Function field of $\mathbb{P}^1$

Question

What is the function field of $\mathbb{P}^1$ when $\mathbb{P}^1$ is viewed as a projective curve? From what I read, I think it should be $k(x)$, but I don't understand how.

Answer 1

The only regular functions on $\mathbb P^1$ are constants, so you have to pass to nontrivial open subsets to get anything interesting. But once you have deleted a single point, you are now looking at $\mathbb A^1$ or a subset thereof, and the ring of regular functions on $\mathbb A^1 \setminus \left\{ a_1,a_2,...,a_n \right\}$ is $k\left[x,\frac{1}{x-a_1}, \frac{1}{x-a_2},...,\frac{1}{x-a_n}\right]$. Now as you vary over all possible finite subsets you get a directed system of rings starting with $k \subset k[x]$ (and the other possible ways to include $k$ into a one variable polynomial ring by deleting a point from $\mathbb P^1$). Taking the colimit over this system you obtain $k(x)$, essentially as a consequence of the universal property of localization.