# Proving that $x^4 - 10x^2 + 1$ is not irreducible over $\mathbb{Z}_p$ for any prime $p$.

So I have seen the similar question and answers on here for $x^4 +1$, but I am having trouble extending anything there to this polynomial... I understand it is fairly trivial with Galois theory, but my class has just barely covered Field Extensions, so suffice it to say we have no Galois theory to play with.

I managed to prove it for the primes such that $p \equiv 1, 7 \pmod 8$, by noting that $2$ is a square modulo those primes and thus $x^4 - 2x^2 +1 = (x^2 -1 + 2qx)(x^2 - 1 - 2qx)$ for those $\mathbb{Z}_p$... however, trying to get a similar result for $3 \pmod 8$ and $5 \pmod 8$ has been stumping me for a long time, I am having a hard time making $q^2 = -1$ and $q^2 = -2$ give me something factorable...

I guess the worst part of all of this is that I don't think this solution is even particularly enlightening, in terms of abstract algebra. It's really just some number theory trickery. I don't think my course has prepared me theoretically for this problem, does anyone have an elementary approach to it?

• – Bill Dubuque Jul 30 '16 at 13:41

## 2 Answers

We have $$x^4-10x^2+1 =(x^2-1)^2-8x^2 = (x^2-1-2bx)(x^2-1+2bx)$$ if $2=b^2$, or $$x^4-10x^2+1 =(x^2+1)^2-12x^2 =(x^2+1-2cx)(x^2+1+2cx)$$ if $3=c^2$. Thus the only cases that remain are those where neither $2$ nor $3$ is a square - but then $6$ is a square and with $6=c^2$, we find $$x^4-10x^2+1 =(x^2-5)^2-24=(x^2-5-2c)(x^2-5+2c).$$

(Note that we do not even need quadratic reciprocity or similar "advanced" stuff - all we need is that the product of non-squares is square, which follows from the fact that $\Bbb F_p^\times$ is cyclic).

• Just to check my reasoning - it follows from the fact the multiplicative group is cyclic because all elements are a power of some generator, and thus the product of two odd powers is even, right? – abhalphiest Jul 30 '16 at 13:44

Another simple explanation comes from the fact that the zeros of $m(x)=x^4-10x^2+1$ are $$x=\pm\sqrt2\pm\sqrt3$$ with all four sign combinations. So if $p$ is a prime, then you get the splitting field of $m(x)$ over $K=\Bbb{F}_p$ by adjoining $\sqrt2$ and $\sqrt3$. Because up to isomorphism the field $K$ has only a single quadratic extension, namely $L=\Bbb{F}_{p^2}$, we immediately see that $m(x)$ splits into linear factors over $L$. This is because $\sqrt2$ and $\sqrt3$ are both elements of $L$. Consequently $m(x)$ splits into quadratics at worst over $K$.

From this way of looking at it it is obvious how to generalize this. Any biquadratic polynomial with zeros of the form $\pm\sqrt{d_1}\pm\sqrt{d_2}$ for some integers $d_1,d_2$ will split into quadratic (or linear) factors modulo $p$ for all primes $p$. From basic field theory we see that it will be irreducible over $\Bbb{Q}$ whenever the field $F=\Bbb{Q}(\sqrt{d_1},\sqrt{d_2})$ is a degree four extension over $\Bbb{Q}$.

See Qiaochu's answer here for more Galois theory.

• Of course, the fact that by adjoining a single square root to $\Bbb{F}_p$ you get all of them also follows from the observation that modulo $p$ the product of two quadratic non-residues is a quadratic residue. – Jyrki Lahtonen Aug 1 '16 at 6:47