# Birational between affine and projective

I'm asked to prove that $\mathbb{A}^n$ is birational to $\mathbb{P}^n$. I know I'm suppose to find a rational function whose inverse is also a rational function. But I have no idea where to start. Maybe it's because I'm having trouble believing this is result is even true. This is because I know that $\mathbb{A}^n$ is birational to $\mathbb{P}^{n}$ with one of the coordinates fixed as $1$.

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What is your working definition of "rational function"? – Brad Aug 5 '13 at 3:04
A map $\phi:X\rightarrow Y$ is called rational if it can be written as $\psi(p)=[F_1(p):\ldots:F_{n+1}(p)]$ for every $p\in X$ for some forms $F_1,\ldots,F_{n+1}$ – user44322 Aug 5 '13 at 3:07
In that definition of rational map, you write $\phi(p)$ in homogeneous coordinates. Are you assuming that $Y$ is a subset of $\mathbb{P}^n$? – Brad Aug 5 '13 at 3:52
@user44322: Dear user, Based on your post, and on your comments below Brad's answer, I think you have some pretty basic confusions about the concepts you are asking about. Why don't you try to think through the case $n = 1$. Ask yourself: what does it mean to give a rational map $\mathbb A^1 \to \mathbb P^1$; i.e. what would the formula for such a map look like. Then ask what it would mean to give a rational map back the other way. Then you can think about what it means for them to be inverse. One thing I should note is that almost anyone who knows algebraic geometry regards the ... – Matt E Aug 5 '13 at 4:04
... statement that $\mathbb A^n$ and $\mathbb P^n$ are birational as being absolutely obvious (with a very strong emphasis on both words), almost by definition. So the fact that you don't see it, and in fact, have trouble believing it, means that you're mental model of the situation is very different to that of a more experienced algebraic geometer. That's okay! --- you're obviously just beginning. But if you try to work out the details of the definitions, and think about your problem, in the case $n = 1$, it might help to bring your intuition more in line with the mainstream. Regards, – Matt E Aug 5 '13 at 4:07

You mention that $\mathbb{A}^n$ is birational to any of the subsets $U$ of $\mathbb{P}^n$ obtained by fixing a coordinate to be 1, e.g. $$U = \{[x_0:x_1:\cdots:x_{n-1}:1]: x_i\in k\}]\subset \mathbb{P}^n$$ where $k$ is the field over which we're working. In fact, they are isomorphic, not just birational: the usual map $\mathbb{A}^n\to U$ given by $(x_0,\ldots,x_{n-1})\mapsto [x_0:\cdots:x_{n-1}:1]$ is an isomorphism. By the definition you gave, this defines a rational map $\phi:\mathbb{A}^n\dashrightarrow\mathbb{P}^n$ (by composing with the inclusion map $U\hookrightarrow \mathbb{P}^n$). See if you can write down a rational map from $\mathbb{P}^n$ to $\mathbb{A}^n$ which is an inverse for $\phi$.
So you are defining $\phi$ as $(x_1,\ldots,x_n)\mapsto [x_1,\ldots,x_n,1]$? – user44322 Aug 5 '13 at 3:17
Sorry, I had an $n$ where I meant $n-1$. I've edited my answer. But yes, what you write there is the same as my $\phi$, just with different indices. – Brad Aug 5 '13 at 3:18
Would it be $[x_0:\ldots:x_{n-1}:x_n]\mapsto (x_0,\ldots,x_{n-1})$? So having an inverse does not mean it has to be isomorphic? – user44322 Aug 5 '13 at 3:25
That's okay: rational maps are not defined everywhere. For example, $z\mapsto \frac{1}{z^2-1}$ is a perfectly fine rational map from $\mathbb{A}^1$ to $\mathbb{A}^1$, even though it is not defined at the points $\pm 1$. – Brad Aug 5 '13 at 4:04
How about $[x_0,\ldots,x_{n-1},x_n]\mapsto (\frac{x_0}{x_n},\ldots,\frac{x_{n-1}}{x_n})$? This is not defined on anything with $x_n=0$ and it will take care of the not well defined. – user44322 Aug 5 '13 at 4:16