Return to Answer

Make an open cover $D^2=\cup_j(U_j\cup V_j)$, for example, by polydisks such that $f$ has zeros only in $U_j$ and $g$ has no zeros in $U_j$. This is possible since zeros of $f$ and $g$ are disjoint.

Solve the 1st Cousin problem with Cousin data $-1/(fg)$ in $U_j$ and $0$ in $V_j$. The solution is a meromorphic function $\phi$ such that $\phi+1/(fg)$ is holomorphic in $U_j$ and $\phi$ is holomorphic in $V_j$. Let $b:=-f\phi$. Then $-b+1/g$ is holomorphic and divisible by $f$ in $U_j$, and thus $b$ is also holomorphic in $U_j$ since $1/g$ is holomorphic in $U_j$. So $b$ is holomorphic everywhere. Now $-b+1/g$ is divisible by $f$ also in $V_j$ since $f$ has no zeros in $V_j$. Then since $-bg+1$ is holomorphic and divisible by $f$, then $a:=(1-bg)/f$ is holomorphic and $af+bg=1$ as required.

In modern texts they refer to H. Cartan's theorem A and B, but the case of polydisk of dimension 2 this was in the original paper of Cousin.

Make an open cover $D^2=\cup_j(U_j\cup V_j)$, for example, by polydisks such that $f$ has zeros only in $U_j$ and $g$ has no zeros in $U_j$. This is possible since zeros of $f$ and $g$ are disjoint.

Solve the 1st Cousin problem with Cousin data $-1/(fg)$ in $U_j$ and $0$ in $V_j$. The solution is a meromorphic function $\phi$ such that $\phi+1/(fg)$ is holomorphic in $U_j$ and $\phi$ is holomorphic in $V_j$. Let $v:=-f\phi$. Then $-v+1/g$ is holomorphic and divisible by $f$ in $U_j$, and thus $v$ is also holomorphic in $U_j$ since $1/g$ is holomorphic in $U_j$. So $v$ is holomorphic everywhere. Now $-v+1/g$ is divisible by $f$ also in $V_j$ since $f$ has no zeros in $V_j$. Then since $-vg+1$ is holomorphic and divisible by $f$, then $u:=(1-vg)/f$ is holomorphic and $uf+vg=1$ as required.

In modern texts they refer to H. Cartan's theorems A and B, but the case of polydisk of dimension 2 this was in the original paper of Cousin.

Make an open cover $D^2=\cup_j(U_j\cup V_j)$, for example, by polydisks such that $f$ has zeros only in $U_j$ and $g$ has no zeros in $U_j$. This is possible since zeros of $f$ and $g$ are disjoint.

Solve the 1st Cousin problem with Cousin data $-1/(fg)$ in $U_j$ and $0$ in $V_j$. The solution is a meromorphic function $\phi$ such that $\phi+1/(fg)$ is holomorphic in $U_j$ and $\phi$ is holomorphic in $V_j$. Let $b:=-f\phi$. Then $-b+1/g$ is holomorphic and divisible by $f$ in $U_j$, and thus $b$ is also holomorphic in $U_J$ since $1/g$ is holomorphic in $U_j$. So $b$ is holomorphic everywhere. Now $-b+1/g$ is divisible by $f$ also in $V_j$ since $f$ has no zeros in $V_j$. Then since $-bg+1$ is holomorphic and divisible by $f$, then $a:=(1-bg)/f$ is holomorphic and $af+bg=1$ as required.

In modern texts they refer to H. Cartan's theorem A and B, but the case of polydisk of dimension 2 this was in the original paper of Cousin.

Make an open cover $D^2=\cup_j(U_j\cup V_j)$, for example, by polydisks such that $f$ has zeros only in $U_j$ and $g$ has no zeros in $U_j$. This is possible since zeros of $f$ and $g$ are disjoint.

Solve the 1st Cousin problem with Cousin data $-1/(fg)$ in $U_j$ and $0$ in $V_j$. The solution is a meromorphic function $\phi$ such that $\phi+1/(fg)$ is holomorphic in $U_j$ and $\phi$ is holomorphic in $V_j$. Let $b:=-f\phi$. Then $-b+1/g$ is holomorphic and divisible by $f$ in $U_j$, and thus $b$ is also holomorphic in $U_j$ since $1/g$ is holomorphic in $U_j$. So $b$ is holomorphic everywhere. Now $-b+1/g$ is divisible by $f$ also in $V_j$ since $f$ has no zeros in $V_j$. Then since $-bg+1$ is holomorphic and divisible by $f$, then $a:=(1-bg)/f$ is holomorphic and $af+bg=1$ as required.

In modern texts they refer to H. Cartan's theorem A and B, but the case of polydisk of dimension 2 this was in the original paper of Cousin.

Make an open cover $D^2=\cup_j(U_j\cup V_j)$, for example, by polydisks such that $f$ has zeros only in $U_j$ and $g$ has no zeros in $U_j$. This is possible since zeros of $f$ and $g$ are disjoint.

Solve the 1st Cousin problem with Cousin data $-1/(fg)$ in $U_j$ and $0$ in $V_j$. The solution is a meromorphic function $\phi$ such that $\phi+1/(fg)$ is holomorphic in $U_j$ and $\phi$ is holomorphic in $V_j$. Let $b=-f\phi$. Then $-b+1/g$ is holomorphic and divisible by $f$ in $U_j$ (holomorphic since $1/g$ is holomorphic in $U_j$). Now $b$ is holomorphic everywhere and $-b+1/g$ is divisible by $f$ in $V_j$ since $f$ has no zeros in $V_j$. Then since $-bg+1$ is holomorphic and divisible by $f$, then $a:=(1-bg)/f$ is holomorphic and $af+bg=1$ as required. is holomorphic, and we have

Make an open cover $D^2=\cup_j(U_j\cup V_j)$, for example, by polydisks such that $f$ has zeros only in $U_j$ and $g$ has no zeros in $U_j$. This is possible since zeros of $f$ and $g$ are disjoint.

Solve the 1st Cousin problem with Cousin data $-1/(fg)$ in $U_j$ and $0$ in $V_j$. The solution is a meromorphic function $\phi$ such that $\phi+1/(fg)$ is holomorphic in $U_j$ and $\phi$ is holomorphic in $V_j$. Let $b:=-f\phi$. Then $-b+1/g$ is holomorphic and divisible by $f$ in $U_j$, and thus $b$ is also holomorphic in $U_J$ since $1/g$ is holomorphic in $U_j$. So $b$ is holomorphic everywhere. Now $-b+1/g$ is divisible by $f$ also in $V_j$ since $f$ has no zeros in $V_j$. Then since $-bg+1$ is holomorphic and divisible by $f$, then $a:=(1-bg)/f$ is holomorphic and $af+bg=1$ as required.

In modern texts they refer to H. Cartan's theorem A and B, but the case of polydisk of dimension 2 this was in the original paper of Cousin.