# Hurwitz genus formula

The following formula, due to Hurwitz, is extremely useful when trying to compute the genus of an algebraic curve. In this entry $K$ is a perfect field^{} (i.e. every algebraic extension^{} of $K$ is separable). Recall that a non-constant map of curves $\psi :{C}_{1}\to {C}_{2}$ over $K$ is separable if the extension of function fields $K({C}_{1})/{\psi}^{\ast}K({C}_{2})$ is a separable extension of fields.

###### Theorem (Hurwitz Genus Formula).

Let ${C}_{\mathrm{1}}$ and ${C}_{\mathrm{2}}$ be two smooth curves defined over $K$ of genus ${g}_{\mathrm{1}}$ and ${g}_{\mathrm{2}}$, respectively. Let $\psi \mathrm{:}{C}_{\mathrm{1}}\mathrm{\to}{C}_{\mathrm{2}}$ be a non-constant and separable map. Then

$$2{g}_{1}-2\ge (\mathrm{deg}\psi )(2{g}_{2}-2)+\sum _{P\in {C}_{1}}({e}_{\psi}(P)-1)$$ |

where ${e}_{\psi}\mathit{}\mathrm{(}P\mathrm{)}$ is the ramification index of $\psi $ at $P$. Moreover, there is equality if and only if either $\mathrm{char}\mathit{}\mathrm{(}K\mathrm{)}\mathrm{=}\mathrm{0}$ or $\mathrm{char}\mathit{}\mathrm{(}K\mathrm{)}\mathrm{=}p\mathrm{>}\mathrm{0}$ and $p$ does not divide ${e}_{\psi}\mathit{}\mathrm{(}P\mathrm{)}$ for all $P\mathrm{\in}{C}_{\mathrm{1}}$.

###### Example.

As an application of the Hurwitz genus formula, we show that an elliptic curve^{} $E:{y}^{2}=x(x-\alpha )(x-\beta )$ defined over a field $K$ of characteristic^{} $0$ has genus $1$. Notice that the fact that $E$ is an elliptic curve over $K$ implies that $0,\alpha $ and $\beta $ are distinct elements of $K$, otherwise $E$ would be a singular curve. We define a map:

$$\psi :E\to {\mathbb{P}}^{1},[x,y,z]\mapsto [x,z]$$ |

and notice that $[0,1,0]$, the “point at infinity” of $E$, maps to $[1,0]$, the point at infinity of ${\mathbb{P}}^{1}$. The degree of this map is $2$: generically every point in ${\mathbb{P}}^{1}$ has two preimages, namely $[x,y,z]$ and $[x,-y,z]$. Moreover, the genus of ${\mathbb{P}}^{1}$ is $0$ and the map $\psi $ is ramified exactly at $4$ points, namely ${P}_{1}=[0,0,1],{P}_{2}=[\alpha ,0,1],{P}_{3}=[\beta ,0,1]$ and the point at infinity. It is easily checked that the ramification index at each point is ${e}_{\psi}({P}_{i})=2$. Hence, the Hurwitz formula reads:

$$2{g}_{1}-2=2(2\cdot 0-2)+\sum _{i=1}^{4}({e}_{\psi}({P}_{i})-1)=-4+4=0.$$ |

We conclude that ${g}_{1}=1$, as claimed.

Title | Hurwitz genus formula |
---|---|

Canonical name | HurwitzGenusFormula |

Date of creation | 2013-03-22 15:57:15 |

Last modified on | 2013-03-22 15:57:15 |

Owner | alozano (2414) |

Last modified by | alozano (2414) |

Numerical id | 6 |

Author | alozano (2414) |

Entry type | Theorem |

Classification | msc 14H99 |

Related topic | RiemannRochTheorem |

Related topic | EllipticCurve |