meromorphic function on projective space must be rational

To define a rational function on complex projective space ${\mathbb{P}}^n$, we just taketwo homogeneous polynomials of the same degree $p$ and $q$ on ${ℂ}^{n+1},$ and wenote that $p/q$ induces a meromorphic function on ${\mathbb{P}}^n.$ In fact, everymeromorphic 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 $\tilde{f}$ on${ℂ}^{n+1}\setminus \{0\}$. Let $p$ and $q$ be two homogeneous polynomialssuch that $q=0$ are the poles and $p=0$ are the zeros of $\tilde{f}.$ We can assume we can take $p$ and $q$ such that if we multiply $\tilde{f}$by $q/p$ we have a holomorphic function outside the origin. Hence $(q/p)\tilde{f}$ extends through the originby Hartogs’ theorem. Further since $\tilde{f}$ wasconstant on complex lines through the origin, it is not hard to see that $(q/p)\tilde{f}$ is homogeneousand 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 $(q/p)\tilde{f}$must be a constant, and the proof is finished.∎