3

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?

emiel647
  • 81
  • 4
    Welcome to MSE. Please use MathJax to format math. In regards to your problem, you're missing the hypothesis that this is a plane curve. For a hint, look at lines through $P$: what can you say about the intersection of the line and your curve? – KReiser Jan 02 '22 at 18:53
  • Neat exercise, I guess the analogous statement is true for hypersurfaces – Nick L Jan 02 '22 at 19:41
  • @KReiser I indeed forgot to say that we are working with a plane curve. We used Bézout's Theorem to say that the intersection of the curve with a line has exactly d points, counted with multiplicity and therefore the curve can only have one singular point, but we have no idea on how to use this to prove that the curve is rational or to find a resolution – emiel647 Jan 04 '22 at 09:32
  • To prove a curve is rational one has to construct a birational map $\mathbb{P}^1 \dashrightarrow X$, using KReiser's hint, do you see how to construct such a map? (Further hint: the space of lines containing $P$ is itself isomorphic to $\mathbb{P}^1$) – Nick L Jan 04 '22 at 18:48
  • I'd be happy to write you an answer if you improve your post with an [edit]: specifically, please use MathJax to format the math in your post, and add some of your thoughts from the comments. – KReiser Jan 04 '22 at 19:21

1 Answers1

3

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.

KReiser
  • 74,746
  • Thanks for your answer! This was very useful! Do you have any idea on how to find a resolution of $X$? – emiel647 Jan 21 '22 at 09:27
  • What do you mean a resolution of $X$? If you mean a resolution of singularities, the map $\Bbb P^1\to X$ constructed in the answer suffices: it is a proper birational map from a smooth variety. – KReiser Jan 21 '22 at 09:51