degree, genus, and birational map

Question

I'm confused on the following:

Suppose we have a polynomial $f(T)$, e.g. $T^d$, with degree $d>2$, then I consider the curve defined by $y-f(x)$. Then there seems an obvious birational map given by $$ z\longmapsto (z,f(z)). $$ On the other hand, the degree-genus formula tells us that $g=\frac{(d-1)(d-2)}{2}>0$, and hence shouldn't birational to $\mathbb{P}^1$.

Question: what's wrong with above argument?

Answer 1

For $d = \deg f > 2$, the curve defined by $y − f(x)$, or more precisely $z^{d-1} y − z^d f(x/z)$ is not smooth at the point $[0:1:0]$ (at infinity). Therefore the degree-genus formula does not apply.