# Is there a shortcut to Miller's algorithm?

Consider an elliptic curve $$E: y^2 = x^3 + ax + b$$ over some finite field $$F_{q^k}$$ and a point $$P$$ on $$E$$ of order $$n$$.

Miller's algorithm tells us how to efficiently construct a rational function $$f_{n,P}(x,y)$$ on $$E$$ with divisor $$div f = n [P] - n [\mathcal{O}]$$.

I have worked through examples and exercises and have understood how and why Miller's algorithm works.

But I cannot answer the following question: Why can't we just take $$f_{n,P} = (y-y_P)^n$$? It obviously has an $$n^{\rm th}$$ order root at $$P$$ and $$n^{\rm th}$$ order pole at $$\mathcal{O}$$. It's also obvious that it doesn't have any other poles or roots, because only the point $$P$$ has $$y$$-coordinate $$y_P$$. What am I missing?

First, $$y-y_P$$ does not have a zero of order $$1$$ at $$P$$ for some choices of $$P$$. For instance, for any point $$P$$ with horizontal tangent line, $$y-y_P$$ vanishes to order two at $$P$$. Next, $$y-y_P$$ has a pole of order three at $$O$$: the line $$V(Y-y_PZ)$$ has three points of intersection with your elliptic curve away from $$Z=0$$; the line $$V(Z)$$ has a triple intersection at $$O$$; thus their ratio has a triple pole at $$O$$.