There are several well-known criteria for a polynomial with integer coefficients to be irreducible over $\mathbb{Z}$, e.g., Eisenstein's criterion. I'm looking for the opposite: other than factorization, is there some sufficient condition to show that a given polynomial must be reducible? Ideally, I'd like some property that depends only on the coefficients appearing in the polynomial, like in Eisenstein's criterion.
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1$\begingroup$ I edited the title to reflect what I understand to be your question. $\endgroup$– Timothy ChowCommented Apr 2, 2018 at 17:04
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3 Answers
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This question seems too vague to have a good answer. I can take any theorem in algebraic number theory whose hypotheses start "let $f$ be an irreducible polynomial" and take the contrapositive. For example
(1) Minkowski's discriminant bound implies that an irreducible polynomial of degree $n$ has discriminant at least $\tfrac{\pi^{n/2} n^n}{2^n n!}$. For example, since $(x-1)(x-2)$ has discriminant $1 < \tfrac{4 \pi}{8}$, it is reducible.
(2) Cebatarov's density theorem states that an irreducible polynomial has, on average, $1$ root in $\mathbb{F}_p$, as we average over $p$. This has effective versions. Therefore, a list of roots of $f(x)$ for enough primes to violate those bounds proves $f$ is reducible.
The latter is actually a plausible strategy for heuristically testing polynomials for irreduciblity, since it is very fast to write and run a script which finds roots of $f$ modulo $p$ for the first $10^4$ primes or so. I don't know how reasonable it is to get up the range where this would constitute an actual proof.
answered Apr 2, 2018 at 17:26
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$\begingroup$ Hi David, thanks, this is very helpful! You've clarified my thinking. $\endgroup$– GautamCommented Apr 27, 2018 at 22:18
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$\begingroup$ One question about the first point, the discriminant have to be taken in absolute value? If not, take the polynomial $x^2+1$, whose discriminant is $b^2-4ac = -4 < 1 < \frac{4\pi}{8}$ but the polynomial is irreducible... $\endgroup$– Wilem2Commented Dec 6, 2021 at 20:02
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Pick your favorite polynomial $q(x).$ If the resultant of $p, q$ is $0,$ then $p$ is reducible.
answered Apr 2, 2018 at 0:52
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1$\begingroup$ Nice criterion (I think one has to propose that q is no multiple of p). $\endgroup$ Commented Apr 2, 2018 at 1:10
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Finding sufficient conditions is wildly open territory; one can easily list conditions that are far far away from necessary. For example conditions that 0 is a root, or 1 is a root, or $-1$ is a root are easily translatable into ones on coefficients (a) constant term is 0 (b) sum of the coefficients is 0 (c) sum of the coefficients of even powers of the variable equals that of odd powers.
So you need to be more careful about what is required which might add additional conditions.
