4
$\begingroup$

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 $p \equiv 3,7 \mod 10$.

  • 9
    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.2012-03-14
  • 0
    @franzlemmermeyer Why don't you work that out as an answer?2012-03-16
  • 0
    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?2012-03-16
  • 0
    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...2012-03-16
  • 0
    Well, may be modulo 30 is enough? I was all sold on conductor divisible by 4 :-)2012-03-16

2 Answers 2