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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?

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  • $\begingroup$ 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 $\langle f+g\rangle$. $\endgroup$ – Thomas Andrews Feb 17 '17 at 21:38
  • $\begingroup$ @ThomasAndrews I have no clue on how to use the hints. I have edited $<f+g>$ to $<f>+<g>$. $\endgroup$ – Kenneth.K Feb 17 '17 at 21:40
  • $\begingroup$ 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?" $\endgroup$ – Thomas Andrews Feb 17 '17 at 21:41
  • $\begingroup$ @ThomasAndrews As the question said that "given a sequence $(z_n)$ of complex numbers$, I want to use this to prove something. $\endgroup$ – Kenneth.K Feb 17 '17 at 21:45
  • $\begingroup$ Have you looked at this question? The answer there gives a fairly good sketch of the proof. $\endgroup$ – Daniel Fischer Feb 17 '17 at 21:45
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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 $$d_h(z)=\min(d_f(z),d_g(z)).$$

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

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