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Let $\alpha$ be a root of a polynomial $a_nx^n + \ldots + a_1x$ with integral coefficients.

I would like to determine $\varepsilon > 0$ depending on $a_1, \ldots, a_n$ so that $|\alpha| < \varepsilon$ implies $\alpha = 0$.

Is it possible to give a "formula" for such an $\varepsilon$ without refering to the complete list of roots?

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    $\begingroup$ Root separation: en.wikipedia.org/wiki/… $\endgroup$ – Matt F. Feb 11 at 16:24
  • $\begingroup$ @MattF. Thank you, that is what I am searching for. Do you know any good source (besides the one mentioned on Wikipedia)? $\endgroup$ – J. Fabian Meier Feb 11 at 20:44
  • $\begingroup$ The accepted solution does not even use the fact that the coefficients are integral. $\endgroup$ – jarauh Feb 12 at 7:55
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Of course we must assume some $a_j \ne 0$. Say $a_j$ is the one with least index. Then you want $\varepsilon$ such that $p(x) = a_n x^{n-j} + \ldots + a_j \ne 0$ for $|x| < \varepsilon$. You may use inequalities such as $|p(x)| \ge |a_j| - \sum_{k=j+1}^{n} |a_k| |x|^{k-j} \ge |a_j| - m \sum_{k=j+1}^n |a_k|$

where $m = \max(|x|^{n-j}, |x|)$. Thus $p(x) \ne 0$ if $m < |a_j|/\sum_{k=j+1}^n |a_k|$.

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    $\begingroup$ In your first sentence, the 1 in $a_1$ is a typo? $\endgroup$ – Gerry Myerson Feb 11 at 20:38
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Everything you wanted to know (and a little more) is in this paper by John Abbott

Abbott, John, Bounds on factors in $\Bbb Z[x]$, J. Symb. Comput. 50, 532-563 (2013). ZBL1295.12010.:

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