Irreducibility problem
 

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)

[QUOTE=fivemack;500663]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)[/QUOTE]

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.

[QUOTE=fivemack;500676]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.[/QUOTE]

[CODE]polisirreducible(4*x^2+27)[/CODE] would check irreducibility anyways.

[QUOTE=fivemack;500676]Sorry, this is total nonsense, I had confused 'is irreducible' and 'has no roots'. The polynomial [[color=red]4*x^6 + 27][/color]] splits as either a product of two cubics or a product of three quadratics or a product of six linear factors modulo all primes.[/QUOTE]Ah, yes, the Galois group manifests itself! The polynomial is irreducible in Q[x]. But the Galois group is D[sub]6[/sub](6), the dihedral group of 6 elements, as a subgroup of S[sub]6[/sub].

An irreducible polynomial of degree n can be irreducible (mod p) for some p, only if its Galois group has a n-cycle; and D[sub]6[/sub](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 [i]normal[/i] extension of Q of degree n, always splits (mod p) into factors all of the [i]same[/i] degree, for all primes p which do not divide the discriminant.