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 $$uf+vg=1$$?

• This seems to be a SCV version of the corona theorem: I am not sure about the state of the art for this result, but by googling you'll get a fairly large number of examples. Jan 17 at 13:04
• Is it implied here that the gcd exists? Because in several variables this need not be the case with your definition. Jan 17 at 13:05
• @Wojowu: I had thought about that — whether there always exist a $\gcd$; thanks for pointing this out. I’d rephrase the problem on that assumption. Jan 17 at 13:11
• @Alexandre Eremenko: If the answer (to my question) is positive, could you write out a proof as answer or suggest a reference instead? Many thanks. Jan 17 at 15:06
• @Daniele Tampieri: The crucial requirement that makes Corona non-trivial is that the functions must be bounded. The answer to the question is positive. Jan 18 at 6:56

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.
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.