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Let $h(x)=x^4+12x^3+14x^2-12x+1$, and let $p>5$ be a prime.

I want to show $h(x)$ factors into 2 quadratics $\mod p$, if $p \equiv 9,11 \mod 20$, while $h(x)$ factors mod $p$ into 4 linear factors, if $p \equiv 1,19 \mod 20$. I can show $h(x)$ is irreducible if $p \equiv 3,7 \mod 10$.

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Show that the splitting field is abelian, compute its discriminant, show that the splitting field is a subfield of the field of 20th roots of unity, and use the decomposition law in cyclotomic extensions. – franz lemmermeyer Mar 14 '12 at 18:40
@franzlemmermeyer Why don't you work that out as an answer? – draks ... Mar 16 '12 at 14:30
As an afterthought to my answer: $$x^4+12x^3+14x^2-12x+1=\big(x^2+(6+2\sqrt5)x-1\big)\big(x^2+(6-2\sqrt5)x-1\big)‌​.$$ From this you get the splitting field. I'm afraid I couldn't show with elementary means that the splitting field would be the real subfield of the 20th cyclotomic field. Another cause for concern is that I think this splits completely modulo $p=29,31,61$, and to a product of two quadratic factors modulo $p=41,71,101$. Has anybody checked this further? – Jyrki Lahtonen Mar 16 '12 at 19:53
After further testing I'm beginning to think that this has conductor 60. Splitting into linear factors happens when $p\equiv1,29,31,59$, and when $p\equiv 11,19,41,40$ we get quadratic factors only. Further evidence comes from $p=3$, where we have $h(x)\equiv(x^2+1)^2$. My number theory is a bit rusty, so I don't remember for sure whether this implies that $3$ ramifies in the splitting field... – Jyrki Lahtonen Mar 16 '12 at 20:48
Well, may be modulo 30 is enough? I was all sold on conductor divisible by 4 :-) – Jyrki Lahtonen Mar 16 '12 at 21:02

For starters consider the following candidate factorization: $$ (x^2+ax-1)(x^2+bx-1)=x^4+(a+b)x^3+(ab-2)x^2-(a+b)x+1. $$ This works if $a+b\equiv 12$ and $ab\equiv 16$. The discriminant of the quadratic $$ (T-a)(T-b)=T^2-(a+b)T+ab=T^2-12T+16 $$ is equal to $20$. So if $20$ is a quadratic residue modulo $p$, then we have a factorization of this kind. As $20=2^2\cdot5$ we are really interested in, whether $5$ is a quadratic residue modulo $p$ or not. By quadratic reciprocity this happens, iff $p$ is a quadratic residue modulo $5$. So we get that a factorization of this kind exists, iff $p\equiv\pm1\pmod5.$ Of course, this is equivalent to $p$ being congruent to either $1,9,11$ or $19$ modulo $20$.

I don't know, if it is easy or difficult to extend this elementary argument and detect, whether these quadratic factors will split further. Also, this argument obviously doesn't rule out the possibility of other quadratic factors, but you seemed to have the cases $p\equiv\pm2\pmod 5$ covered already.

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pari tells us that $h$ generates the same splitting field as $g(x) = x^4 - x^3 - 4x^2 + 4x + 1$, which has discriminant $1125 = 3^2 5^3$. Its Galois group is cyclic, hence the field must be a subfield of the 15th roots of unity by Kronecker-Weber. The cyclotomic decomposition law tells us that

  • $p$ splits completely if and only if $p \equiv \pm 1 \bmod 15$,

  • $p$ splits into two prime factors if and only if $p \equiv \pm 4 \bmod 15$,

  • $p$ is inert if and only if $p \equiv \pm 2, \pm 7 \bmod 15$.

By a classical theorem of Kummer, this reflects the splitting of the polynomial modulo $p$ except for divisors of the dirscriminant of $h$.

You can show that the splitting field of $h$ is cyclotomic without pari and Kronecker-Weber by solving the quartic with radicals, but this requires some efforts.

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Wow. What's wrong with this answer? – franz lemmermeyer Mar 25 '12 at 16:18

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