As title says, I'm asked to prove that the polynomial $x^4-6x^2+4$ is irreducible in $\mathbb{Q}[x]$. This is what I tried so far:

• I've noticed it has no roots in $\mathbb{Q}$, since if it had then they would be of the form $a/b$ where $a|4$ and $b|1$ (and none of them satisfy the polynomial).

• I've also tried considering the polynomial (actually its image via ring morphism from $\mathbb{Z}[x]$ to $\mathbb{Z_p}[x]$) in $\mathbb{Z_2}$,$\mathbb{Z_3}$,$\mathbb{Z_5}$,$\mathbb{Z_7}$ but it turns out it is reducible over all of these rings. Also wolframalpha told me it is reducible in many other $\mathbb{Z_p}$.

• Finally I tried applying Eisenstein's criterion to the shifted polynomial $p(x+1), p(x+2), p(x+3)$ but had no success.

Seems to me that I've used all the tools that I've got. But maybe there are some consideration about its roots in a splitting field that can be done in order to prove its irreducibility, so I'm asking for your help. Thank you very much indeed.

  • 1
    $\begingroup$ You can easily compute the roots of your polynomial... If $x_1,...,x_4$ are it's root, prove that $(x-x_1)(x-x_i)$ for $i=2,3,4$ is not in $\mathbb Q[X]$ and the claim follow. $\endgroup$
    – Surb
    Nov 15 '20 at 11:16
  • $\begingroup$ See also here. $\endgroup$ Nov 15 '20 at 11:40
  • $\begingroup$ You could write $x^4-6x^2+4=(x^2+ax+b)(x^2+cx+d)$ with integers $a,b,c,d$ and compare the cofficients. Then notice that given system of equations has no solution in integers. (also notice that by Gauss Lemma in this case irreducibiliy over rationals is equivalent to irreducibility over integers) $\endgroup$
    – Sil
    Nov 15 '20 at 12:12

This polynomial is trickier to work with, because Eisenstein and mod $p$ thinking are both not helpful, as you say.

Instead, one can easily compute the roots of the given polynomial. Just putting $y=x^2$ , and solve $y^2-6y+4 = 0$ i.e. $y = 3 \pm \sqrt 5$. Thus, we get that the four roots of $x$ are $\pm \sqrt{3 \pm \sqrt 5}$.

Now, none of these roots are rational, so the factor theorem tells you that there can't be a degree one factor. However, if there is a degree two factor, say $x^2+\alpha x + \beta$, then two of the above numbers are its roots, and in particular the sum and product of those two numbers must be rational (using Vieta's formulas).

Therefore, the question reduces to :

Are there two numbers among $\pm\sqrt{3\pm \sqrt 5}$ such that both their product and sum are rational?

The answer to that is a clear no : for the product to be rational, we must have the pair to be either $\sqrt{3 \pm \sqrt 5}$ or $- \sqrt{3 \pm \sqrt 5}$ (easy to see) , but in either case the sum of the two said numbers is irrational.

For more on proving irreducibility when you can find the roots of the polynomial explicitly, read up on "proving irreducibility via Galois theory". Basically, if you want to prove that a polynomial is irreducible, then you find some elements of $\mathbb C$ such that all the roots of this polynomial are contained in the field containing $\mathbb Q$ and all those elements (for example, in this case $\mathbb Q$ and $\sqrt{3 + \sqrt 5}$ alone, would have worked). Now Galois theory will tell you that this field is an extension of $\mathbb Q$ of degree its minimal polynomial, and finding the degree can be done using the tower property of fields, for example.

  • $\begingroup$ Thank you very much indeed, your comment definitely solved my problem. $\endgroup$ Nov 15 '20 at 11:42
  • $\begingroup$ You are welcome, @MatteoRossi $\endgroup$ Nov 15 '20 at 13:09

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