Show that $x^4-x^2+1$ is irreducible over $\mathbb{Q}$

Solution 1:

There are no rational roots, so no linear factors.

If $p(x)$ is a factor of $x^4-x^2+1$ then $p(-x)$ is, too.

If $x^4-x^2+1 = (x^2+a)(x^2+b)$ then $x^2-x+1=(x+a)(x+b)$. Show that $x^2-x+1$ is irreducible.

On the other hand, you'd have to have $x^4-x^2+1=(x^2+ax+b)(x^2-ax+b)$ where $b^2=1$ and $a\neq 0$. This means that $x^4+(2b-a^2)x^2+b^2 = x^4-x^2+1$.

So you need $2b-a^2=-1$. If $b=-1,$ then this means $a^2=-1$, and if $b=1$ then $a^2=3$.

Solution 2:

Another approach which I saw Robert Israel use here would be to note that $x^4-x^2+1$ takes on prime values for $x=\pm2,\pm3,\pm4,\pm5$ and $\pm9$. That's ten points, so that one of the quadratic factors would have to take on the value $\pm 1$ at least five times. Finally one of the quadratic factors would have to take on either $+1$ or $-1$ at least 3 times which is impossible for a quadratic, since a non-constant polynomial that takes the same value three times must have degree at least three.

Solution 3:

$$(x^4-x^2+1)(x^2+1) = x^6+1$$ implies: $$ x^4-x^2+1 = \frac{x^6+1}{x^2+1} = \frac{(x^{12}-1)(x^2-1)}{(x^6-1)(x^4-1)} = \Phi_{12}(x) $$ hence the LHS is irreducible over $\mathbb{Q}$ since it is the minimal polynomial of $\exp\left(\frac{2\pi i}{12}\right)$.
The irreducibility of cyclotomic polynomials is a well-known fact, proved here.

Solution 4:

Let $r(x)$ be the resolvent cubic of your polynomial. Then $r(x)=x^3-2x^2-3x$. The roots of $r(x)$ are $-1$, $0$, and $3$, none of which is the square of a non-zero rational number. Furthermore, your polynomial has no rational root and the coefficient of $x$ in $r(x)$ is not a perfect square in $\mathbb Q$. Therefore your polynomial is irreducible in $\mathbb{Q}[x]$.