Let $C \subset \mathbb{P}^2$ be a smooth plane curve of degree $d$, and $\pi: C \to \mathbb{P}^1$ the projection from a point $Q \notin C$. A generic line through $Q$ meets $C$ in $d$ points, so $\deg \pi = d$.

Concretely, for $C: y^2 = x^3 + 1$ (a genus-1 curve of degree 3 in its Weierstrass embedding), the projection $\pi(x,y) = x$ has $\deg \pi = 2$: the generic fiber consists of the two points $(x_0, \pm\sqrt{x_0^3 + 1})$.

The map $f: \mathbb{P}^1 \to \mathbb{P}^1$ given by $f([x:y]) = [x^n : y^n]$ (or $t \mapsto t^n$ in affine coordinates) has degree $n$. The function field extension is $k(t) \hookrightarrow k(s)$ where $s^n = t$, which has degree $n$.

The generic fiber over $t_0 \neq 0, \infty$ consists of $n$ distinct points (the $n$-th roots of $t_0$). Over $t_0 = 0$ and $t_0 = \infty$, the fiber has a single point with multiplicity $n$.

For $f: \mathbb{P}^1 \to \mathbb{P}^1$, $t \mapsto t^2$ (degree 2):

• Over $Q = a^2 \neq 0, \infty$: the fiber is $\{a, -a\}$, two distinct points with $e_a = e_{-a} = 1$. Unramified.
• Over $Q = 0$: the fiber is $\{0\}$ with $e_0 = 2$. Ramified! The uniformizer at $Q = 0$ is $t$, and $f^*t = s^2$ where $s$ is the coordinate on the source. So $v_0(s^2) = 2$.
• Over $Q = \infty$: similarly $e_\infty = 2$.

The branch locus is $\{0, \infty\} \subset \mathbb{P}^1$. The sum checks: $e_a + e_{-a} = 1 + 1 = 2 = \deg f$ (generic fiber), and $e_0 = 2 = \deg f$ (ramified fiber).

For $f: \mathbb{P}^1 \to \mathbb{P}^1$, $t \mapsto t^3$ (degree 3):

• Over $Q \neq 0, \infty$: three distinct preimages, each with $e = 1$.
• Over $Q = 0$: single preimage $P = 0$ with $e_0 = 3$. Totally ramified.
• Over $Q = \infty$: single preimage $P = \infty$ with $e_\infty = 3$. Totally ramified.

Branch locus: $\{0, \infty\}$. The ramification divisor is $R = 2 \cdot [0] + 2 \cdot [\infty]$ with $\deg R = 4$.

The degree-$n$ Chebyshev polynomial $T_n: \mathbb{P}^1 \to \mathbb{P}^1$ satisfies $T_n(\cos\theta) = \cos(n\theta)$. As a map of degree $n$:

• Over $Q = \pm 1$: the preimages of $1$ are $\cos(2\pi k/n)$ for $k = 0, \ldots, n-1$, but some coincide. Specifically, $T_n$ has ramification index $2$ at each of $\cos(\pi k/n)$ for $k = 1, \ldots, n-1$ (the interior critical points).
• Over generic $Q$: $n$ distinct preimages.

For $T_2(t) = 2t^2 - 1$: branch points at $\pm 1$. Over $Q = 1$: preimages at $t = \pm 1$ with $e = 1$ each (since $T_2(\pm 1) = 1$ and $T_2'(\pm 1) = \pm 4 \neq 0$). Over $Q = -1$: single preimage $t = 0$ with $e = 2$ (since $T_2(0) = -1$ and $T_2'(0) = 0$). So the branch locus is $\{-1\}$.

Let $k$ have characteristic $p > 0$. The Artin--Schreier cover $f: C \to \mathbb{P}^1$ is defined by $y^p - y = x^{-m}$ for $\gcd(m, p) = 1$.

This is a degree-$p$ cyclic cover (Galois group $\mathbb{Z}/p$), totally ramified above $x = \infty$ with ramification index $e = p$. The different exponent at the unique point above $\infty$ is $d = (p-1)(m+1)$, which is strictly greater than $e - 1 = p - 1$ when $m \geq 2$.

For $p = 2$, $m = 1$: the curve $y^2 - y = 1/x$ over $\mathbb{F}_2$. Here $e = 2$, $d = 1 \cdot 2 = 2 > 1 = e - 1$. Wild!

For $p = 3$, $m = 2$: the curve $y^3 - y = x^{-2}$ over $\mathbb{F}_3$. Here $e = 3$, $d = 2 \cdot 3 = 6 > 2 = e - 1$. The genus of $C$ is larger than it would be with tame ramification.

Let $f: C \to \mathbb{P}^1$ be a degree-2 cover with $g_D = 0$. The Hurwitz formula gives:

$2g_C - 2 = 2(0 - 2) + \deg R = -4 + \deg R.$

Each ramification point has $e_P = 2$ (totally ramified), so contributes $e_P - 1 = 1$ to $R$. If there are $r$ ramification points, $\deg R = r$, and:

$g_C = \frac{r - 2}{2}.$

So $r$ must be even, and:

• $r = 2$: $g_C = 0$ (an isomorphism after base change).
• $r = 4$: $g_C = 1$ (elliptic curve).
• $r = 6$: $g_C = 2$ (genus 2, all genus-2 curves are hyperelliptic).
• $r = 2g + 2$: $g_C = g$ (general hyperelliptic curve).

For a hyperelliptic curve $C: y^2 = f(x)$ where $f$ has degree $2g + 2$ with distinct roots $\alpha_1, \ldots, \alpha_{2g+2}$, the map $\pi: C \to \mathbb{P}^1$, $(x,y) \mapsto x$ has degree 2.

The ramification points are exactly the Weierstrass points: the $2g + 2$ points $(\alpha_i, 0)$ where $f(\alpha_i) = 0$. At each of these, the fiber is a single point with $e = 2$.

When $\deg f = 2g + 1$ instead (one root "at infinity"), the point at infinity is also a ramification point, giving $r = 2g + 2$ total. The Hurwitz formula still gives $g_C = g$.

For $g = 2$, $C: y^2 = x^5 - 1$: the 6 ramification points are $(e^{2\pi i k/5}, 0)$ for $k = 0,\ldots,4$ plus the point at infinity. The 6 Weierstrass points are the fixed points of the hyperelliptic involution $(x,y) \mapsto (x,-y)$.

Let $f: E_1 \to E_2$ be an isogeny of elliptic curves (both genus 1). The Hurwitz formula gives:

$0 = \deg(f) \cdot 0 + \deg R$

so $\deg R = 0$, meaning $f$ is unramified everywhere. This confirms the general fact: a homomorphism of abelian varieties is etale (unramified) if and only if it is an isogeny, which is automatic for isogenies between elliptic curves in characteristic 0 (or when the degree is coprime to the characteristic).

A cyclic cover of degree $n$ is $f: C \to \mathbb{P}^1$ given by $y^n = \prod_{i=1}^r (x - \alpha_i)^{a_i}$ where $0 < a_i < n$ and $\gcd(a_i, n) = 1$ ensures the cover is totally ramified above $\alpha_i$.

At each ramification point above $\alpha_i$, the ramification index is $e = n$ and (in the tame case, $\operatorname{char} k \nmid n$) the contribution to $R$ is $n - 1$.

By Hurwitz: $2g_C - 2 = n(2 \cdot 0 - 2) + r(n-1) = -2n + r(n-1)$, giving:

$g_C = \frac{(n-1)(r-2)}{2} \quad \text{(if all } a_i = 1 \text{)}.$

For $n = 3$, $r = 4$ (four branch points): $g_C = \frac{2 \cdot 2}{2} = 2$.

For $n = 4$, $r = 3$: $g_C = \frac{3 \cdot 1}{2}$, which is not an integer! This means the ramification pattern with all $a_i = 1$ and $r = 3$ is not compatible with $n = 4$ unless we include the point at infinity, making $r = 4$ and $g_C = 3$.

The Fermat curve $F_n: x^n + y^n = z^n$ in $\mathbb{P}^2$ is smooth of degree $n$, so $g = \frac{(n-1)(n-2)}{2}$.

The projection $\pi: F_n \to \mathbb{P}^1$, $[x:y:z] \mapsto [x:z]$ (i.e., $t = x/z$) has degree $n$. Setting $s = y/z$, we have $s^n = 1 - t^n$.

The ramification occurs where $\frac{ds}{dt}$ is undefined, i.e., where $ns^{n-1} \frac{ds}{dt} = -nt^{n-1}$, so $\frac{ds}{dt} = -t^{n-1}/s^{n-1}$, which blows up when $s = 0$. This happens at $t = \zeta$ where $\zeta^n = 1$ (the $n$-th roots of unity).

Over each branch point $t = \zeta$, the fiber is a single point $(x,y,z) = [\zeta : 0 : 1]$ with $e = n$. There are $n$ such branch points.

Hurwitz check: $2g - 2 = n \cdot (-2) + n(n-1)$ gives $2g - 2 = n^2 - 3n = (n-1)(n-2) - 2$, which is $g = \frac{(n-1)(n-2)}{2}$, matching the genus formula for plane curves.

For a degree-2 cover $f: C \to \mathbb{P}^1$ branched at $\{p_1, \ldots, p_{2g+2}\}$:

The monodromy group is $S_2 = \mathbb{Z}/2$. Each branch point contributes a transposition $(1\,2)$. The monodromy representation is:

$\rho: \pi_1(\mathbb{P}^1 \setminus \{p_1, \ldots, p_{2g+2}\}) \to \mathbb{Z}/2, \quad \gamma_{p_i} \mapsto (1\,2).$

The fundamental group relation $\gamma_{p_1} \cdots \gamma_{p_{2g+2}} = 1$ requires $(1\,2)^{2g+2} = \mathrm{id}$, which holds since $2g+2$ is even.

The cover is always Galois (since $S_2$ is abelian), with Galois group $\mathbb{Z}/2$. The nontrivial automorphism is the hyperelliptic involution $(x,y) \mapsto (x,-y)$.

Consider $f: C \to \mathbb{P}^1$ given by $y^3 = x(x-1)$ (a degree-3 cyclic cover). The branch points are $\{0, 1, \infty\}$, and the monodromy representation maps into $S_3$:

$\rho(\gamma_0) = (1\,2\,3), \quad \rho(\gamma_1) = (1\,2\,3), \quad \rho(\gamma_\infty) = (1\,2\,3)^{-1} \cdot (1\,2\,3)^{-1} = (1\,2\,3).$

Wait -- the relation is $\gamma_0 \gamma_1 \gamma_\infty = 1$, so $\rho(\gamma_\infty) = (1\,2\,3)^{-2} = (1\,2\,3)$. The image is $\mathbb{Z}/3 \subset S_3$, confirming the cover is Galois with group $\mathbb{Z}/3$.

Hurwitz: $2g - 2 = 3(-2) + 3 \cdot 2 = 0$, so $g = 1$. The curve $C$ is an elliptic curve.

A Belyi map is a finite morphism $f: C \to \mathbb{P}^1$ unramified outside $\{0, 1, \infty\}$.

Belyi's theorem: A smooth projective curve $C$ over $\mathbb{C}$ admits a Belyi map if and only if $C$ is defined over $\overline{\mathbb{Q}}$.

Example 1: The identity $\mathrm{id}: \mathbb{P}^1 \to \mathbb{P}^1$ is the simplest Belyi map (degree 1, no ramification).

Example 2: $f(t) = 4t(1-t)$ is a degree-2 Belyi map. Branch points: $f'(t) = 4 - 8t = 0$ at $t = 1/2$, so $f(1/2) = 1$. The map is ramified at $t = 1/2$ (above $1$) and at $t = \infty$ (above $\infty$), with $e = 2$ at each. Unramified outside $\{0, 1, \infty\}$ since the only critical value is $1$ (and $\infty$).

Example 3: The Belyi map $f(t) = \frac{(t^2 - t + 1)^3}{t^2(t-1)^2}$ gives a degree-6 cover of $\mathbb{P}^1$ by $\mathbb{P}^1$ (genus 0). The ramification type over $0$ is $[2, 2, 2]$, over $1$ is $[3, 3]$, and over $\infty$ is $[2, 2, 2]$. This corresponds to the dessin of a tetrahedron.

The elliptic curve $E: y^2 = x(x-1)(x-\lambda)$ has a natural degree-2 map $\pi: E \to \mathbb{P}^1$ via $(x,y) \mapsto x$, branched at $\{0, 1, \lambda, \infty\}$.

To make a Belyi map, we compose with a rational function $g: \mathbb{P}^1 \to \mathbb{P}^1$ that sends $\{0, 1, \lambda, \infty\}$ into $\{0, 1, \infty\}$. For instance, if $\lambda = -1$, we can use $g(x) = x^2$, which sends $\{0, 1, -1, \infty\} \to \{0, 1, 1, \infty\}$.

The composition $g \circ \pi: E \to \mathbb{P}^1$ is then a degree-4 Belyi map on the curve $y^2 = x(x-1)(x+1) = x^3 - x$.

The extension $\mathbb{Z}[i]/\mathbb{Z}$ (i.e., $\mathbb{Q}(i)/\mathbb{Q}$) is a "degree-2 cover" of $\operatorname{Spec} \mathbb{Z}$:

• $p = 2$: $(1+i)^2 = 2i$, so $2 = -i(1+i)^2$. Ramified ($e = 2$). The prime $2$ is the unique branch point.
• $p \equiv 1 \pmod{4}$: $p = \mathfrak{p} \bar{\mathfrak{p}}$ splits into two primes. Split (unramified, two preimages).
• $p \equiv 3 \pmod{4}$: $p$ remains prime in $\mathbb{Z}[i]$. Inert (unramified, one preimage of degree 2).

This mirrors a double cover of curves: branch points (ramified primes), unramified points splitting into 2 preimages, and (unlike the geometric case) inert primes where the residue field doubles.

Let $C: y^2 = (x - 1)(x - 2)(x - 3)(x - 4)(x - 5)(x - 6)(x - 7)(x - 8)$ over $\mathbb{C}$.

The projection $\pi(x,y) = x$ is a degree-2 map to $\mathbb{P}^1$ with $r = 8$ branch points (the roots of the polynomial, all with $e = 2$). By Hurwitz: $g = (8 - 2)/2 = 3$.

The ramification divisor is $R = \sum_{i=1}^{8} P_i$ where $P_i = (i, 0)$, with $\deg R = 8$.

The canonical divisor is $K_C = \pi^* K_{\mathbb{P}^1} + R \sim \pi^*(-2[\infty]) + R$. Since $\deg K_C = 2g - 2 = 4$ and $\deg R = 8$, $\deg \pi^* K_{\mathbb{P}^1} = 2 \cdot (-2) = -4$, and indeed $-4 + 8 = 4$.

The Klein quartic $C: x^3 y + y^3 z + z^3 x = 0$ in $\mathbb{P}^2$ is a smooth plane curve of degree 4, so $g = 3$.

It has the remarkable automorphism group $\operatorname{Aut}(C) \cong \operatorname{PSL}_2(\mathbb{F}_7)$ of order 168, achieving the Hurwitz bound $|\operatorname{Aut}(C)| \leq 84(g-1) = 168$.

The quotient map $f: C \to C/\operatorname{Aut}(C) \cong \mathbb{P}^1$ has degree 168. By Hurwitz:

$2(3) - 2 = 168(0 - 2) + \deg R \implies \deg R = 4 + 336 = 340.$

The ramification has three orbits of branch points in $\mathbb{P}^1$:

• Above one branch point: 24 points with $e = 7$, contributing $24 \cdot 6 = 144$.
• Above another: 42 points with $e = 4$, contributing $42 \cdot 3 = 126$.
• Above the third: 56 points with $e = 3$, contributing $56 \cdot 2 = 112$. -- but wait, let us recheck: we need $24 \cdot 7 = 168$, $42 \cdot 4 = 168$, $56 \cdot 3 = 168$. Contribution to $\deg R$: $24 \cdot 6 + 42 \cdot 3 + 56 \cdot 2 = 144 + 126 + 112 = 382$. But we need $\deg R = 340$.

The correct ramification data is: above three branch points, with orbit sizes $24, 84, 56$ and ramification indices $7, 2, 3$:

• 24 points with $e = 7$: contribution $24 \cdot 6 = 144$.
• 84 points with $e = 2$: contribution $84 \cdot 1 = 84$.
• 56 points with $e = 3$: contribution $56 \cdot 2 = 112$.

Total: $144 + 84 + 112 = 340$. Check: $24 \cdot 7 = 168$, $84 \cdot 2 = 168$, $56 \cdot 3 = 168$. Correct!

The curve $C: y^3 = x(x-1)$ is a degree-3 cyclic cover of $\mathbb{P}^1$ via $(x,y) \mapsto x$.

Branch points: $x = 0$ and $x = 1$ (where $y^3 = 0$, a single root with $e = 3$), and $x = \infty$ (where the equation becomes $y^3 = x^2$ in suitable coordinates, again $e = 3$).

So $r = 3$ branch points, each totally ramified with $e = 3$. By Hurwitz:

$2g - 2 = 3(-2) + 3(3-1) = -6 + 6 = 0.$

Thus $g = 1$: this is an elliptic curve, not genus 2!

To get genus 2, we need more ramification: $C: y^3 = x(x-1)(x-2)(x-3)$. Here the branch points are $\{0, 1, 2, 3\}$, but each root appears with exponent 1, and since $\gcd(1,3) = 1$ each is totally ramified. The exponent sum is $4 \equiv 1 \pmod{3}$, so $\infty$ is also a branch point (since the total degree $4 \not\equiv 0 \pmod{3}$), but there we need a separate analysis. Ultimately this gives a curve of genus $g = \frac{(3-1)(5-2)}{2} = 3$ (with 5 branch points).

Let $C$ be a smooth plane quartic ($g = 3$). Projecting from a point $P \in C$ gives $\pi_P: C \to \mathbb{P}^1$ of degree 3 (a line through $P$ meets $C$ in $4$ points, one of which is $P$, leaving $3$ others).

By Hurwitz: $2(3) - 2 = 3(-2) + \deg R$ gives $\deg R = 10$. So there are $10$ ramification points (each with $e = 2$, generically), corresponding to the lines through $P$ that are tangent to $C$ at some other point.

Geometrically: $10 = 4 \cdot 3 - 2 =$ the number of tangent lines from $P$ to $C$ other than the tangent at $P$ itself (by the class formula for plane curves).