mersenneforum.org Irreducibility problem
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 2018-11-21, 15:00 #1 fivemack (loop (#_fork))     Feb 2006 Cambridge, England 11000110011102 Posts 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)
2018-11-21, 16:35   #2
fivemack
(loop (#_fork))

Feb 2006
Cambridge, England

2×52×127 Posts

Quote:
 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.

2018-11-21, 18:16   #3
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 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.
Code:
polisirreducible(4*x^2+27)
would check irreducibility anyways.

2018-11-22, 02:24   #4
Dr Sardonicus

Feb 2017
Nowhere

29×113 Posts

Quote:
 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.

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