# Analytic functions with integer coefficients with prescribed zeros

I would like to ask the following question. Maybe some of you can help me at least with a hint.

Let $\alpha, \beta\in B(0,1)$ (unit ball), with $\alpha\neq \beta, \overline{\beta}$. I would like to prove that there exists a function $f\in \mathbb{Z}[[z]]$ analytic in the unit ball and such that $f(\beta)=0$ and $f(\alpha)\neq 0$.

Yesterday, Pietro Majer was able to help me in the case $\alpha,\beta\in \mathbb{R}$. In fact, I was able to adapt Pietro's idea for solving the case $\beta\in (-1,1)$ and $\alpha\in B(0,1)$. However, I can not deal with a complex $\beta$ since his idea depends on writing $1/\beta$ in basis $1/\beta^2$.

This question is related to one of my works on transcendental functions with integer coefficients with prescribed arithmetic behavior in algebraic points.

Thanks in advance,

• Pick the first few integers so that substituting $\alpha$ you get something far from 0; and substituting $\beta$ you get something close to 0. Now pick the remaining coefficients, only using a term when you halve the remaining distance to 0. You can obtain a uniform bound on the size of the coefficients (depending on the argument and modulus of $\beta$). Provided the initial terms gave you something very large for $\alpha$ and very small for $\beta$, you will ensure that the $\beta$ sum converges to 0, while you will not have added enough to ensure that the $\alpha$ sum converges to 0. – Anthony Quas Nov 18 '16 at 20:46
• Thanks for the hint @AnthonyQuas. For example, I can find $a$ and $b$ integers such that $|a+b\beta|<|\beta|$ and $|a+b\alpha|>1/4$. I do not know if this is enough, but could you please explain me better your idea in "Now pick the remaining coefficients, only using a term when you halve the remaining distance to 0. You can obtain a uniform bound on the size of the coefficients (depending on the argument and modulus of $\beta)"? Thanks in advance – Diego Nov 18 '16 at 21:08 • The earlier question to which OP refers is mathoverflow.net/questions/254957/… – Gerry Myerson Nov 18 '16 at 22:13 • I'm suggesting that you get the sum of the first terms much smaller for$\beta$than for$\alpha$. Write$f$for the polynomial giving the first terms, of degree$n$say. Now$-f(\beta)$is your "target" - it's what you want the remainder of the series to sum to. Is there an integer multiple of$\beta^{n+1}$that gets you closer to the target? If so, add/subtract it; if not move on to the next power of$\beta$etc. You should be able to prove that there is a constant$M$such that the distance from the target when you are adding multiples of$\beta^n$is at most$M|\beta|^n$. – Anthony Quas Nov 18 '16 at 22:37 ## 1 Answer David Harbater wrote a paper devoted to this ring (denoted by$\mathbf{Z}\{t\}$) and similar ones, see "Convergent arithmetic power series", Amer. J. Math., 106 (1984), no. 4, pp. 801-846. In particular, in Lemma 1.5, he proves that for each$r \in (0,1)$and each$\lambda \in \mathbf{C}$of absolute value at most$r$, there exists$f\in \mathbf{Z}\{t\}$with a zero of order 1 at$\lambda$and$\bar\lambda$and no other zero on the closed disk of center 0 and radius$r$. • Thanks Jérôme! Do you know if this function$f$is analytical in$B(0,1)$? Or only in$B(0,r)$? I was not able to find his paper. Do you have it? Thanks in advance! – Diego Nov 19 '16 at 11:20 • I need the function analytic in$B(0,1)$and with$|f(z)|$bounded for something explicit (like$A/(1-B|z|)$). – Diego Nov 19 '16 at 11:40 • Yes, the function$f$converges on$B(0,1)\$. For the bound, I do not know and would have to get back to the proof. Send me an e-mail if you want Harbater's paper. – Jérôme Poineau Nov 19 '16 at 12:14
• Perfect! My e-mail is dyhegu2@gmail.com. Thanks in advance! – Diego Nov 19 '16 at 14:14