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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,

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    $\begingroup$ 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. $\endgroup$
    – Anthony Quas
    Commented Nov 18, 2016 at 20:46
  • $\begingroup$ 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 $\endgroup$
    – Diego
    Commented Nov 18, 2016 at 21:08
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    $\begingroup$ The earlier question to which OP refers is mathoverflow.net/questions/254957/… $\endgroup$
    – Gerry Myerson
    Commented Nov 18, 2016 at 22:13
  • $\begingroup$ 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 $M|\beta|^n$. $\endgroup$
    – Anthony Quas
    Commented Nov 18, 2016 at 22:37

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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 $r$.

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  • $\begingroup$ 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! $\endgroup$
    – Diego
    Commented Nov 19, 2016 at 11:20
  • $\begingroup$ I need the function analytic in $B(0,1)$ and with $|f(z)|$ bounded for something explicit (like $A/(1-B|z|)$). $\endgroup$
    – Diego
    Commented Nov 19, 2016 at 11:40
  • $\begingroup$ 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. $\endgroup$
    – Jérôme Poineau
    Commented Nov 19, 2016 at 12:14
  • $\begingroup$ Perfect! My e-mail is [email protected]. Thanks in advance! $\endgroup$
    – Diego
    Commented Nov 19, 2016 at 14:14

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