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I have a question about the following exercise from Hartshorne's book 'Algebraic geometry':

Let $X$ be a curve of genus $g$. Show that there is a finite morphism $f:X\rightarrow \mathbb P^1$ with degree $\leq g+1$.

My idea is the following: We choose $g+1$ points $P_i$ in $X$. This gives us by a previous exercise (4.1.2) a rational function $r=\frac g h$ with poles at the $P_i$ and nowhere else. Now we define the map on closed points to be $x \mapsto [h(x):g(x)]$. As this map is non-constant, it is finite.

The fibre of $f^{-1}([1:0] )$ contains exactly the $P_i$ and hence the degree of $f$ is smaller than g+1. What obstructs us from choosing less than g+1 points in the beginning?

Sincerely

slin0

slin0
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    I'm a bit confused about your degree calculation. Yes, there are at most $g+1$ points above $[1:0]$, but they could be ramification points, since in IV.1.2, your rational function had possibly higher order poles. – Takumi Murayama Mar 30 '16 at 16:02

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Let $P$ be a point on $X$. Consider the divisor $D = (g+1)[P]$ on $X$. Let's compute a lower bound for the dimension of $\mathrm{H}^0(X,D)$.

By Riemann-Roch, $$\dim \mathrm{H}^0(X,D) = (g+1)+ 1- g+ \dim \mathrm{H}^1(X,D) \geq 2 + \dim \mathrm{H}^1(X,D) \geq 2.$$ Thus, there exists a non-constant $f$ in $\mathrm{H}^0(X,D)$.

Any non-constant $f$ in $\mathrm{H}^0(X,D)$ gives a finite morphism $f:X\to \mathbb P^1$ of degree at most the degree of $D$. Thus, as $\deg(D) = g+1$, there is a finite morphism $X\to\mathbb P^1$ of degree at most $g+1$.

Ariyan Javanpeykar
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  • Nice answer. Can it actually be that $\deg f < \deg D$ ? – Heitor Fontana Apr 01 '16 at 09:03
  • @HeitorFontana Yes, you have inclusions of vector spaces $H^0(X,\mathcal O_X) \subset H^0(X,\mathcal O_X(P)) \subset \ldots \subset H^0(X,\mathcal O_X((g+1)P)$. It could very well be that $f$ lies in some proper subspace. The reason it works with "g+1" is because it is guaranteed in this case that the Riemann-Roch space is of dimension at least two. O the other hand, there are certainly better bounds for the gonality of a genus $g$ curve (by Brill-Noether theory). I think $g/2+1$ is one. – Ariyan Javanpeykar Apr 01 '16 at 11:04
  • Ah yes. I got confused because I was erroneously thinking, e.g. in the case $X=\Bbb{P}^n$, of $H^0(\mathcal{O}(d))$ as the vector space of homogeneous polynomials of degree $d$. With this identification the inclusions you write, which are formally correct, become wrong. But this, of course, is just due to this choice of isomorphism. – Heitor Fontana Apr 01 '16 at 13:19
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    Why is it true that the degree of $f$ is at most the degree of $D$? – ponchan Jan 04 '23 at 19:47
  • @AriyanJavanpeykar I'll repeat the question of ponchan, which is the single most important thing in the reasoning. Why is it true that the degree of $f$ is at most the degree of $D$ ? – raisinsec Jun 16 '23 at 09:07
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    @raisinsec We have $D=(g+1)P$ and $D + \mathrm{div}(f) \geq 0$. Write $\mathrm{div}(f)$ as $\mathrm{div}0(f) - \mathrm{div}{\infty}(f)$, where $\mathrm{div}0(f)$ is the divisor of zeroes of $f$ and $\mathrm{div}{\infty}(f)$ is the divisor of poles of $f$. Since $\mathrm{div}{\infty}(f)$ and $\mathrm{div}{0}(f)$ have disjoint support, the effectivity of $D + \mathrm{div}(f) $ entails that $\mathrm{div}{\infty}(f) \leq (g+1) P$, so that $\deg(f) = \deg \mathrm{div}{\infty}(f) \leq \deg( (g+1) P) = g+1.$ – Ariyan Javanpeykar Jun 16 '23 at 13:33