izak

riemann sphere's field of fractions

To prove that the global meromorphic functions on the Riemann sphere $\mathbb{P}^1$ (the field of fractions for this variety) are the rational functions, we see: The Riemann sphere $\mathbb{P}^1$ is the complex projective line, which can be thought of as the complex plane $\mathbb{C}$ compactified by adding a point at infinity, denoted $\infty$. It is covered by two coordinate charts: $U_0 = \mathbb{P}^1 \setminus \{\infty\}$, with coordinate $z$ (the affine coordinate on $\mathbb{C}$), $U_1 = \mathbb{P}^1 \setminus \{0\}$, with coordinate $w = \frac{1}{z}$ (the coordinate near $\infty$). A global meromorphic function on $\mathbb{P}^1$ is a function that is holomorphic except for isolated poles on $\mathbb{P}^1$. A rational function is a ratio of two polynomials: $ f(z) = \frac{P(z)}{Q(z)}, $ where $P(z)$ and $Q(z)$ are polynomials in $z$. Such a function is meromorphic on $\mathbb{C}$ because it is holomorphic except at the zeros of $Q(z)$, where it has poles. To check its behavior at $\infty$, we use the coordinate $w = \frac{1}{z}$: $ f(z) = \frac{P(z)}{Q(z)} = \frac{P(1/w)}{Q(1/w)}. $ This expression is meromorphic at $w = 0$ (i.e., $z = \infty$) because $P(1/w)$ and $Q(1/w)$ are polynomials in $1/w$. Thus, rational functions are global meromorphic functions on $\mathbb{P}^1$. Now, let $f$ be a global meromorphic function on $\mathbb{P}^1$. We show that $f$ must be a rational function. On $U_0 = \mathbb{C}$, $f$ is meromorphic, so it can be written as: $ f(z) = \frac{P(z)}{Q(z)}, $ where $P(z)$ and $Q(z)$ are polynomials, and $Q(z) \neq 0$ except at the poles of $f$. Near $\infty$ (i.e., in $U_1$), $f$ must also be meromorphic. Using the coordinate $w = \frac{1}{z}$, we write: $ f(z) = f\left(\frac{1}{w}\right) = \frac{P(1/w)}{Q(1/w)}. $ For this to be meromorphic at $w = 0$, the degrees of $P(z)$ and $Q(z)$ must satisfy $\deg(P) \leq \deg(Q)$. Otherwise, $f(z)$ would have an essential singularity at $\infty$, which is not allowed for a meromorphic function. Thus, $f(z)$ is a ratio of polynomials with $\deg(P) \leq \deg(Q)$, meaning $f(z)$ is a rational function. Every global meromorphic function on $\mathbb{P}^1$ is a rational function, and every rational function is a global meromorphic function on $\mathbb{P}^1$. Therefore, the global meromorphic functions on $\mathbb{P}^1$ are exactly the rational functions. So, the global meromorphic functions on $ \mathbb{P}^1 $ are exactly the rational functions.