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I want to show, that $X^4-4X^2-1$ is irreducible over $\mathbb{Q}[X]$. Since there are no roots in $\mathbb{Q}$ it has to be:

$(X^4-4X^2-1)=(X^2+aX+b)(X^2+cX+d)$

Comparision of the coefficients shows that this can not hold over $\mathbb{Q}$. But I am searching for an easier way to show this which uses less calculation.

I tried this:

$X^4-4X^2-1=X^4-4X^2+4-5=(X^2-2)^2-5=(X^2-2-\sqrt{5})(X^2-2+\sqrt{5})$

Which is obviously not in $\mathbb{Q}[X]$ anymore. But would this calculation be enough to show, that $X^4-4X^2-1$ is irreducible. How do I know, that this is the only possible way to factor it?

Thanks in advance.

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    $\begingroup$ This doesn't answer the exact question being asked here, but as a tangential aside, this polynomial is irreducible $\pmod{p}$ for a small $p$ and hence irreducible over $\mathbb{Q}$. $\endgroup$
    – Kaj Hansen
    Sep 26 '17 at 22:59
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    $\begingroup$ Look up the rational root theorem. $\endgroup$
    – Doug M
    Sep 26 '17 at 23:00
  • $\begingroup$ @KajHansen: But using the reduction theorem (which I think you are refering too) not involve a similar calculation, which might be easier? $\endgroup$
    – Cornman
    Sep 26 '17 at 23:04
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    $\begingroup$ @DougM: The rational root theroem should not work here, since the degree of the polynomial is 4. The non existence of a root does not mean the polynomial is irreducible. $\endgroup$
    – Cornman
    Sep 26 '17 at 23:05
  • $\begingroup$ @Cornman, the nice thing about it is that the number of irreducible quadratic polynomials in this particular $\mathbb{F}_p$ is quite small--small enough to just try them all. On the other hand, the infinitude of polynomials in $\mathbb{Q}[x]$ makes the current endeavor tricky. $\endgroup$
    – Kaj Hansen
    Sep 26 '17 at 23:05
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Your polynomial has no rational roots. Therefore, if it was reducible in $\mathbb{Q}[x]$, then it would have to be the product of two quadratic polynomials $p_1(X),p_2(X)\in\mathbb{Q}[X]$. Each of them either has to real roots or two complex non-real roots.

The roots of your polynomial are $\pm\sqrt{\sqrt5+2}$ and $\pm i\sqrt{\sqrt5-2}$. Of these, only the last two are complex non-real. But$$\left(X-i\sqrt{\sqrt5-2}\,\right)\left(X+i\sqrt{\sqrt5-2}\,\right)=X^2-2+\sqrt5\notin\mathbb{Q}[X].$$Therefore, your polynomial is irreducible.

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  • $\begingroup$ Therefore my factorisation would be enough to show that the polynomial is irreducible over $\mathbb{Q}[X]$? $\endgroup$
    – Cornman
    Sep 26 '17 at 23:11
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    $\begingroup$ @Cornman I don't see why. The fact that your decomposition does not work doesn't imply that no decomposition works. For instance, $$x^4-5x^2+6=\left(x^2-\left(\sqrt2+\sqrt3\right)x+\sqrt6\right)\left(x^2+\left(\sqrt2+\sqrt3\right)x+\sqrt6\right),$$but $x^4-5x^2+6$ is reducible, since it is equal to $(x^2-3)(x^2-2)$. $\endgroup$ Sep 26 '17 at 23:16
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Let's write $f(x)=x^4-4x^2-1$. Note that since $f(x)=f(-x)$, the roots of $f$ come in positive and negative pairs; say the roots are $a$, $-a$, $b$, and $-b$. Up to renaming the roots, then, there are two cases of factorizations with integer coefficients that $f$ could have. We could have $f(x)=g(x)h(x)$ where $g(x)=(x-a)(x+a)=x^2-a^2$ and $h(x)=(x-b)(x+b)=x^2-b^2$. But that would mean that $x^2-4x-1$ factors as $(x-a^2)(x-b^2)$, and we know $x^2-4x-1$ is irreducible.

The other possibility is that we have a factorization like $f(x)=g(x)h(x)$ with $g(x)=(x-a)(x-b)=x^2-(a+b)x+ab$ and $h(x)=(x+a)(x+b)=x^2+(a+b)x+ab$. In this case, though, $ab$ would be an integer and the constant term of $f(x)$ would be the square of $ab$. Since the constant term is $-1$, this is impossible.

More generally, a similar argument shows that if $e(x)\in\mathbb{Z}[x]$ is irreducible and monic of degree $n$, then $f(x)=e(x^2)$ is irreducible unless its constant term is $(-1)^n$ times a perfect square. See the answers to For which monic irreducible $f(x)\in \mathbb Z[x]$ , is $f(x^2)$ also irreducible in $\mathbb Z[x]$? for more details.

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