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This is a follow-up question to this one: Rational functions on curve

In that setting, assume $X=\mathbb{P}^1$ and $f$ an isomorphism, so that we are looking at automorphisms of $\mathbb{P}^1$. Denoting the coordinate ring of $\mathbb{A}^1$ as $k[x]$, we have that the rational functions on $\mathbb{P}^1$, denoted by $K(\mathbb{P}^1)$, is the field of fractions of $k[x] = \mathcal{O}_{\mathbb{P}^1}(\mathbb{A}^1)$. I am trying to show that $f^*:K(\mathbb{P}^1) \rightarrow K(\mathbb{P}^1)$ induced by $f$, sends $x$ to $f^{\vee}$ (defined in the previous question - seems clear theoretically but maybe in this practical example there is a direct calculation to show it) $and$ that there exist $a,b,c,d$ s.t. $f^{\vee}$ can be written as $\left(\frac{ax +b}{cx+d}\right)$.

This last part should be doable using divisors - and maybe the RR theorem - but I'm at a loss as to how exactly. Perhaps it would be helpful to me to know what properties a function like $f^{\vee}$ must have on $\mathbb{P}^1$ that can be deduced by using divisors and the Riemann-Roch theorem. Maybe from those properties it is clear that it must be possible to write it that way.

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    $\begingroup$ If $f : \mathbb{P}^1 \to \mathbb{P}^1$ is an automorphism then $f^* \mathscr{O}(1)$ must be an invertible sheaf whose global sections are a vector space of dimension 2 - which can only be $\mathscr{O}(1)$. $\endgroup$ Dec 9, 2018 at 17:44
  • $\begingroup$ Hi Daniel, I don't know exactly what an invertible sheaf is and I struggle to give meaning to $\mathcal{O}(1)$ too. $\endgroup$
    – Dalamar
    Dec 9, 2018 at 17:46

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As in the last question, you seem to keep jumping to divisors and RR. These are very useful and important things, but they're not necessary here either.

Let $f:\newcommand\PP{\mathbb{P}}\PP^1_k\to\PP^1_k$ be an automorphism. Fix homogeneous coordinates $(x:y)$ on $\Bbb{P}^1$. Then by construction of $f^\vee$, the zero set of $f^\vee$ is the set of points of $\Bbb{P}^1$ mapping to $(0:1)$ and the pole set of $f^\vee$ is the set of points of $\Bbb{P}^1$ mapping to $(1:0)$. Since $f$ is an automorphism, both of these sets are singletons. Since we know that the rational functions on $\Bbb{P}^1$ are fractions of homogeneous polynomials with numerator and denominator of the same degree. This immediately tells us that $f^\vee$ (expressed in lowest terms, using that $k[x,y]$ is a UFD) has both numerator and denominator of degree $1$. Hence $f^\vee = \frac{ax+by}{cx+dy}$. In the open affine with $y=1$, this is the rational function $\frac{ax+b}{cx+d}$.

If you regard the rational functions on $\PP^1_k$ as being the rational functions on the open affine with $y=1$, then the definition of $f^\vee$ is that it is the pullback of the regular function $x$ regarded as a rational function on all of $\PP^1$. Thus the induced map of function fields is indeed $x\mapsto \frac{ax+b}{cx+d}$.

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    $\begingroup$ @Dalamar rational functions aren't defined on all of $\Bbb{P}^1$. They're defined on open subsets of it. The ring of rational functions on an open subset of $\Bbb{P}^1$ (say $\Bbb{A}^1$) is exactly the same as the ring of rational functions on all of $\Bbb{P}^1$. $\endgroup$
    – jgon
    Dec 9, 2018 at 18:08
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    $\begingroup$ As for why the numerator and denominator have to be of degree 1, it's because they have to have a single root (in $\Bbb{P}^1$) each, and your field is algebraically closed, so the only way that can happen is if they have degree 1. $\endgroup$
    – jgon
    Dec 9, 2018 at 18:09
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    $\begingroup$ Lastly the divisor of $f^\vee$ would be $Z-P=(-b:a)-(-d:c)$ $\endgroup$
    – jgon
    Dec 9, 2018 at 18:10
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    $\begingroup$ @Dalamar yes, that's right $\endgroup$
    – jgon
    Dec 9, 2018 at 18:10
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    $\begingroup$ @Dalamar Well, you would know that it has a single pole and a single zero, since $f$ is supposed to be an automorphism. Thus it would have to be of the form $Z-P$ where $Z$ and $P$ are the unique points that are the zero and pole of $f^\vee$ respectively. $\endgroup$
    – jgon
    Dec 9, 2018 at 18:15

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