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Suppose $f_1,f_2,\cdots,f_k$ are entire functions without common zeros.Suppose each $f_i$ has finite number of zeros.Prove that there exist entire functions $g_1,\cdots,g_k$ such that $$\sum_{i=1}^kf_ig_i=1$$ Is it still true without the assumption that each $f_i$ has finite number of zeros?

This argument seems very strange to me and only approach I can think of is Weierstrass factorization theorem that I can show $f_i=P_ie^{h_i}$ where $P_i$ is a polynomial and $h_i$ is entire function.To each each $g_i$ I can impose term $e^{-h_i}$ to cancel the exponential term in $f_i$,then choose $g_i$ to be polynomial with coefficients undetermined and let every $f_ig_i$ has same degree,then the problem reduced to be a algebraic system of undetermined coefficients.

The problem about my approach is that it's not very "analysis",first it's not clear why we need the functions without common zeros.Secondly if we don't have condition of finite zeros,then the factorization will not produce a polynomial and such method will fail.

I suppose there's a beautiful complex analysis solution behind this and any help will be great.

Daniel S.
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  • You need the functions to be without common zeros because if $z_0$ is a common zero of $f_i$, then $\sum f_i g_i$ evaluated at $z_0$ will be 0, not 1. – Ted May 16 '14 at 05:09

2 Answers2

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Consider an arbitrary non-empty open domain $D\subset \mathbb C$ and the corresponding ring (without zero divisors) of holomorphic functions $\mathcal O(D)$.
That ring is never noetherian but is a Bézout: every finitely generated ideal is principal.
This is an easy consequence of the not easy Weierstrass theorem which allows you to construct holomorphic functions on $D$ vanishing with preassigned multiplicities on any closed discrete subset $F\subset D$.

That said, your problem is now easily solved: given $f_1,\cdots,f_k\in \mathcal O(D)$ there exists $g\in \mathcal O(D)$ generating the same ideal as the $f_i $'s : $\langle g\rangle=\langle f_1,\cdots,f_k \rangle$.
It the $f_1,\cdots,f_k$ have no common zero, then $g$ necessarily has no zero at all and is thus invertible in the ring $\mathcal O(D)$; in other words $\langle g\rangle=\langle 1\rangle \in \mathcal O(D)$.
Thus $\langle f_1,\cdots,f_k \rangle=\langle g\rangle=\langle 1\rangle$ and we can write $\sum_{i=1}^kf_ig_i=1$ for some $g_1,\cdots,g_k\in \mathcal O(D)$ .

Remarks
1) That the $f_i$'s have only finitely many zeros is irrelevant, and you needn't assume $D=\mathbb C$ either.
2) Actually the result is still true if $D$ is a non compact Riemann surface or even a Stein manifold of arbitrary dimension.
You then use Cartan-Serre's Theorem A instead of Weierstrass's theorem in one variable .
[Actually I learned the result for Stein manifolds in several complex variables before I heard of Weierstrass's theorem in one variable: mathematical education can be very ahistorical!]

Georges Elencwajg
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  • This is a beautiful algebraic solution and thank you very much! – Daniel S. May 18 '14 at 07:32
  • Dear Daniel, you are welcome. – Georges Elencwajg May 18 '14 at 07:38
  • But pardon me for not familiar with Algebra,the key argument here seems to be that finitely generated ideal is principal.Suppose $I=<f_1,\cdots,f_k>$, $R$ be the ring of holomorphic functions on $D$,then exists $g$ s.t. $I=RgR$,can you clarify why can Weierstrass Theorem support this argument? – Daniel S. May 18 '14 at 07:47
  • Dear Daniel, yes every finitely generated ideal is principal.The idea is that you take the union of the set of zeros of the $f_i$' and attach to each such zero the minimum of the order of the $f_i$'s at it. A Weierstrass function $g$ having a zero with that multiplicity will then generate the ideal. A detailed proof can be found in Theorem 5.20 of our friend Pete Clark's great online notes – Georges Elencwajg May 18 '14 at 08:26
  • I've tried your idea but it seems only prove the part $I\subset RgR$,my question mostly lies in the other half$RgR\subset I$. – Daniel S. May 18 '14 at 08:46
  • It is easy to see that $g$ is a GCD of the $f_i$'s. By induction it suffices to prove the case where there are only two $f_i$'s, and you have to prove that $g\in <f_1,f_2>$ .Dividing by $g$ you reduce to Pete Clark's Step 1 in the link above. By the way, you should write $Rg$, not $RgR$ in commutative rings. – Georges Elencwajg May 18 '14 at 09:58
  • You can also look up Remmert's Classical Topics in Complex Function Theory, Springer, GTM Volume 172 Chapter 6, §3.3, page 138. – Georges Elencwajg May 18 '14 at 10:24
  • Thanks!I looked it up and got it.You are very kind!God bless you. – Daniel S. May 18 '14 at 12:39
  • You are welcome, dear Daniel. – Georges Elencwajg May 18 '14 at 14:15
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We prove that there are $g_1$ and $g_2$ such that $f_1 g_1+f_2g_2=1$. By Mittag-Leffler, there is a meromorphic $M$ with same singularities of $1/f_2$ such that $h=M-1/f_1f_2$ is meromorphic only at the zero of $f_1$ since $f_1,f_2$ have no common zeros. Let $g_1=Mf_2$ and $g_2=-hf_1$ be entire, $$f_1g_1+f_2g_2=1$$

Let $f_1 ,\cdots,f_k$ be entire without common zeros, suppose there are entire $g_1 ,\cdots,g_k$ such that $\sum^k f_ig_i=1$.

From the case $n=2$, there is $G_1$ and $G_2$ such that $1\cdot G_1+f_{k+1}G_2=1$. From the induction assumption there are are entire $g_1 ,\cdots,g_k$ such that $\sum^k f_ig_i=1$. By choosing $q_1=g_1G_1,\cdots,q_k=g_kG_1$ and $q_{k+1}=G_2$ then $\sum^{k+1} f_iq_i=1$.

mnmn1993
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