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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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    $\begingroup$ 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. $\endgroup$
    – Peter Mueller
    Commented Aug 17, 2015 at 18:03
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    $\begingroup$ 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. $\endgroup$
    – Vesselin Dimitrov
    Commented Aug 20, 2015 at 8:02
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    $\begingroup$ 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. $\endgroup$
    – Vesselin Dimitrov
    Commented Aug 20, 2015 at 9:06
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    $\begingroup$ 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 $\ell < i < d$. $\endgroup$
    – Vesselin Dimitrov
    Commented Aug 20, 2015 at 9:13
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    $\begingroup$ Sorry, $n = d = \deg{f}$. I mixed up the two notations. $\endgroup$
    – Vesselin Dimitrov
    Commented Aug 20, 2015 at 9:21

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