# Irreducibility using Fermat's Little Theorem

Prove that $x^4 + x^2 - 1$ is irreducible

My attempt:

I will try to prove that $x^4 + x^2 - 1$ is irreducible over $\mathbb{Z} / 3\mathbb{Z}$. Is this approach valid:

By Fermat's little theorem, $x^4 \equiv x^2 \pmod{3}$. Therefore, $x^4 + x^2 - 1 \equiv 2x^2- 1 \pmod{3}$. It can be checked that $2x^2 - 1$ has no linear factors over $\mathbb{Z} / 3\mathbb{Z}$ and hence it is irreducible.

Is this correct?

• It is true that your polynomial is irreducible $\pmod 3$ but you can't prove it that way. After all, $x^3$ is obviously reducible but, as you remark, it is $\equiv x \pmod 3$. – lulu Mar 19 '18 at 13:15
• Yes I've edited it. It was a typo. Thank you. – Confuse Mar 19 '18 at 13:16
• Note: despite the comments to the contrary, you can not show irreducibility this way. Again, $x^3$ is obviously reducible but $x$ isn't. – lulu Mar 19 '18 at 13:17
• Yes that makes sense. So what is the source of the error. – Confuse Mar 19 '18 at 13:18
• The error is that the polynomials $x^4$ and $x^2$ are not the same. To show irreducibility you need to rule out all possible factorings. – lulu Mar 19 '18 at 13:23

Despite the comments to the contrary, you can not show irreducibility this way. After all, the cubic $x^3$ is manifestly reducible $\pmod 3$ but $x^3\equiv x\pmod 3$. Keep in mind that congruences of that form only mean that they take same values in the finite field of three elements. $x^3$ and $x$ are not the same polynomial.

It is true that $x^4+x^2-1$ is irreducible $\pmod 3$ but I don't see any way to show other than by looking for factors.

Your proof is not entirely correct. Or rather, it's not complete. You have shown that the polynomial doesn't have any roots in $\Bbb Z_3$, (and hence no roots in $\Bbb Z$), and therefore no linear factors. But that only implies irreducibility if the degree of the original polynomial is $3$ or less. (you could've reached this result without reducing modulo $3$ by the rational root theorem: neither of $\pm 1$ is a root, and therefore there are no rational roots.)

Beware: Even though Fermat's little theorem says that the function $x\mapsto x^4$ is the same as the function $x\mapsto x^2$ for $x\in \Bbb Z_3$, the polynomials $x^4$ and $x^2$ are distinct.

You still have to show that it doesn't factor into two second-degree polynomials.

Trying to complement the other answer by explaining why $f(x)=x^4+x^2-1$ is, indeed, irreducible over the prime field $\Bbb{Z}_3$.

With a few basic facts about finite fields in place one easy way to see this is to observe that $f(x)=g(x^2)$, where $g(x)=x^2+x-1$. The key facts that I need are

• $g(x)$ is irreducible over $\Bbb{Z}_3$. This is easy. It has no zeros in $\Bbb{Z}_3$ and for a quadratic that's enough.
• $g(x)$ is not a factor of $x^4-1$ in $\Bbb{Z}_3[x]$. This is also easy, for the factorization of $x^4-1$ into irreducibles is $(x^4-1)=(x^2+1)(x+1)(x-1)$.

So by the first bullet the zeros of $g(x)$ are elements of the field of nine elements $\Bbb{F}_9$. Because the non-zero elements of that field form a cyclic group of order eight, by the second bullet, the zeros of $g(x)$ are roots of unity of order eight.

Because $2\mid 8$ this is enough to conclude that any root $\alpha$ of $f(x)=g(x^2)$ (in some extension field) must be of order sixteen: the order is obviously a factor sixteen, but it cannot be a proper factor, because otherwise the order of $\alpha^2$ would be a proper factor of eight.

So $\alpha$ cannot be an element of $\Bbb{F}_9$ (the order of an element cannot exceed the order of the group). All the quadratic polynomials in $\Bbb{Z}_3[x]$ have their zeros in $\Bbb{F}_9$, so $f(x)$ has no such quadratic factors, and is thus irreducible.