4
$\begingroup$

I have a question on rational and birational maps: Is the map $\mathbb{P}^1\rightarrow \mathbb{P}^2, (x:y) \mapsto (x:y:1)$ rational? Birational? If birational what is its inverse? Same questions for map $\mathbb{P}^1 \rightarrow \mathbb{P}^2, (x:y) \mapsto (x:y:0).$ My guess is that both aren't birational and that both are rational, but would like to hear another opinion.

Thank you

2 Answers 2

5

The first map you define is not even a map (except if you consider only the set structure and if the base field is $\Bbb F_2$). In $\Bbb P^1$, the points $[ x : y ]$ and $[ \lambda x : \lambda y ]$ are the same for all non zero $\lambda$ in the base field. So their image must be the same. But obviously all the $[\lambda x : \lambda y : 1 ]$ are not equal.

The second one is a morphism, defined everywhere, but it's not birational since its image is not dense in $\Bbb P^2$.

  • 1
    @math-visitor: $\mathbb{P}^1(\mathbb{F}_2)$ is, roughly speaking, the set of points on the projective line that have coordinates in $\mathbb{F}_2$. The zero set to the homogeneous polynomial $x^2 + xy + y^2$ is an example of a point of $\mathbb{P}^1_{\mathbb{F}_2}$ that is not an element of $\mathbb{P}^1(\mathbb{F}_2)$. (this point in $\mathbb{P}^1_{\mathbb{F}_2}$ corresponds to the conjugate pair of points $(\alpha : 1)$ and $(\alpha + 1 : 1)$ in $\mathbb{P}^1(\mathbb{F}_4)$, where $\mathbb{F}_4 = \mathbb{F}_2(\alpha)$)2012-10-03
4

The first formula you give doesn't even define a set-theoretic map. The second defines a morphism and a fortiori a rational map.
But it is not birational because birational maps can exist only between varieties of the same dimension.

NB
Consider the open subset $U\subset \mathbb P^2$ consisting of points with coordinates $[x:y:1]$.
It is an affine variety (isomorphic to $\mathbb A^2$) and thus the only morphism $\mathbb{P}^1 \rightarrow \mathbb P^2$ with image included in $U$ are the constant ones $\mathbb{P}^1 \rightarrow \mathbb P^2: [x:y]\mapsto [a:b:1]$

  • 0
    It's a pleasure, Hollow$d$ead.2012-06-12