# [Math] $X^4-4X^2-1$ irreducible over $\mathbb{Q}[X]$

abstract-algebrafield-theoryirreducible-polynomialspolynomials

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?

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.