Let $f(x)=a_nx^n+\cdots+a_0 \in \mathbb{Z}[x]$ be an integer polynomial in one variable. Recall that the height $H(f):=\textrm{max}\,|a_n|$ is the largest coefficient. Consider the set of polynomials of fixed degree which divide a non-zero polynomial of bounded height: $$S_{d,H}:=\{g\in\mathbb{Z}[x]\,\big{|}\,\deg{g}=d\textrm{ and }g\mid f\textrm{ for some non-zero}f\textrm{ with }H(f)\leq H\}.$$ Crucially, there are no conditions on the degree of $f$. Is $S_{d,H}$ a finite set for all $d$ and $H$? If so, is there a known bound in the literature for $\textrm{max}_{g\in S_{d,H}} H(g)$ in terms of $H$ and $d$?
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1$\begingroup$ I think this should follow from the inequalities between Mahler measure, which is multiplicative, and height which is not. Of course these are rather loose, but they exist. $\endgroup$– Watson LaddCommented May 27, 2021 at 4:39
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$\begingroup$ See, e.g., Laurent Panaitopol and Doru Stefanescu, Bounds for heights of integer polynomial factors, Journal of Universal Computer Science, vol. 1, no. 8 (1995), 603-613, citeseerx.ist.psu.edu/viewdoc/… $\endgroup$– Gerry MyersonCommented May 27, 2021 at 12:52
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$\begingroup$ One comment worth making: I'm seeking a bound like $H^{100d}$ i.e. exponential in the height, with exponent linear in $d$. $\endgroup$– Philip EngelCommented May 27, 2021 at 18:15
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$\begingroup$ $H^{100d}$ is polynomial in $H$ and exponential in $d$, not exponential in $H$. $\endgroup$– Emil JeřábekCommented Jun 16, 2023 at 11:48
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This questions seems to be answered on another entry on MathOverflow
But the formula (Gelfond's formula) in that entry is different from the one found in a paper by Mahler. The Mahler paper shows a base of 2 rather than e, and has an additional term; it says that Gelfond showed that 2 was optimal. But it ends with a comment "It would therefore have great interest to find the exact maxima..."
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1$\begingroup$ The bounds in the linked question depend on the degree of $f$, which the OP here explicitly wants to avoid. So it does not answer the question. $\endgroup$ Commented Jun 16, 2023 at 11:47
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3$\begingroup$ On second thoughts, Pietro Majer’s argument actually gives a bound that only depends on the degree of $g$ (I have no idea why it was formulated in terms of $n$ rather than $m$ on the last line). So, yes: $H(g)\le H(H+2)^d$, and $S_{d,H}$ is finite for all $d$ and $H$. $\endgroup$ Commented Jun 17, 2023 at 21:18
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$\begingroup$ @user2183078 It would be very nice if you could address the comments in editing your answer. $\endgroup$ Commented Jun 17, 2023 at 22:41