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Let $f_1,f_2,\ldots, f_n$ be $n$ entire functions, and they don't have any common zero as a whole (not in pairs), then can we assert that there exist $n$ entire functions $g_1,g_2,\ldots,g_n$,such that $F=f_1g_1+f_2g_2+\cdots+f_ng_n$ is zero free? We know that if $f_1,\ldots,f_n$ are known to be polynomials, the conclusion follows from Bezout equation and induction. But things become complicated when infinite products get involved.

(The original formulation of this problem is: Each finitely generated ideal in the ring of entire functions must be principal, which evidently can be reduced to the problem above.)

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I've added LaTeX formatting to your question, and made a few very minor edits. – Zev Chonoles Aug 21 '11 at 1:56

This was originally proved in

O. Helmer, "Divisibility properties of integral functions", Duke Math. J. 6(1940), 345-356.

Notice that the result holds for the ring of holomorphic functions over any open connected subset of $\mathbb C$. See the book "Classical topics in complex function theory" by R. Remmert (GTM 172) for details and history.

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The result also holds on Stein spaces, for example on non-compact Riemann surfaces. However, it may be of some interest to note that if you delete the origin from $\mathbb C^2$, you obtain a complex manifold $X$ on which the result no longer holds. Indeed, the coordinate functions $z,w$ have no common zero on $X$ but it is impossible to write $1=zf(z,w)+wg(z,w) \; [*]$ with $f,g$ holomorphic on $X$. If you had such an equation Hartogs would extend holomorphically $f$ and $g$ through zero and you would get a contradiction $1=0$ by evaluating $[*]$ at the origin. Of course, $X$ is not Stein. – Georges Elencwajg Aug 21 '11 at 10:55

A proof of the fact that the ring of holomorphic functions on a connected open subset of the complex plane is a Bezout domain can be found in $\S 5.3$ of my commutative algebra notes. The proof uses some standard theorems in complex analysis: Weierstrass Factorization and Mittag-Leffler. If I remember correctly, the discussion is taken from Rudin's Real and Complex Analysis.

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before reading the section §5.3 of your notes, I did not think that the ring of holomorphic functions could have so many interesting properties. I knew the local theory as treated in Gunning-Rossi, but not the global one. Thanks and congratulations for your notes! – Andrea Aug 21 '11 at 18:06
@Andrea: thanks for your comment; it is much appreciated. For those who are motivated by it to look at my notes, let me say though that the one other nontrivial fact about the ring $\operatorname{Hol}(U)$ that I prove is that it is a domain with fraction field the meromorphic functions on $U$. Aside from that there is some business about the "order function" of a meromorphic function (which is what an algebraic geometer would call the "divisor" of $f$), but that is somewhere between a proof technique and a point of view, rather than a result of independent interest (in my opinion...). – Pete L. Clark Aug 21 '11 at 20:44

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