20181121, 15:00  #1 
(loop (#_fork))
Feb 2006
Cambridge, England
1100011001110_{2} 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) 
20181121, 16:35  #2  
(loop (#_fork))
Feb 2006
Cambridge, England
2×5^{2}×127 Posts 
Quote:


20181121, 18:16  #3  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
Code:
polisirreducible(4*x^2+27) 

20181122, 02:24  #4  
Feb 2017
Nowhere
29×113 Posts 
Quote:
An irreducible polynomial of degree n can be irreducible (mod p) for some p, only if its Galois group has a ncycle; and D_{6}(6), being a 6element group that isn't cyclic, hasn't got a 6cycle. 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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