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One can see that algebraic numbers are dense in the complex plane by just looking at quadratic polynomials. I am interested in a "higher order" density of algebraic numbers.

More specifically: is it known that if $D_1,\ldots, D_n \subset \mathbb{C}$ are disjoint disks, then there is an irreducible polynomial in $\mathbb{Z}[x]$ having at least one root in each $D_i$?

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    $\begingroup$ Just an idea: what about picking $n$ rationals $q_i$ wherever you like, then looking at $P(X)=\prod_i{(X-q_i)}+\epsilon$. Can't we find a small rational $\epsilon$ such that $P(X)$ is irreducible and its roots within some given $\delta$ of the $q_i$'s? $\endgroup$
    – Yaakov Baruch
    Commented Jan 21, 2015 at 15:42
  • $\begingroup$ The irreducibility for most $\epsilon=1/n$, say, is given by Hilbert's irreducibility theorem. $\endgroup$
    – Lior Bary-Soroker
    Commented Jan 22, 2015 at 19:38

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The idea by Yaakov Baruch works. Take any $q_i\in D_i\cap \mathbb Q[i]$ and take the polynomial $$ g(x)=\prod_{i=1}^n (x-q_i)(x-\overline q_i). $$ It has rational coefficients.

If a polynomial $f(x)$ of degree $2n$ differs coefficientwise sufficiently small from $g(x)$, then $f(x)$ has a root in each $D_i$ --- e.g., by Rouché's theorem. It remains to make $f(x)$ irreducible.

To perform this, one may use Eisenstein's criterion. Multiply $g(x)$ by a sufficiently large integer $N$ divisible by all the denominators of its coefficients. Then change its coefficients by 0 or 1 so that the leading one is odd, all others are even, and the last one is not divisible by 4. Finally, divide back by $N$.

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  • $\begingroup$ That's nice. To make the question harder, say I am interested in algebraic integers instead of algebraic numbers. Any ideas for that? $\endgroup$
    – Balazs Strenner
    Commented Jan 21, 2015 at 20:44
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    $\begingroup$ Yes with algebraic integers: Take g(x),N as above, then let f(x)=x^(1+deg g) + (2^M)*N *g(x) + 2, where M>>0 is a large integer. Eisenstein's criterion gives irreducibility of f(x) and Rouche's theorem (applied to f(x)/(2^M *N)) gives that f(x) has (algebraic integer) zeros near zeros of g(x). $\endgroup$
    – David Lampert
    Commented Feb 15, 2015 at 18:16

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