I am struggling to understand the concept of rational maps. In Hartshorne, the definition of rational maps is as follows
Let $X$ and $Y$ be varieties. A rational map $\phi:X\rightarrow Y$ is an equivalence class of pairs $\langle U,\phi_{U} \rangle$ where $U$ is a nonempty open subset of $X$, $\phi_{U}$ is a morphism of $U$ to $Y$, and where $\langle U,\phi_{U} \rangle$ and $\langle V,\phi_{V} \rangle$ are equivalent if $\phi_{U}$ and $\phi_{V}$ agree on $U\cap V$. The rational map $\phi$ is dominant if for some (and hence every) pair $\langle U,\phi_{U} \rangle$, the image of $\phi_{U}$ is dense in $Y$.
I get what this definition is saying, but It seems so obscure and doesn't allow me to look at a map and say whether it is rational or not. Also, why is it called rational… It doesn't feel rational (fractions).
Apparently, I have seen somewhere (very briefly, so this may be wrong) that $\mathbb{P}^1$ is birational to $\mathbb{A}^1$. If I were to try to prove this is map I would go for is $\psi:\mathbb{A}^1\rightarrow \mathbb{P}^1$, $\phi(x)=[x:1]$. The inverse map I would then go for is $\theta:\mathbb{P}^1\rightarrow \mathbb{A}^1,\theta([a:b])=\frac{a}{b}$. This seems to be wrong since it fails if $b=0$… This is my issue, are the maps even the right ones and even if they are, how can they be seen to be rational using the definition above.
What I would really like are some examples of varieties which are birationally equivalent to each other and really explain why the maps between these spaces are rational.
Another exercise in Hartshorne is to show that any conic in $\mathbb{P}^2$ is birational to $\mathbb{P}^n$ for some $n$. I would like to eventually justify this myself but could someone explain how the specific conic (or another conic, if you so please) $x_1x_0+x_0x_2+x_2^2=0$ is birational to $\mathbb{P}^n$ for some $n$ emphasizing on the explanation of how the maps between these two varieties are rational.
As you can probably tell, I am not very good at this subject so please refrain from saying things like "it's trivial" or "it's obvious" as it will probably won't be obvious to me (of course though, if its the issue of showing something like a bijection, that's all good just please don't dance around the details of rational maps…). Also, I do not know schemes, so please no answers appealing to much higher results than the ones presented in Hartshorne up until this point of the book.
Thanks in advance.
Best Answer
A rational map is a morphism that is defined on an open subset, which is almost everywhere since varieties are irreducible. They are called rational because they mirror the 'rational functions' $\{ f / g \mid f, g \in k[x] \}$ which are defined everywhere except $g(x) = 0$. The example of a biration map between $\mathbb{A}^1$ and $\mathbb{P}^1$ is fine because $\mathbb{P}^1 \to \mathbb{A}^1$ doesn't have to be defined at $1:0$.
Varieties are so rigid that rational maps whose images are dense form a category (since composition now is well defined.) Like we know previously that the category of affine varieties is equivalent to the opposite category of fin. gen. $k$-algebras, we can show the category of varieties with dominant rational maps is equivalent to the opposite category of fin. gen. extensions of $k$ (their function fields).