5

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

agt
  • 4,772

2 Answers2

7

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 $\Bbb P^2$.

Lierre
  • 4,420
  • 2
    I would say it's not a map on $\mathbb{P}^1_{\mathbb{F}_2}$ either, although it is a function on the set $\mathbb{P}^1(\mathbb{F}_2)$. –  Jun 12 '12 at 06:59
  • @Hurkyl I understand that $\mathbb{P}_{F_2}^1$ is the projective line over the field of 2 elements but what does this $\mathbb{P}^1(\mathbb{F}_2)$ mean? – math-visitor Jun 12 '12 at 08:10
  • When you are defining maps between two projective varieties in general, when do you need to restrict to their affine charts to define a map and when is it allowed to define a map as a whole (for example, like the above second example)? – math-visitor Jun 12 '12 at 08:17
  • @Hurkyl — Thanks, corrected. – Lierre Jun 12 '12 at 09:27
  • 1
    @math-visitor — As a set $\Bbb P^1$ is defined by a quotient set. So to define a map from on $\Bbb P^1$, you have to give a function of $x$ and $y$ which depends only on the point $[x : y]$ that they represent. – Lierre Jun 12 '12 at 09:32
  • 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 $\mathbb{F}_4 = \mathbb{F}_2(\alpha)$) –  Oct 03 '12 at 19:46
5

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 $\mathbb{P}^1 \rightarrow \mathbb P^2: [x:y]\mapsto [a:b:1]$

Georges Elencwajg
  • 150,790