Irreducibility using Fermat's Little Theorem

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

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.