Polynomial

**ZetaX** · March 24th 2009, 11:51 AM

I have the polynomial $f(x) = x^4-10x^2+1$ over $\mathbb Q$ . It can be easily seen that Klein's Four group is the Galois group of this polynomial.
How can I show that the corresponding polynomial
$g(x) \in \mathbb F_p[x]$ is irreducible in $\mathbb F_p[x]$ for every prime $p$ .

**Moo** · March 24th 2009, 12:06 PM

Oh man, please, it's polynomial !

**ThePerfectHacker** · March 27th 2009, 11:26 AM

This is not true. If you take $p=3$ , $x^4 - 10x^2 + 1 = (x^2+1)^2$ . I think you want to prove that this polynomial is always reducible as opposed to irreducible. One way to proved is by Dedekind's theorem (again) and by an understanding of transitive subgroups of $S_4$ . The transitive subgroup $V$ (Klein four group) consists of an identity element and $2$ -cycle products ( $V=\{\text{id},(12)(34),(13)(24),(14)(23)\}$ ). Therefore, mod any $p$ the polynomial $x^4-10x^2+1$ either splits or factors into quadradic factors. Therefore, it is always reducible.

**ZetaX** · March 28th 2009, 12:17 AM

Sorry what I meant was reducible not irreducible, it was a mistake.

**ThePerfectHacker** · March 28th 2009, 02:53 PM

Originally Posted by ZetaX

I have the polynomial $f(x) = x^4-10x^2+1$ over $\mathbb Q$ . It can be easily seen that Klein's Four group is the Galois group of this polynomial.
How can I show that the corresponding polynomial
$g(x) \in \mathbb F_p[x]$ is reducible in $\mathbb F_p[x]$ for every prime $p$ .

Here is another way of proving this which is elementary. Let $p$ be prime with $p\geq 7$ .

Notice the following factorizations:
$x^4-10x^2+1 = (x^2 + 2\sqrt{2}x-1)(x^2-2\sqrt{2}x-1)$ [1]
$x^4-10x^2+1 = (x^2 + 2\sqrt{3}x+1)(x^2 - 2\sqrt{3}x+1)$ [2]
$x^4-10x^2+1 = (x^2 - 5 + 2\sqrt{6})(x^2 - 5 - 2\sqrt{6})$ [3]

If $2$ is a square root in $\mathbb{F}_p$ then we can use the factorization in [1] where we replace $\sqrt{2}$ by a square root of $2$ modulo $p$ . If $3$ is a square root in $\mathbb{F}_p$ then we can use the factorization in [2] where we replace $\sqrt{3}$ by a square root of $3$ modulo $p$ . If neither $2,3$ are squares mod $p$ then it means $(2/p),(3/p)=-1$ (Legendre symbol). However, this means $(6/p) = (2/p)(3/p) = (-1)(-1) = 1$ . Therefore, $6$ would be a square in $\mathbb{F}_p$ and so we can use the factorization in [3].

Therefore, $\overline{f(x)}\in \mathbb{F}_p[x]$ is always reducible for any $p$ .

**ZetaX** · March 29th 2009, 10:04 AM

why should p be greater than or equal to 7?

**ThePerfectHacker** · March 29th 2009, 10:10 AM

Originally Posted by ZetaX

why should p be greater than or equal to 7?

Because I just wanted $(2,p)=(3,p) = 1$ if $p=2,3$ then they are not relatively prime anymore. The cases $p=2,3$ can be checked seperately. It seems I did not need $p\geq 7$ but rather $p\geq 5$ .

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