# meromorphic function on projective space must be rational

To define a rational function on complex projective space ${\mathbb{P}}^{n}$, we just take two homogeneous polynomials of the same degree $p$ and $q$ on ${\u2102}^{n+1},$ and we note that $p/q$ induces a meromorphic function on ${\mathbb{P}}^{n}.$ In fact, every meromorphic function on ${\mathbb{P}}^{n}$ is rational.

###### Theorem.

Let $f$ be a meromorphic function on ${\mathrm{P}}^{n}$. Then $f$ is rational.

###### Proof.

Note that the zero set^{} of $f$ and the pole set are analytic subvarieties of ${\mathbb{P}}^{n}$
and hence algebraic by Chow’s theorem. $f$ induces a meromorphic function $\stackrel{~}{f}$ on
${\u2102}^{n+1}\setminus \{0\}$. Let $p$ and $q$ be two homogeneous polynomials
such that $q=0$ are the poles and $p=0$ are the zeros of $\stackrel{~}{f}.$ We can assume we can take $p$ and $q$ such that if we multiply $\stackrel{~}{f}$
by $q/p$ we have a holomorphic function^{} outside the origin. Hence $(q/p)\stackrel{~}{f}$ extends through the origin
by Hartogs’ theorem. Further since $\stackrel{~}{f}$ was
constant on complex lines through the origin, it is not hard to see that $(q/p)\stackrel{~}{f}$ is homogeneous^{}
and hence a homogeneous polynomial, by the same argument as in the proof of Chow’s theorem.
Since it is not zero outside the origin, it can’t be zero at the origin, and hence $(q/p)\stackrel{~}{f}$
must be a constant, and the proof is finished.
∎

Title | meromorphic function on projective space must be rational |
---|---|

Canonical name | MeromorphicFunctionOnProjectiveSpaceMustBeRational |

Date of creation | 2013-03-22 17:52:13 |

Last modified on | 2013-03-22 17:52:13 |

Owner | jirka (4157) |

Last modified by | jirka (4157) |

Numerical id | 5 |

Author | jirka (4157) |

Entry type | Theorem |

Classification | msc 51N15 |

Classification | msc 32A20 |

Related topic | ChowsTheorem |