I'm reading a book on elliptic curves, and they say:
for any function $f$ on $E$, we always have $\deg((f)) = 0$ [where $(f)$ is the multiset of divisors with positive multiplicity for roots, negative multiplicity for poles]. Roughly speaking, this property follows from observing that the degree of the affine equation that solves for the zeros of $f$ on $E$ matches the degree of the projective equation that determines the multiplicity of $f$ at $\mathcal{O}$, i.e. the projective version of $f$ is $g/h$ where $g$ and $h$ both have the same degree as $f$.
Can someone give a bit more detail to this? I'm familiar with projective coordinates and the homogenization
\begin{align*} x &\rightarrow X/Z \\ y &\rightarrow Y/Z \end{align*}
I think that this fact has something to do with the fact that the term with the highest degree in $f(x,y)$ will have a pole of the same degree at $Z = 0$ (i.e. the point at infinity, aka $\mathcal{O}$), because of the homogenization above. I'm having trouble formalizing this though, and haven't been able to find an answer elsewhere online.
Edited to add: I'll give an example. Given a (non-vertical) line $y = mx+b$, the line will intersect the curve at three roots (counting multiplicity). Projectifying and using $y = mx+b$, we get, plugging back into the Weierstrass equation for $E$:
$$ (\frac{mX + bZ}{Z})^2 = (\frac{X}{Z})^3 + a(\frac{X}{Z}) + b$$
which has a pole of order $3$ at $Z=0$. Unfortunately, it's not clear to me how to do something analogous for an arbitrary rational function $f$.
This may be using tools you do not know, but as Lord Shark suggested, the result holds more generally on any compact Riemann surface (or nonsingular algebraic curve). The Residue Theorem tells you that the sum of the residues of the meromorphic $1$-form $\dfrac{df}f$ is $0$, and — as usual with the argument principle — this is #(zeroes) $-$ #(poles).