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,



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> 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 $uf+vg=1$?
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Bezout’s identity for analytic functions of several variables
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](https://en.wikipedia.org/wiki/Cousin_problems) 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.


