This is Exercise 5.3 (a) in Undergraduate Algebraic Geometry by Reid.

Does the map $\phi: \mathbb{P^2\rightarrow \mathbb{P^1}}$ defined by $\phi(x,y,z)=(x,y)$ define a rational map? Determine $\text{dom}(\phi)$, determine if $\phi$ is birational, and if so describe the inverse map.

My answer:

It is rational since $\phi$ sends $(x,y,z)$ to $(\frac{x}{y},1)$ if $y\ne 0$ and both $\frac{x}{y}$ and $1$ are rational functions defined on $\mathbb{P}^2$.

Its domain is $\mathbb{P}^2 \backslash \{(0,0,1)\}$.


I believe it is not birational. But I think I have some conceptual misunderstanding here. I think there exists a birational map between a $1$ dimensional affine piece of $\mathbb{P}^2$ and $\mathbb{P}^1$, which is defined by: $$\phi(x,y,z)=(x,y)=(x/y,1)\\ \phi^{-1}(x,y)=(x,y,y)$$

The domain of this function $\phi$ is then restricted to $\{(x,1,1)\in\mathbb{P}^2\}$.

To check whether they are inverse: $$\phi\circ\phi^{-1}(x,y)=\phi(x,y,y)=(x,y)\\ \phi^{-1}\circ \phi(x,y,z)=\phi^{-1}(x,y)=(x,y,y)=(x,y,z)$$

The last equality is because $\phi$ is restricted to the domain $\{(x,1,1)\}$.

But how can we be sure $\phi$ is not birational? By definition it is birational as long as we can find an open subset of $\mathbb{P}^2$ which is in the domain of $\phi$ and its inverse is defined on an open subset of $\mathbb{P}^1$. How can we say there is not one?

Thanks for any help!

  • 3
    $\begingroup$ "I think there exists a birational map between the affine piece $A^2_3$ of $P^2$ and $P^1$." What do you think that map is? $\endgroup$
    – Schemer
    Feb 4 '16 at 19:24
  • $\begingroup$ @Schemer: Thanks for your reply! I edited my question. So it shouldn't be $A^2_3$ but instead just a $1$D piece of $P^2$. I doubt it is correct though. Could you check for me? $\endgroup$
    – KittyL
    Feb 4 '16 at 19:31
  • $\begingroup$ Your inverse is not an inverse of the given $\phi$, right? You’d need to find one. $\endgroup$
    – Lubin
    Feb 4 '16 at 19:50
  • $\begingroup$ @Lubin: Thanks for your reply! Sorry I still don't quite understand. I edited and checked the identity in my question. Could you check again? $\endgroup$
    – KittyL
    Feb 4 '16 at 19:50
  • $\begingroup$ @Lubin: I couldn't find one that gives back the $z$, so I restricted the domain of $\phi$. Is it not allowed to restrict the domain? I thought for rational functions, you can do that. $\endgroup$
    – KittyL
    Feb 4 '16 at 19:52

I think it’ll help if we restrict our view to the finite plane, i.e. setting $Z=1$, using the equivalence $(x,y,z)\leftrightarrow(x/z,y/z,1)$, which is good except on the “line at infinity” given by $Z=0$. You’ve done that in your exposition. Now, to modify slightly what you’ve said, the given map $\phi$ has the effect of sending $(x,y,1)$ to $(x,y)\leftrightarrow(1,y/x)\in\Bbb P^1$.

What have you done? You’ve excluded the origin from consideration, but otherwise defined a map that takes any point $(a,b)$ in the finite-plane part of $\Bbb P^2$, and sends it to the slope of the line through the origin that contains your point, namely $Y=\frac baX$. If you want to restrict to the real projective plane and line, you can think of this as mapping your plane to a fixed circle centered at the origin. (This is all right: the real projective line is equivalent in every way to the real unit circle centered at the origin.) Now, your proposed inverse would have to spread out the range of the given function $\phi$, namely that circle, back all over the plane. In other words, stretch out the thin, one-dimensional circle, to the two-dimensional plane. No way this can be done! (If you restrict the domain of the original function to a closed and not open subset, then all bets are off, because this destroys the original setup.)

So as you correctly state, the original function $\phi$ is rational all right, but it has no rational inverse, just as your original intuition led you to believe.

What’s the moral of the story? Algebraic Geometry really genuinely is Geometry, and you should not put your geometric intuition away into the closet.

  • $\begingroup$ I see. So the point is the inverse has to map back to the original domain. For some reason projective geometry is not that intuitive to me. Thank you for the clear explanation! $\endgroup$
    – KittyL
    Feb 4 '16 at 20:41

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