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The complete reference is Cor 7.24 of this notes: https://www.mathematik.uni-kl.de/~gathmann/de/alggeom.php

We want to prove the statement "Let $X$ be a connected complete variety, then $\mathcal{O}_X(X)=k$."
The proof goes like this: A global regular function $\varphi\in\mathcal{O}_X(X)$ gives a morphism $\varphi:X\to\mathbb{A}^1$. Then we extend the target to $\mathbb{P}^1=\mathbb{A}^1\cup\{\infty\}$ and the image $\varphi(X)$ of the new map does not contain the point at infinity. But (by the corollary I will write down below), since $X,\mathbb{P}^1$ are varieties and $X$ is complete, we have $\varphi(X)$ must be a complete closed subvariety of $\mathbb{P}^1$, which is a set of finite points. Then as the image of a connected space, $\varphi(X)$ must be connected, and hence must be a single point.
My question is, why does the author extend the target to $\mathbb{P}^1$? If we do not do that and stick to $\mathbb{A}^1$, then we still have the same result since, $\mathbb{A}^1$ is an affine variety hence also a variety so that we are able to use the corollary below.
Corollary. Let $f:X\to Y$ be a morphism between varieties, if $X$ is complete then $f(X)$ is a complete closed subvariety of $Y$.
Also, I am not so sure how do we conclude that, the proper closed subsets of $\mathbb{P}^1$ are finite sets. Because in this case, when we are discussing $V_p(I)$, we have $I\subseteq k[x_0,x_1]$. Then we cannot proceed in the same way as we did to $\mathbb{A}^1$. Any help is appreciated! Thanks in advance.

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    $\begingroup$ We must use $Y=\mathbb{P}^1$ to ensure that the image of $f:X \to Y$ is not all of $Y$. To see that proper closed subsets of $\mathbb{P}^1$ are finite point sets just cover $\mathbb{P}^1$ with two affine $U_i \cong \mathbb{A}^1$. $\endgroup$ Mar 21, 2023 at 13:34
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    $\begingroup$ @JürgenBöhm the corollary cited above is enough to show that the image of $f$ is not all of $\Bbb A^1$, as $\Bbb A^1$ is not complete. $\endgroup$
    – KReiser
    Mar 21, 2023 at 16:00
  • $\begingroup$ @KReiser You are right, I overlooked the word "complete" in the statement of the corollary. $\endgroup$ Mar 21, 2023 at 17:12

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