Bezout’s identity for analytic functions of several variables In (single-variable) complex analysis, given analytic functions $f$ and $g$ with no common zeros, one can find analytic functions $u$ and $v$ such that $uf+vg=1$. I’d like to know if the same holds in several variables; as a simple case, specifically,

Let $f,g\colon\mathbb{D}^2\to\mathbb{C}$ be analytic (in the bi-disc $\mathbb{D}^2$) with no common zeros. Does there exist analytic functions $u,v\colon \mathbb{D}^2\to\mathbb{C}$ such that $uf+vg=1$?

 A: 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.
A: For the record, let me translate Alexandre Eremenko's answer in modern terms (after all, this is why sheaf theory was invented). The hypothesis implies an exact sequence
$$0\rightarrow \mathscr{O}_{\mathbb{D}}\xrightarrow{\ (-v,u)\ } \mathscr{O}_{\mathbb{D}}^2\xrightarrow{\ (u,v)\ } \mathscr{O}_{\mathbb{D}}\rightarrow 0$$hence, using $H^1(\mathbb{D}^2,\mathscr{O})=0$ (Cartan's theorem B), the map $(u,v):H^0(\mathbb{D}^2,\mathscr{O})^2\rightarrow H^0(\mathbb{D}^2,\mathscr{O})$ is surjective.
