Find the splitting field of $x^4-4x^2+1$ over $\mathbb{Q}$

Question

I proved that the polynomial $x^4-4x^2+1$ is irreducible over $\mathbb{Q}$ , so if $a$ is a root of the polynomial, i got:

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

I don't know how to prove that the quadratic polynomial is reducible in order to prove that $\mathbb{Q(a)}$ is the splitting field. Thank you all.

Answer 1

It turns out that $\frac 1 a(=4a - a^3)$ is also a root of the polynomial.

Thus it suffices to check that $x^2 + a^2 - 4 = (x - \frac 1 a)(x + \frac 1 a)$, which is easy.

Answer 2

If $a$ is a root of $f(x)=x^4-4x^2+1$, then $f(1/a)=\frac{1}{a^4}-\frac{4}{a^2}+1=\frac{1}{a^4}(1-4a^2+a^4)=\frac{f(a)}{a^4}=0$, as $a$ is a root. So, your roots are $\pm a$ and $\pm\frac{1}{a}$. Thus your splitting field is $\mathbb{Q}(a)$. Based on GreginGre's comment, we can explicitly find $a$ as $a=\sqrt{2+\sqrt{3}}$.

Answer 3

You can also complete the square with varying the middle term, $$ 0=x^4−4x^2+1=(x^2−1)^2−2x^2=(x^2+1)^2−6x^2 $$ and use the binomial identities to factorize further. These real factorizations then allow a direct computation of the roots via the quadratic solution formula or a standard quadratic completion.

first variant $$0=x^2\pm\sqrt2x-1=(x\pm\sqrt{1/2})^2-3/2$$

second variant $$ 0=x^2\pm\sqrt6+1=(x\pm\sqrt{3/2})^2-\frac12 $$

Both ways lead to the same root set $\pm\sqrt{1/2}\pm\sqrt{3/2} \in\Bbb Q[\sqrt2,\sqrt3]$.