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Using a CAS, I know that $x^6 + 24x - 20$ is irreducible over $\mathbb{Q}$, but I am struggling to prove it. Eisestein's Criterion doesn't apply, and it is reducible modulo every prime I've tested. I'm not familiar with much else beyond these tests, so how would I go about showing this?

JaniBill
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  • Which primes have you tested? – TomGrubb Dec 02 '17 at 21:02
  • @ThomasGrubb I just quickly ran through all primes through than 97. – JaniBill Dec 02 '17 at 21:08
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    FWIW I'm fairly sure the Galois group of this sextic is $PSL_2(\Bbb{F}_5)\simeq A_5$ acting on the six zeros the same way it does on the six points of the projective line. That group has no elements of order six explaining the absence of primes $p$ such that it would remain irreducible in $\Bbb{F}_p[x]$. – Jyrki Lahtonen Dec 02 '17 at 22:10

2 Answers2

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Modulo $3$ it splits into irreducible quadratics, while modulo $7$ there are irreducible factors of degrees one and five. These are incompatible so the polynomial is irreducible in $\mathbb{Z}[X]$.

user16394
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    The only thing I was aware of with the modulo p test is that if the polynomial is irreducible modulo some prime p, then it is also irreducible in $\mathbb{Q}$. What does the decomposition of the modulo p polynomial into irreducible factors tell us? – JaniBill Dec 02 '17 at 21:12
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    @JaniBill They tell you the possible degrees of factors in $\mathbb{Z}[X]$. Reducing mod $3$ gives you the possible degrees $2,4,6$; reducing mod $5$ gives you the possible degrees $1,5,6$. – user16394 Dec 02 '17 at 21:13
  • Did you mean modulo seven instead of modulo five? Modulo five we have $$x^6+24x-20\equiv x^6-x=x(x-1)^5.$$ Modulo seven Mathematica says it has an irreducible quintic factor. – Jyrki Lahtonen Dec 02 '17 at 21:26
  • @user16394 Okay thanks. Modulo 5 that quintic is reducible as $f_5 = x^6 + 4x = x(x + 4)^5$ by the way. What you said still works modulo 7 though because $f_7$ splits into irreducible factors as $(x + 3)(x^5 + 4x^4 + 2x^3 + x^2 + 4x + 5)$. I would still need to prove that quintic is irreducible though, and I'm not quite sure how to do that. – JaniBill Dec 02 '17 at 21:27
  • @JyrkiLahtonen You're right, thanks. JaniBill: Irreducibility over a finite field is not so hard... you can test this with the euclidean algorithm, for example... if you have trouble with this then it could be a separate question – user16394 Dec 02 '17 at 21:31
  • @user16394 Sorry I'm a bit new to these things, how could you use the Euclidean algorithm to show irreducibility? EDIT: Just saw you said to post it as a separate question, I'll do that. – JaniBill Dec 02 '17 at 21:34
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    @JaniBill Take gcd with polynomials $x^{p^n} - x$ to find factors of degree dividing $n$ – user16394 Dec 02 '17 at 21:37
  • It suffices to check that the quintic has no common factors with $x^{49}-x$. All the irreducible quadratics are factors of that. Only the calculation of the first remainder (in Euclid's algorithm) involves higher degree polys :-) – Jyrki Lahtonen Dec 02 '17 at 21:47
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In a related post астон вілла олоф мэллбэрг linked to a useful criterion

Here is a lovely lemma by (Prof.) Ram Murty:

Let $f(x) = a_mx^m + ... + a_1x + a_0$ be a polynomial of degree $m$ in $\mathbb Z[x]$. Let $H = \displaystyle\max_{0 \leq i \leq m-1} \left|\frac{a_i}{a_m}\right|$. If $f(n)$ is prime for some $n \geq H+2$, then $f(x)$ is irreducible in $\mathbb Z[x]$.

Link : http://cms.dm.uba.ar/academico/materias/2docuat2011/teoria_de_numeros/Irreducible.pdf

Here $H=24$ we search for $f(n)$ prime for $n\ge 26$

It happens that $f(59)=42180535037$ is prime, so $f$ is irreductible in $\mathbb Z[x]$.

zwim
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  • https://math.stackexchange.com/questions/2226070/how-can-i-show-that-the-polynomial-p-x5-x3-2x2-2x-1-is-irreducibl?rq=1 – zwim Dec 02 '17 at 21:24
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    This is nice, but it does beg the question: Is it really easier to prove that $f(59)$ is a prime than to prove those factorizations modulo three and seven?? :-) – Jyrki Lahtonen Dec 02 '17 at 21:30
  • OP apparently is a fervent user of CAS but not familiar with criterions other than Eisenstein, so this one might serve him. Of course I agree that as a manual tool it is almost useless unless numbers involved are let say $<10000$. I would say it gives something tangible, compared to the bare return TRUE/FALSE when invoking IsIrreductiblePolynomial(f) ? – zwim Dec 02 '17 at 21:42
  • Fair enough :-) – Jyrki Lahtonen Dec 02 '17 at 21:45
  • @zwim This is one part of a larger problem I've been assigned, so I've been using Sage just to investigate what I need to show. Ultimately I need to prove it with old fashioned pen and paper using basic tests, but we breezed over most of the relevant material in class, so I'm pretty weak on it. I definitely wouldn't be permitted to use as niche a criteria as this, but it is a mathematically interesting answer nonetheless. – JaniBill Dec 02 '17 at 21:49