# Thread: Prove x^4 - x^2 + 1 is reducible over F_p for all p

1. ## Prove x^4 - x^2 + 1 is reducible over F_p for all p

Prove that $\displaystyle x^4 - x^2 + 1$ is reducible over $\displaystyle \mathbb{F}_p$ for every prime $\displaystyle p$.

I know $\displaystyle x^4 - x^2 + 1$ is the 12th cyclotomic polynomial and know a few properties of such polynomials, but I'm not sure if that really helps here. I can also show that $\displaystyle p^2 -1$ is divisible by 12 for $\displaystyle p>3$ but can't seem to make use of that (although have been given a hint that this would be useful).

2. Originally Posted by Boysilver
Prove that $\displaystyle x^4 - x^2 + 1$ is reducible over $\displaystyle \mathbb{F}_p$ for every prime $\displaystyle p$.

I know $\displaystyle x^4 - x^2 + 1$ is the 12th cyclotomic polynomial and know a few properties of such polynomials, but I'm not sure if that really helps here. I can also show that $\displaystyle p^2 -1$ is divisible by 12 for $\displaystyle p>3$ but can't seem to make use of that (although have been given a hint that this would be useful).

== In $\displaystyle \mathbb{F}_2\,,\,\,x^4-x^2+1=(x^2-x+1)^2$ , and in $\displaystyle \mathbb{F}_3\,,\,\,x^4-x^2+1=\left(x^2+1\right)^2$ . All the following is based on Gauss' Quadratic Reciprocity Law:

== Sometimes it helps to factor out these things over the reals: in $\displaystyle \mathbb{R}\,,\,\,x^4-x^2+1=\left(x^2+\sqrt{3}\,x+1\right)\left(x^2-\sqrt{3}\,x+1\right)$ , so in every field in which 3 is a quadratic

residue we get the above factorization , and this happens in $\displaystyle \mathbb{F}_p\,,\,\,with\,\,p=1\!\!\!\pmod 3\,,\,\,or\,\,p=2\!\!\!\pmod 3\,\,and\,\,also\,\,p=3\!\!\!\pmod 4$. This last kind of primes

may be expressed more simply by $\displaystyle p=3+4k\,,\,\,k=2\!\!\!\mod 3$. Examples: $\displaystyle p=11\,,\,23\,,\,47$ , etc.

So the primes that haven't been considered are those that are $\displaystyle 1\!\!\!\pmod 4$ but not $\displaystyle 1\!\!\!\pmod 3$ , like $\displaystyle 17\,,\,29\,,\,41$ , etc., and also primes $\displaystyle =3\!\!\!\pmod 4$ which

are $\displaystyle =1\!\!\!\pmod 3$ , like $\displaystyle 19\,,\,43$ , etc., but...but then nothing: unless I made some mistake somewhere, the pol. is irreducible over $\displaystyle \mathbb{Z}_{17}$ since (1) 3 is not a quadratic

residue modulo 17 (and thus the pol. doesn't have a actorization in two quadratic pol's. as above), and (2) it's easy to check (this pol. is an even function...) that

no element in $\displaystyle \mathbb{Z}_{17}$ is a root of the pol.

Check this.

Tonio

3. I didn't really follow the argument completely, but in any case the polynomial does factor over $\displaystyle \mathbb{F}_{17}$ as $\displaystyle (x^2+4x-1)(x^2-4x+1)$...

4. Originally Posted by Boysilver
I didn't really follow the argument completely, but in any case the polynomial does factor over $\displaystyle \mathbb{F}_{17}$ as $\displaystyle (x^2+4x-1)(x^2-4x+1)$...

No, it doesn't: $\displaystyle (x^2+4x-1)(x^2-4x+1)=x^4-16x^2-1\neq x^4-x^2+1\!\!\!\!\pmod{17}$

Tonio

5. Sorry, you're right. I meant $\displaystyle (x^2+4x-1)(x^2-4x-1)$.

6. Originally Posted by Boysilver
Sorry, you're right. I meant $\displaystyle (x^2+4x-1)(x^2-4x-1)$.

Now this is true, and it's exactly the case I missed to check! In the decomposition in two quadratics $\displaystyle x^4-x^2+1=\left(x^2+\sqrt{3}\,x+1\right)\left(x^2-\sqrt{3}\,x+1\right)$ I missed the case when the two free coefficients are -1, so it can also be in general

$\displaystyle x^4-x^2+1=(x^2+ax-1)(x^2-ax-1)$ , where $\displaystyle a^2=-1$, and thus we already covered all the primes $\displaystyle =1\!\!\!\!\pmod 4$ (among them, of course, 17...!)

Resuming, we have that our pol. is reducible $\displaystyle \mod p$ if:

0) $\displaystyle p=2\,,\,3$ ;

1) $\displaystyle p=1\!\!\!\!\pmod 4$ ;

2) $\displaystyle p=1\!\!\!\!\pmod 3$ ;

3) $\displaystyle p=2\!\!\!\!\pmod 3\mbox{ and also }=3\!\!\!\!\pmod 4$ .

And that covers all the possible cases for a prime, because if $\displaystyle p=3\!\!\!\!\pmod 4\mbox{ but }p\neq 2\!\!\!\!\pmod 3\mbox{ then }p=1\!\!\!\!\pmod 3$ , which is already covered in case (2), so Q.E.D.!

Great eye-opener example with $\displaystyle p=17$, boysilver: thanx.

Tonio