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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

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2 Answers 2

up vote 4 down vote accepted

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 $\Bbb P^2$.

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I would say it's not a map on $\mathbb{P}^1_{\mathbb{F}_2}$ either, although it is a function on the set $\mathbb{P}^1(\mathbb{F}_2)$. –  Hurkyl Jun 12 '12 at 6:59
    
Thank you for the answer. –  user17090 Jun 12 '12 at 7:22
    
@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 8:10
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@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 9:32
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@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 $\mathbb{F}_4 = \mathbb{F}_2(\alpha)$) –  Hurkyl Oct 3 '12 at 19:46
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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 $\mathbb{P}^1 \rightarrow \mathbb P^2: [x:y]\mapsto [a:b:1]$

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Thank you for the answer. –  user17090 Jun 12 '12 at 7:22
    
It's a pleasure, Hollowdead. –  Georges Elencwajg Jun 12 '12 at 8:18
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