## Polynomial on each variable

Problem: Suppose ${f\in \mathcal{O}(\mathbb{C}^2)}$. If ${f(z_1,z_2)}$ is a polynomial of ${z_1}$ for every fixed ${z_2}$ and a polynomial of ${z_2}$ for every fixed ${z_1}$. Then ${f}$ must be a polynomial.

Proof: Since ${f}$ is holomorphic on the ${\mathbb{C}^2}$, we have the power series expansion

$\displaystyle f(z_1,z_2)=\sum_{i,j\geq 0}a_{ij}z_1^iz_2^j=\sum_{i\geq 0}a_i(z_1)z_2^i$

where ${a_i(z_1)}$ is a holomorphic function on ${\mathbb{C}}$ for any ${i\geq 0}$. Suppose there exists a sequence ${i_1,i_2,\cdots}$ with ${\lim_{k\rightarrow \infty}i_k=\infty}$ satisfying ${a_{i_k}(z_1)\neq 0}$. Since the set of roots of ${a_{i_k}}$ are countable in ${\mathbb{C}}$, then all the union of these sets are also countable. Therefore, ${\exists\, w\in\mathbb{C}}$ such that ${a_{i_k}(w)\neq 0}$ for all ${k\geq 0}$. Then ${f(w,z_2)}$ can never be a polynomial. So

$\displaystyle f(z_1,z_2)=\sum_{i=0}^Ma_i(z_1)z_2^i=\sum_{i\geq 0}b_i(z_2)z_1^i$

where every ${b_i}$ is polynomial of ${z_2}$. Repeating the above proof we get this is also a finite sum, which means ${f}$ is a polynomial. $\Box$

Remark: Exercise from Shabat, 1.10. Idea comes from Hanlong Fang.