# The ring of entire functions on $\mathbb{C}$ is a Bézout domain

Prove that the ring of entire functions on $\mathbb{C}$ is a Bézout domain (You may assume that, given a sequence $(z_n)$ of complex numbers with no limit point and a specification of the Taylor coefficients at $z_n$ up to some finite degree, there is a holomorphic function $f$ on $\mathbb{C}$ with, for each $z_n$, the specified Taylor coefficients).

Given two principal ideals, $<f>$ and $<g>$, in the ring of entire functions on $\mathbb{C}$. Then $f$ and $g$ will have corresponding sequences $z_n$ and $z'_n$. Then $f+g$ will have a sequence consisting $z_n$ and $z'_n$. Then $f+g$ by the assumption is a holomorphic function $h$.

Is $<f>+<g>=<h>$?

I have found the same question here A problem about generalization of Bezout equation to entire functions but I cannot understand the answer. Would you mind explaining the problem specific to the case of only 2 entire functions?

• No idea what you are trying to say about $h$. You also never write about $\langle f\rangle +\langle g\rangle$, which is the ideal in question, and is different from $\langle f+g\rangle$. – Thomas Andrews Feb 17 '17 at 21:38
• @ThomasAndrews I have no clue on how to use the hints. I have edited $<f+g>$ to $<f>+<g>$. – Kenneth.K Feb 17 '17 at 21:40
• What hints? I'm asking for clarification. What are the sequence $z_n$ and $z_n'$?, for example? Functions don't "have sequences." What do you mean by "will have a sequence?" – Thomas Andrews Feb 17 '17 at 21:41
• @ThomasAndrews As the question said that "given a sequence $(z_n)$ of complex numbers$, I want to use this to prove something. – Kenneth.K Feb 17 '17 at 21:45 • Have you looked at this question? The answer there gives a fairly good sketch of the proof. – Daniel Fischer Feb 17 '17 at 21:45 ## 1 Answer Hint: The principal ideal of an entire function$f$is the set of entire functions$g$such that if$f(z)=0$has a root at$z_0$of degree$d$, then$g(z)=0$has a root at$z_0$of degree at least$d$. So you need to find an$h$such that for each$z_0$with$f(z_0)=0$with degree$d_1$and$g(z_0)=0$with degree$d_2$, then$h(z_0)=0$with degree$\min(d_1,d_2)$.$f+g$works if$d_1\neq d_2$for all$d$and$f(z)\neq g(z)$when$f(z)\neq 0$. A more verbose way of saying this is defining, for each (non-zero) entire$f$, the function$d_f(z)$as the smallest integer$d$such that$f^{(d)}(z)\neq 0$. Then a non-zero$h$,$h\in\langle f\rangle$if and only if$d_h(z)\geq d_f(z)$for all$z$. Given$\langle f\rangle+\langle g\rangle=\langle h\rangle$, this means that you need to find$h\$ to have the property that $$d_h(z)=\min(d_f(z),d_g(z)).$$

• is it "principal ideal"? – Vim Feb 17 '17 at 21:57
• Yep, brain death. – Thomas Andrews Feb 17 '17 at 21:58