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

Question

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.

You can easily find roots of this polynomial and thus confirm the factorization as a product of two quadratics and each of them has some irrational coefficients. — Paramanand Singh, Sep 02 '17 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 $(x^{2}+\sqrt{3}x+1)(x^{2}-\sqrt{3}x+1)$. — Paramanand Singh, Sep 02 '17 at 18:47
@ParamanandSingh but the sqrt of 3 is not rational so this is not showing what is wanted. — , Sep 02 '17 at 18:48
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. — Jyrki Lahtonen, Sep 02 '17 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} $. — Paramanand Singh, Sep 02 '17 at 18:50
As Peter said this is the 12th cyclotomic polynomial. They are all irreducible, but the proof is not quite trivial in general. In the present case you can still find the zeros in closed form, and, following Paramanand Singh's advice, check that no combination gives rise to a factor with rational coefficients. Edit: I corrected the link to point at Paramanand's blog, because that is more thorough. — Jyrki Lahtonen, Sep 02 '17 at 18:52
Finding enough primes of the form $x^4-x^2+1$ also shows the irreducibility. In the range $[1,100]$ there are already $29$ $x$-values such that $x^4-x^2+1$ is prime , more than enough. Does anyone know, how many primes we exactly need depending on the degree to ensure the irreducibility ? — Peter, Sep 02 '17 at 18:54
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 $p$. — Jyrki Lahtonen, Sep 02 '17 at 18:55
The factorization over $\mathbb{R}$ that @Paramanand Singh: gave should take care of it. — orangeskid, Sep 02 '17 at 18:57
@JyrkiLahtonen, see also On the Reducibility of Cyclotomic Polynomials over Finite Fields. — lhf, Sep 02 '17 at 20:16
@lhf Thanks for the reference. I have covered that too many times here already, so I should hold back :-). For a while at least. — Jyrki Lahtonen, Sep 02 '17 at 20:59
@lhf: I apologize about the knee-jerk movement. The link actually is interesting. Thank you! — Jyrki Lahtonen, Sep 02 '17 at 21:06

Thomas Andrews · Accepted Answer · 2017-09-02T18:59:53.473

11

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 $a^2=3$.

edited Sep 02 '17 at 18:59

answered Sep 02 '17 at 18:58

Thomas Andrews

177,126

The polynomial is $x^4-x^2+1$, not $x^4+x^2+1$. – Dave Sep 02 '17 at 18:59
Whoops, yes, fixed. Same argument applies, and actually works for $x^4-x^2+1$. :) – Thomas Andrews Sep 02 '17 at 19:01

score 8 · Answer 2 · answered Sep 02 '17 at 19:46

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.

Jack D'Aurizio · Answer 3 · 2017-09-02T21:34:16.023

4

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

edited Sep 02 '17 at 21:34

answered Sep 02 '17 at 19:40

Jack D'Aurizio

353,855

2

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. – Jyrki Lahtonen Sep 02 '17 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. – Jack D'Aurizio Sep 02 '17 at 21:12
1

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. – Jyrki Lahtonen Sep 02 '17 at 21:27

score 2 · Answer 4 · answered Sep 02 '17 at 19:16

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 $\mathbb{Q}[x]$.

score 1 · Answer 5 · answered Apr 27 '18 at 06:50

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.

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

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