Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

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

2 Answers 2

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.

share|improve this answer
3  
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.

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.