Is there a classification of regular maps $\mathbb{P}^1(k)\to\mathbb{A}^1(k)$?

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If $\mathbb{P}^1(k)$ and $\mathbb{A}^1(k)$ are the projective line and affine line, respectively, over an algebraically closed field $k$, is there any known classification of the regular maps $\mathbb{P}^1(k)\to\mathbb{A}^1(k)$?

I searched a bit on google and through my books, but can't find any mention of any classification or description. Is it possible to determine all such maps? Thanks.

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Maps into $\mathbb{A}^1$ are just regular functions, and it's likely that you've shown that the only functions defined on all of $\mathbb{P}^1$, or more generally on a connected projective variety, are constant. In the setting of complex manifolds this is just the maximum principle. Algebraically, you're looking for $f(x) \in k[x]$ and $g(y) \in k[y]$ such that \[ f(x) = g(1/x) \] inside of $k[x, x^{-1}]$.

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$\Bbb{P}^1(k)$ is connected, and regular map from connected projective variety to affine line must be constant.

Let $(a:b)\in \Bbb{P}^1(k)$, then regular function at $(a:b)$ is $$ f(x:y)=\frac{p(x,y)}{q(x,y)}$$ where $ p,q\in k[x,y]$ are homogeneous, have same degree and $q(a,b)\ne 0$. Let $f=p'/q'$ be another representation of $f$. Then $$ p/q=p'/q'$$ and $pq'=p'q$. We can assume $ p,q$ and $p',q'$ are relatively prime. It follows $q$ divide $q'$. Therefore if $q$ is not a constant, there is a point $p\in \Bbb{P}^1(k)$ s.t. $f$ is not defined at $p$. A contradiction.