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I would like to see criteria for the irreducibility of a polynomial over $\mathbb{Z}$ based (mainly) on the location of the roots of the polynomial in the complex plane. An example of such a criterion is:

Let $f(z) \in \mathbb{C}[z]$ be a nonconstant monic polynomial with integral coefficients, and let $D = \{z \in \mathbb{C} : |z| < 1\}$ be the open unit disk in the complex plane. Suppose that $f$ has exactly one root not in $D$, and that its constant term is nonzero. Then $f$ is irreducible in $\mathbb{Q}[z]$.

Furthermore, the irreducibility of several specific polynomials can be established from an approximate knowledge of the complex roots, as carried out by Selmer in his work "On the irreducibility of certain trinomials" for polynomials such as $X^n - X - 1$ (see also Is $x^{n}-x-1$ irreducible?).

Feel free to point at similar and related results.

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Pablo
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    Interesting question! It is funny that the $p$-adic analogue of this question has obvious answers. For instance the Eisenstein criterion can be seen as a sufficient condition on the $p$-adic roots. – Peter Mueller Aug 17 '15 at 18:03
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    You can mix the type of criterion you mention on the complex roots with $p$-adic conditions a la Eisenstein-Dumas. Here is the simplest example: If $f = a_dX^d+ \cdots + a_0 \in \mathbb{Z}[X]$ has (i) $a_d = \pm p^a$, (ii) $a_{d-1} \not\equiv 0 \mod{p}$, (iii) $a_0 \neq 0$, and (iv) all complex roots lying in $|z| < 1$, then $f$ is irreducible. – Vesselin Dimitrov Aug 20 '15 at 08:02
  • @VesselinDimitrov could you post this as an answer or write some explanation here because a proof is not obvious for me. Also, I will be glad to hear about more criteria of this type (you said this is the simplest example). – Pablo Aug 20 '15 at 08:55
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    Actually this example is very easy. If $f = gh$ then condition (i) yields that both $g$ and $h$ have $\pm p$-power leading coefficients. Then, by (ii), either $g$ or $h$ (say, the former) has leading coefficient $\pm 1$. But by (iii) and (iv), this is impossible unless $g$ is constant. – Vesselin Dimitrov Aug 20 '15 at 09:06
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    Likewise, if $f$ has leading coefficient a prime and all its zeros in $|z| < 1$, then it must be irreducible. And here is a generalization (due to Petkov) of the example I gave, which you can prove in a similar way from the Eisenstein-Dumas theorem: An integer $f$ with $a_0 \neq 0$ and having all its complex roots in $|z| < 1$ is irreducible if, for some prime $p$ and index $\ell < d$: (i) $a_d = \pm p^s$; (ii) $p \nmid a_{\ell}$; (iii) $(n-\ell,s) = 1$; (iv) $\mathrm{ord}_p a_i \geq s \frac{i-\ell}{n-\ell}$ for all $\ell < i < d$. – Vesselin Dimitrov Aug 20 '15 at 09:13
  • @VesselinDimitrov what are $n,d$? – Pablo Aug 20 '15 at 09:16
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    Sorry, $n = d = \deg{f}$. I mixed up the two notations. – Vesselin Dimitrov Aug 20 '15 at 09:21
  • @VesselinDimitrov Is there some criterion for irreducibilty in case that the largest root of the polynomial is real? Such a real number is called a Perron number (see https://en.wikipedia.org/wiki/Perron_number). These numbers, as well as Pisot numbers, have been studied a lot. – Pablo Aug 20 '15 at 09:27

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