Automorphisms of the projective space.

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.

• 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 $\mathscr{O}(1)$. Dec 9, 2018 at 17:44
• Hi Daniel, I don't know exactly what an invertible sheaf is and I struggle to give meaning to $\mathcal{O}(1)$ too. Dec 9, 2018 at 17:46

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 $$x\mapsto \frac{ax+b}{cx+d}$$.

• @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 $\Bbb{P}^1$.
– jgon
Dec 9, 2018 at 18:08
• 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.
– jgon
Dec 9, 2018 at 18:09
• Lastly the divisor of $f^\vee$ would be $Z-P=(-b:a)-(-d:c)$
– jgon
Dec 9, 2018 at 18:10
• @Dalamar yes, that's right
– jgon
Dec 9, 2018 at 18:10
• @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.
– jgon
Dec 9, 2018 at 18:15