+20T^6. \tag{7} $$
The substitution $\,X = Y = 1\,$ and $\,T = 0\,$ satisfies this
equation and represents the point at infinity. It also gives
$\,f = 71.\,$
In a precise algebraic sense, the poles "cancel out".

Note that the advantage of equation $(3)$ is that it is homogeneous.
That is, $\,(X,Y,Z)\,$ represents the same point as
$\,(cX,cY,cZ)\,$ for any nonzero $\,c.\,$ For equation $(4)$ the
corresponding property is that the tuple $\,(X,Y,T)\,$ represents
the same point as $\,(c^2X,c^3Y,cZ)\,$ and this difference must be
clearly understood.
