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

This is a degree-$p$ cover, totally ramified at $\infty$ with a single point $P_\infty$ above $\infty$. The ramification index is $e_{P_\infty} = p$.

The different at $P_\infty$ is $d_{P_\infty} = (p-1)(m+1)$, which exceeds $e_{P_\infty} - 1 = p - 1$ by $(p-1)m$.

By the generalized Hurwitz formula:

$2g_C - 2 = p(2 \cdot 0 - 2) + (p-1)(m+1) = -2p + (p-1)(m+1).$

So $g_C = \frac{(p-1)(m-1)}{2}$. For example, with $p = 2$ and $m = 3$: $g_C = 1$ (an elliptic curve).

Let $f: \mathbb{P}^1 \to \mathbb{P}^1$ be $f(t) = t^n$ (a degree-$n$ map). Here $g_C = g_D = 0$.

Ramification: The map ramifies at $t = 0$ (with $e_0 = n$, since $t \mapsto t^n$) and at $t = \infty$ (with $e_\infty = n$). All other points are unramified.

$\deg R = (n - 1) + (n - 1) = 2(n - 1).$

Hurwitz check: $2(0) - 2 = n(2(0) - 2) + 2(n - 1)$, i.e., $-2 = -2n + 2n - 2 = -2$. Verified.

Let $C: y^2 = f(x)$ where $f$ has degree $2g + 1$ or $2g + 2$ with distinct roots, defining a double cover $\pi: C \to \mathbb{P}^1$ via $(x, y) \mapsto x$.

Case $\deg f = 2g + 1$: The branch points are the $2g + 1$ roots of $f$ in $\mathbb{A}^1$ plus $\infty$ (since the cover is totally ramified at infinity when $\deg f$ is odd). Total branch points: $2g + 2$. Each has $e = 2$, so $\deg R = (2g + 2) \cdot 1 = 2g + 2$.

Case $\deg f = 2g + 2$: The $2g + 2$ roots of $f$ are the branch points (the point at infinity is unramified). Again $\deg R = 2g + 2$.

Hurwitz check: $2g_C - 2 = 2(2 \cdot 0 - 2) + 2g + 2 = -4 + 2g + 2 = 2g - 2$, so $g_C = g$. The genus matches the definition.

Let $C \subset \mathbb{P}^2$ be a smooth plane quartic ($g = 3$). Projection from a point $P \notin C$ gives a map $\pi_P: C \to \mathbb{P}^1$ of degree $4$.

Hurwitz formula: $2(3) - 2 = 4(2(0) - 2) + \deg R$, so $4 = -8 + \deg R$, giving $\deg R = 12$.

For a general projection, the map has $12$ simple ramification points (each with $e = 2$), consistent with $\deg R = 12$. These correspond to the tangent lines from $P$ to $C$, of which there are generically $12$ by the class formula for a degree-$4$ plane curve.

An elliptic curve $E: y^2 = x^3 + Ax + B$ over a field with $\operatorname{char} \neq 2$ admits a degree-$2$ map $E \to \mathbb{P}^1$ via $(x, y) \mapsto x$.

The branch points are the three roots of $x^3 + Ax + B$ plus $\infty$, giving $4$ branch points, each with $e = 2$.

$\deg R = 4 \cdot (2 - 1) = 4.$

Hurwitz check: $2(1) - 2 = 2(2(0) - 2) + 4$, i.e., $0 = -4 + 4 = 0$. Verified.

Consider $f: C \to \mathbb{P}^1$ defined by $y^3 = (x - a_1)(x - a_2) \cdots (x - a_r)$ where the $a_i$ are distinct and $3 \nmid r$ (so the cover is totally ramified at $\infty$ as well).

Each root $a_i$ gives a totally ramified point with $e = 3$, contributing $e - 1 = 2$ to $\deg R$. If $3 \nmid r$, then $\infty$ is also totally ramified with $e = 3$, contributing $2$ more.

$\deg R = 2r + 2 \quad (3 \nmid r), \qquad \deg R = 2r \quad (3 \mid r).$

Hurwitz: $2g_C - 2 = 3(-2) + \deg R$.

• If $r = 4$ (so $3 \nmid 4$): $\deg R = 10$, $2g_C - 2 = -6 + 10 = 4$, so $g_C = 3$.
• If $r = 3$: $\deg R = 6$, $2g_C - 2 = -6 + 6 = 0$, so $g_C = 1$ (an elliptic curve).
• If $r = 6$: $\deg R = 12$, $2g_C - 2 = -6 + 12 = 6$, so $g_C = 4$.

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

The projection $\pi: C \to \mathbb{P}^1$ via $[x:y:z] \mapsto [x:z]$ has degree $n$ (for a fixed ratio $x/z = \lambda$, we solve $y^n = z^n - x^n$).

The map $\pi$ ramifies at the $n$-th roots of unity $\lambda^n = 1$ (where $y^n = 0$, giving a single point with $e = n$) and at $\infty$ (where $x^n + y^n = 0$, giving $n$ points with $e = 1$ if $k$ contains $n$-th roots of $-1$). Let us verify more carefully: over $\lambda$ with $\lambda^n = 1$, the equation $y^n = z^n(1 - \lambda^n) = 0$ gives $y = 0$, a single preimage with $e = n$. There are $n$ such branch points.

$\deg R = n \cdot (n - 1) = n(n-1).$

Hurwitz check: $2 \cdot \frac{(n-1)(n-2)}{2} - 2 = n(-2) + n(n-1)$, i.e., $(n-1)(n-2) - 2 = -2n + n^2 - n = n^2 - 3n$. The left side is $n^2 - 3n + 2 - 2 = n^2 - 3n$. Verified.

Let $k = \overline{\mathbb{F}_p}$ and $C$ a smooth curve of genus $g$. The absolute Frobenius $F: C \to C$ sends $(x, y) \mapsto (x^p, y^p)$. This is a purely inseparable morphism of degree $p$.

The Hurwitz formula does not directly apply in the naive form since $F$ is inseparable (not separable). However, we can factor the $p$-power map $\mathbb{P}^1 \to \mathbb{P}^1$ via $t \mapsto t^p$ as a purely inseparable map. In this case, every point has ramification index $p$ but the differential $d(t^p) = p \cdot t^{p-1} dt = 0$, so the formula via differentials gives $f^*\omega = 0$ and we need the generalized framework.

For the relative Frobenius $F_{C/k}: C \to C^{(p)}$, the different is $\mathfrak{D} = (p-1)K_C + (\text{contribution from the inseparable structure})$. This is why separability is assumed in the standard Hurwitz formula.

Let $\phi: E_1 \to E_2$ be a separable isogeny of degree $n$ between elliptic curves ($g_{E_1} = g_{E_2} = 1$).

Hurwitz formula: $2(1) - 2 = n(2(1) - 2) + \deg R$, giving $0 = 0 + \deg R$, so $\deg R = 0$.

This means $\phi$ is unramified everywhere. Every separable isogeny between elliptic curves is an unramified (etale) cover. This is consistent with the group-theoretic viewpoint: an isogeny is a group homomorphism, and its fibers are cosets of $\ker(\phi)$, which are all translates of each other and hence uniformly distributed.

Any genus-$2$ curve $C$ admits a degree-$2$ map $f: C \to \mathbb{P}^1$ (every genus-$2$ curve is hyperelliptic).

Hurwitz: $2(2) - 2 = 2(2(0) - 2) + \deg R$, so $2 = -4 + \deg R$, giving $\deg R = 6$.

Since $n = 2$, each ramification point has $e = 2$, contributing $1$ to $\deg R$. So there are exactly $6$ branch points in $\mathbb{P}^1$. These are precisely the Weierstrass points of $C$ (their images under $f$), and their preimages are the $6$ Weierstrass points of $C$.

A Belyi map is a finite morphism $f: C \to \mathbb{P}^1$ ramified over at most three points (conventionally $0, 1, \infty$). By Belyi's theorem, a smooth curve over $\overline{\mathbb{Q}}$ admits a Belyi map if and only if it is defined over $\overline{\mathbb{Q}}$.

For a Belyi map of degree $n$ on a curve of genus $g$, the Hurwitz formula reads:

$2g - 2 = -2n + \deg R.$

Concrete example: Consider the "dessin d'enfant" given by a degree-$4$ Belyi map $f: C \to \mathbb{P}^1$ with ramification type $(4)$ over $0$, $(2, 2)$ over $1$, and $(3, 1)$ over $\infty$. Then:

$\deg R = (4-1) + (2-1) + (2-1) + (3-1) + (1-1) = 3 + 1 + 1 + 2 + 0 = 7.$

So $2g - 2 = -8 + 7 = -1$, giving $g = 1/2$. This is impossible! Therefore, no such Belyi map exists. The ramification data must satisfy $\deg R \equiv 0 \pmod{2}$ (since $2g - 2 + 2n$ is always even).

Valid example: Degree $3$, type $(3)$ over $0$, $(2, 1)$ over $1$, $(3)$ over $\infty$: $\deg R = 2 + 1 + 0 + 2 = 5$, $2g - 2 = -6 + 5 = -1$... still fractional. Let us try type $(3)$ over $0$, $(2, 1)$ over $1$, $(2, 1)$ over $\infty$: $\deg R = 2 + 1 + 1 = 4$, $2g - 2 = -6 + 4 = -2$, $g = 0$. This is valid, and describes a rational Belyi function.

Let $E$ be an elliptic curve and $f: C \to E$ a degree-$2$ cover.

Hurwitz: $2g_C - 2 = 2(2 \cdot 1 - 2) + \deg R = 0 + \deg R$, so $g_C = 1 + \frac{\deg R}{2}$.

Since each ramification point for a double cover has $e = 2$ (contributing $1$), the number of branch points equals $\deg R$. This must be even (by the Hurwitz formula parity constraint), so:

• $\deg R = 0$: $g_C = 1$. Unramified double cover; $C$ is again an elliptic curve.
• $\deg R = 2$: $g_C = 2$. A genus-$2$ curve double-covering $E$ with $2$ branch points.
• $\deg R = 4$: $g_C = 3$. A genus-$3$ curve double-covering $E$ with $4$ branch points.
• $\deg R = 2k$: $g_C = 1 + k$.

If $f: C \to D$ is unramified (etale), then $\deg R = 0$ and the Hurwitz formula gives:

$2g_C - 2 = n(2g_D - 2),$

i.e., $g_C - 1 = n(g_D - 1)$.

Consequences:

• If $g_D = 0$: then $g_C - 1 = -n < 0$, impossible for $n > 1$. So $\mathbb{P}^1$ has no nontrivial etale covers (it is simply connected).
• If $g_D = 1$: then $g_C = 1$ for any $n$. Etale covers of elliptic curves are elliptic curves (the $n$-isogenies).
• If $g_D \geq 2$: then $g_C = n(g_D - 1) + 1 \geq n + 1$. A degree-$2$ unramified cover of a genus-$2$ curve has genus $3$.

The Klein quartic $C: x^3 y + y^3 z + z^3 x = 0$ in $\mathbb{P}^2$ is a smooth curve of genus $3$.

Its automorphism group is $G = \operatorname{PSL}(2, \mathbb{F}_7)$ of order $168 = 84(3 - 1) = 84 \cdot 2$.

The quotient $C/G \cong \mathbb{P}^1$, and the quotient map has three branch points with ramification types $(2, 3, 7)$, achieving the minimum $\mu = 1/42$. This is the unique curve of genus $3$ that achieves the Hurwitz bound.