Closed connected subset of $\mathbb{P}_k^1$

Question

I was reading Vakil's proof that if $X$ is a connected, reduced, proper scheme over an algebraically closed field $k$, then $\Gamma(X, \mathscr{O}_X) = k$. He defined an image $X \to \mathbb{P}_k^1$, and showed that the set theoretic image is connected and closed. Why does this then imply that the image must be either a closed point or all of $\mathbb{P}_k^1$?

Answer 1

Well, $\mathbb{P}^1_k$ does not have a very large variety of closed subsets. Every closed subset is either a finite set of closed points or the entire space. A finite set of closed points has the discrete topology, so the only way it can be connected is if it has just one point. So the only connected closed points are single closed points and the entire space.

(Here I assume you do not consider the empty set to be connected; indeed, the result is not true if you allow $X$ to be empty.)

Answer 2

In fact Vakil defines a morphism $X\to\mathbb A^1_k\hookrightarrow\mathbb P^1_k$. A closed subset of $\mathbb A^1_k$ is given by an ideal $I\subset k[x].$ Now $k[x]$ is a principal ideal domain, and by the Fundamental theorem of algebra we have $I=(0)$ or $I=((x-a_1)\cdot\ldots\cdot (x-a_n))$. The image cannot be the whole $\mathbb A^1_k$, because $\mathbb A^1_k$ is not closed in $\mathbb P^1_k$, therefore it is a point.