A question about factorization and entire function 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.
 A: 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!]
A: 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 $\sum^{k+1} f_iq_i=1$.
