Irreducibility problem

[fivemack]

4*x^6+27 is irreducible modulo about 5/6 of all primes, so I'm a little confused as to why msieve square root is consistently failing to find an irreducible prime for C244_131_108 for which that's the algebraic side.

(it picks 53, and indeed polrootsmod(4*x^6+27,53) = [], but so far I've had twelve 'Newton iteration failed to converge' in succession; I'm running the rest of the 36 dependencies but not terribly confidently)

[fivemack]

Originally Posted by fivemack

4*x^6+27 is irreducible modulo about 5/6 of all primes, so I'm a little confused as to why msieve square root is consistently failing to find an irreducible prime for C244_131_108 for which that's the algebraic side.

(it picks 53, and indeed polrootsmod(4*x^6+27,53) = [], but so far I've had twelve 'Newton iteration failed to converge' in succession; I'm running the rest of the 36 dependencies but not terribly confidently)

Sorry, this is total nonsense, I had confused 'is irreducible' and 'has no roots'. The polynomial splits as either a product of two cubics or a product of three quadratics or a product of six linear factors modulo all primes.

[science_man_88]

Originally Posted by fivemack

Sorry, this is total nonsense, I had confused 'is irreducible' and 'has no roots'. The polynomial splits as either a product of two cubics or a product of three quadratics or a product of six linear factors modulo all primes.

polisirreducible(4*x^2+27)

would check irreducibility anyways.

[Dr Sardonicus]

Originally Posted by fivemack

Sorry, this is total nonsense, I had confused 'is irreducible' and 'has no roots'. The polynomial [4*x^6 + 27]] splits as either a product of two cubics or a product of three quadratics or a product of six linear factors modulo all primes.

Ah, yes, the Galois group manifests itself! The polynomial is irreducible in Q[x]. But the Galois group is D₆(6), the dihedral group of 6 elements, as a subgroup of S₆.

An irreducible polynomial of degree n can be irreducible (mod p) for some p, only if its Galois group has a n-cycle; and D₆(6), being a 6-element group that isn't cyclic, hasn't got a 6-cycle.

But an irreducible polynomial of degree n which defines a normal extension of Q of degree n, always splits (mod p) into factors all of the same degree, for all primes p which do not divide the discriminant.