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Let $K$ be algebraically closed field. Let $f_k, f_{k-1} \in K[x_1,...,x_n]$ coprime homogeneous polynomial of degree $k$ and $k-1$ respectively. I want to prove that:

The variety $$ X = \left\{ (x_0 : x_1 : ... : x_n) \in \mathbb{P}^n \mid f_k(x_1,...,x_n) + x_0 f_{k-1}(x_1,...,x_n) = 0 \right\}$$ is rational.

A projective variety is called rational if it is birational equivalent to $\mathbb{P}^m$.

I have an idea and it is to show that $K(V) \cong K(x_0,...,x_m)$ (Proposition 2.53 in Klaus Hulek, Elementary Algebraic Geometry). The fact that $f_k$ and $f_{k-1}$ are coprime means the following:

  1. They are not zero simultaniously for any $(x_1,...,x_n) \ne (0,...,0)$.
  2. There exists $a,b \in K[x_1,...,x_n]$ such that $a f_k + b f_{k-1} = 1$ so $\langle f_k , f_{k-1} \rangle = \langle 1 \rangle = K[x_1,...,x_n]$.

These two facts and Proposition 2.53 imply that I need to show somehow that the algebra of functions over $X$ is isomorphic to $K(x_1,...,x_n)$ but I don't how to it for this problem.

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up vote 3 down vote accepted

You need to show that there is some open subset of $X$ such that $X \simeq \mathbb{P}^{n-1}$.

HINT Consider the open subset $U$ where $f_{k-1}(x_1,\cdots,x_n) \neq 0$. And note that $X$ is precisely the set of points with $x_0=-f_k/f_{k-1}$.

Can you write up a bijection from $U$ to some open subset of $\mathbb{P}^{n-1}$?

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Thank you. The hint did the trick. But here are few corrections: I think it should be $x_0 = -f_k/f_{k-1}$ (sign) and that the bijection is between $U$ to $W \subset \mathbb{P}^{n-1}$ where $W = \{ (x_1 : ... : x_n ) | f_{k-1}(x_1,...,x_n) \ne 0 \}$. $W$ is open in $\mathbb{P}^{n-1}$ and then I apply Proposition 2.53 from Hulek. – LinAlgMan May 5 '13 at 15:06
@LinAlgMan Yes, you are correct - thank you for your corrections. – Fredrik Meyer May 5 '13 at 15:09

Consider the open subset $W=\mathbb P^{n-1}\setminus V(f_{k-1}) $ defined by $f_{k-1}\neq 0$ and the open subset $U=X\setminus V(f_{k-1}) \subset X$ .
The two mutually inverse isomorphisms $$ W\to U: ( x_1 : \cdots : x_n) \mapsto(-\frac {f_k(x_1 : \cdots : x_n)}{f_{k-1}(x_1 : \cdots : x_n)} : x_1 : ... : x_n) \\ U\to W: (x_0 : x_1 : ... : x_n) \mapsto ( x_1 : \cdots : x_n) $$
show that $X$ is rational.

A riddle
Where have I used that $f_{k-1}$ and $f_k$ have no common factor?

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I have added this answer to Frederik's excellent one (whose notations I followed and which I upvoted) in order to exhibit explicit bijections (for my records, even if nobody else is interested) and more importantly so as to propose my riddle :-) – Georges Elencwajg May 6 '13 at 0:02

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