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

My attempts:

I cannot apply the Eisenstein's criteria here, because there is no prime number that divides the constant term i.e. $1$ Taking a translation of the form $x \rightarrow x+a$ does not solve this issue either.

Next, I tried the mod tests: $\operatorname{mod}2$ doesn't work since $x^4-x^2+1=(x^2+x+1)(x^2-x+1)$, similarly in $\operatorname{mod}3$ $x^4-x^2+1=(x^2+1)^2$. Now I can go on and maybe eventually find a $\operatorname{mod}p$ that works, but that is very time consuming, specially in examinations.

So, I'll use the rational root test. The possibilities for roots are $\pm 1$ and it is easy to see that neither is a root.

The only possibility left then are quadratic factors, say, $(x^2+ax+b)(x^2+cx+d)=x^4-x^2+1$

This gives me a set of equations $bd=1, a+c=0, b+d+ac=-1$. So either $b=d=1$, in which case $a=\pm \sqrt3 \notin \mathbb{Q}$, or $b=d=-1$, which gives $a=\pm i \notin \mathbb{Q}$.

So such factorization is not possible and hence the given polynomial is irreducible.

Is this solution correct? Also, is there an easier way to solve this? Thank you.

• The polynomial is the $12$ th cyclotomic polynomial. Sep 2, 2017 at 18:44
• Or better note that the given polynomial can be written as $x^{4}+2x^{2}+1-3x^{2}$ and thus factorize it as $(x^{2}+\sqrt{3}x+1)(x^{2}-\sqrt{3}x+1)$. Sep 2, 2017 at 18:47
• For finding its roots I recommend writing it in the form $(x^6+1)/(x^2+1)$. Assuming you know how to use complex polar form to find all the roots. Sep 2, 2017 at 18:49
• @RoddyMacPhee: Since the factorization is unique (via fundamental theorem of algebra) it is clear that the factorization is not possible over $\mathbb{Q}$. Sep 2, 2017 at 18:50
• Curiously, this polynomial is not irreducible modulo any prime. That is because its Galois group has no elements of order four (the Galois group is Klein four). Alternatively, the fact $12\mid (p^2-1)$ for all primes $p>3$ implies that $\Phi_{12}$ has a zero in $\Bbb{F}_{p^2}$ and hence a quadratic factor modulo $p$. Sep 2, 2017 at 18:55

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 $a^2=3$.

• The polynomial is $x^4-x^2+1$, not $x^4+x^2+1$.
– Dave
Sep 2, 2017 at 18:59
• Whoops, yes, fixed. Same argument applies, and actually works for $x^4-x^2+1$. :) Sep 2, 2017 at 19:01

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.

$$(x^4-x^2+1)(x^2+1) = x^6+1$$ 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.

• You were may be the fourth or fifth viewer to make the observation that this is $\Phi_{12}$. So without a proof this is IMO worth a comment only. Sep 2, 2017 at 21:08
• @JyrkiLahtonen: a proof of what? The irreducibility of cyclotomic polynomials should be well-known, and the equality between $\Phi_{12}$ and $x^4-x^2+1$ is proved through the shown decomposition, matching the decomposition given by the Moebius inversion formula. Sep 2, 2017 at 21:12
• Yes, it is well known, but apparently not to the asker, so it is not useful to just bluntly state that this polynomial is irreducible. Not even a link. And the proof is not as simple as you claim. Of course, by assuming that the Galois group is what it is, the proof does become trivial. Sep 2, 2017 at 21:27

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 $\mathbb{Q}[x]$.

You can use wonderful criterion of Murty's (see Theorem 1):

Let $f(x)=a_mx^m+a_{m-1}x^{m-1}+\dots+a_1x+a_0$ be a polynomial of degree $m$ in $\mathbb{Z}[x]$ and set $$H=\max_{0\leq i\leq m-1} |a_i/a_m|.$$ If $f(n)$ is prime for some integer $n\geq H+2$, then $f(x)$ is irreducible in $\mathbb{Z}[x]$.

In this case it works because $f(3)=73$ is a prime.