@def maxtoclevel=3

Algebraic Curves Lecture 9:

The 9th lecture of Algebraic Curves by Karl Christ

Coverings

Coverings

Let $f:Y\to X$ be a map of curves, $p\in Y$ , $q = f(p)$ .

The ramification index at $p$ is $v_p(t)$ .

If $v_p(t) = 1$ we say that $f$ is unramified at $p$ . Otherwise $p$ is a ramification point of $f$ , in which case we say that $q$ is a branch point.

The ramification divisor of $f$ is

\begin{aligned} R = \sum_{p\in Y} (v_p(t) - 1)\cdot p. \end{aligned}

Implicit in this definition is the fact that $f$ has finitely many ramification points. Our first theorem is the Riemann-Hurwitz formula.

Theorem: In this situation in characteristic

0

\begin{aligned} K_Y = f^*K_X + R \end{aligned}

and in particular

\begin{aligned} 2g(Y) - 2 = d(2g(X) - 2) + \sum_{p\in Y}(v_p(t) - 1) \end{aligned}

where

d = \deg(t)

Remark: We say that $f$ is separable if the induced field extension $K[Y]/K[X]$ is separable. In positive characteristic

If $f$ is not separable then it is ramified along all of $Y$
Even if it is separable the formula might fail if the ramification is not tame (meaning the ramification index is divisible by the characteristic of the base field).

Proof. (Sketch of an analytic proof.) Let $\omega$ be a meromorphic 1-form on $X$ the target curve. Pulling it back by $f$ gives us a 1-form on $Y$ . Away from branch points, $f$ is a local isomorphism (here we have a drawing of a stack of analytic disks in $Y$ over an analytic disk around the point $q$ in $X$ ) and thus any zero or pole of $\omega$ is given $d$ zeros or $d$ poles on $Y$ .

At a ramification point $p$ , $f$ is given analytic locally be $z\mapsto z^{v_p(t)}$ . On the other hand, $\omega$ is locally given by $dt = dz^{v_p(t)} = (v_p(t))z^{v_p(t)-1}dz$ on $Y$ .

\square

We can see where this fails when the ramification is not tame; if the characteristic of the base field divides $v_p(t)$ then $\omega$ vanishes on $Y$ .

Cor: There is no nontrivial étale cover

Y\to \mathbb P^1

Proof. Recall that étale = flat + unramified. Any finite surjective map of nonsingular varieties is flat, but a cover

f:Y\to \mathbb P^1

will always fail to be unramified. Indeed, suppose

f

is étale. Riemann-Hurwitz says that

\begin{aligned} 2g(Y) - 2 = -2\cdot d + \deg(R) = -2d \end{aligned}

which implies

g(Y) = 0, d = 1

. The important part is that

\deg(f) = d=1

, hence

f

is an isomorphism on function fields. hence

f

is an isomorphism. The only étale map is therefore trivial.

\square

Cor: If

f:Y\to X

is a cover then

g(Y) \geq g(X)

Proof.

If $g(X) = 0$ then there is nothing to show. Otherwise, we get by Riemann-Hurwitz that

\begin{aligned} &2g(Y) - 2 = d\cdot (2g(X) - 2) + \deg(R) \\ &\implies g(Y) = d(g(X) - 1) + \frac{\deg(R)}{2} + 1 \\ &\implies g(Y) = g(X) + (d-1)(g(X) - 1) + \frac{\deg(R)}{2}. \end{aligned}

g(X) > 0

then

(d-1)(g(X) - 1)\geq 0

and

\frac{\deg(R)}{2}\geq 0

so the claim follows.

\square

Cor:

\deg(R)

is even.

Cor: If

g\geq 2

and

g(X) = g(Y) = g

then any cover

f:Y\to X

is an isomorphism.

Proof.

2g - 2 = d\cdot (2g - 2) + \deg(R)

which implies

d = 1

and hence

f

is an isomorphism.

\square

Example: Take

f:\mathbb P^1\xrightarrow{2:1} \mathbb P^1

. Then

\begin{aligned} -2 = 2\cdot (-2) + \deg(R) \implies \deg(R) = 2 \end{aligned}

and so

f

is ramified at two points with ramification index

2

[x:y] \mapsto [x^2:y^2]

Cor: (Hurwitz bound). Suppose

X

has genus

\geq 2

and finitely many automorphisms. Then

|\operatorname{Aut}(X)| \leq 84(g - 1)

Remark:

This bound is sharp in the sense that it is achieved for infinitely many $g$ , but not all of them. For example, up to $g = 6$ it is achieved only in $g = 3$ by the Klein quartic:

\begin{aligned} \{x^3y + y^3z + z^3 x = 0\} \subseteq \mathbb P^2 \end{aligned}

The upshot is that if you fix $g$ , there is a bound on the automorphisms of ALL curves of that genus. A priori, it might be the case that for fixed genus $g$ you could find curves of that genus whose automorphism group has arbitrarily large size.
The Hurwitz bound fails in positive characteristic.

Proof. You can check that $X/\operatorname{Aut}(X) = Y$ is again a curve. So you get a covering $X\to Y$ of degree $\operatorname{Aut}(X)|$ . Ramification points of this cover are the points with non-trivial stabilizers.

Denote by $r_{i,j}$ the ramification points lying over a branch point $p_i$ for $1\leq i\leq b$ . (A picture of a line representing the base curve $Y$ with a tick for point $p_i$ lying in $Y$ . Above $p_i$ is a loopy curved line representing $X$ so that each point $r_{i,j}$ is a ramification point. Each $r_{i,1}, r_{i,2}, r_{i,3}$ is a place where the fiber is "tangent to the curve $X$ ") Over $p_i$ all points in the fiber $f^{-1}(p_i)$ have the same ramification index $f_i$ . If $r_i = |f^{-1}(p_i)|$ then $f_i\cdot r_i = |\operatorname{Aut}(X)|$ . Then Riemann-Hurwitz says

\begin{aligned} 2g(X) - 2 &= |\operatorname{Aut}(X)| \cdot (2g(Y) - 2) + \sum_{i}r_i\cdot (f_i - 1) \\ &= |\operatorname{Aut}(X)| \cdot (2g(Y) - 2) + \sum_{r_{i,j}}(f_i - 1) \end{aligned}

\square