Irreducible plane curve of degree $d>2$ with a point $P$ of multiplicity $d-1$ is rational We have to prove that an irreducible plane curve $X$ of degree $d>2$ with a point $P$ of multiplicity $d-1$ is rational and we need to find a resolution of this $X$.
To prove that $X$ is rational, we can construct a birational map $\mathbb{P}^1 \mapsto X$, using the point $P$. However I have no clue on how to find this birational map.
Also to find a resolution of this curve, I have no idea on how to start.
Does someone know to proceed?
 A: Up to an automorphism of $\Bbb P^2$ we may assume that our point of multiplicity $d-1$ is $[0:0:1]$. Let $F$ be the equation of our curve $X$ after this automorphism, and let $f$ be its dehomogenization with respect to $z$. Since $[0:0:1]$ is of multiplicity $d-1$, we must have $f=f_{d-1}(x,y)+f_d(x,y)$ where $f_{d-1}$ and $f_d$ are homogeneous of degree $d-1$ and $d$, respectively, and neither are zero - $f_{d-1}$ by the multiplicity hypothesis, and $f_d$ by the hypothesis that our curve is irreducible.
Let's find the intersection of our curve with a line of the form $y=tx$. Plugging this in to $f$, we find $x^{d-1}p_{d-1}(1,t)+x^dp_d(1,t)=0$ which factors as $x^{d-1}(p_{d-1}(1,t)+xp_d(1,t))=0$. So the intersection of our curve with the line $y=tx$ is the point $(0,0)$ with multiplicity $d-1$ and the point $(-\frac{p_{d-1}(1,t)}{p_d(1,t)},-\frac{tp_{d-1}(1,t)}{p_d(1,t)})$. Therefore we get a regular map $\Bbb A^1\setminus \{t\mid p_d(1,t)=0\}\to X$ which by the curve-to-projective extension theorem extends to a map $\Bbb P^1\to X$. This is what you're looking for.
