Minimal polynomial and splitting field

Question

Let $f(x) ∈ \mathbb{Q}[x]$ be the minimal polynomial of $\alpha =\sqrt{2 + \sqrt{2}}$. I wish to find $f(x) = $ min$_\mathbb{Q}(α)$ and show that the splitting field of $f(x)$ is $\mathbb{Q}(α)$. I computed $f(x)= x^4 − 4x^2 + 2$ but I am not sure how to proceed.

Answer 1

First to prove that the polynomial is irreducible you can use the Eisenstein criterion for $p=2$.

For the second part you can notice that the roots are $\pm\sqrt{2\pm\sqrt{2}}$. To prove that the $\mathbb{Q}(\alpha)$ is the splitting field you can use the fact that $\pm \frac{\sqrt{2}}{\alpha}$ is a root of the polynomial. Indeed:

$$\left(\pm\frac{\sqrt{2}}{\alpha}\right)^4 - 4\left(\pm\frac{\sqrt{2}}{\alpha}\right)^2 + 2 = \frac{4}{\alpha^4} - \frac{8}{\alpha^2} + 2 = \frac{2}{\alpha^4}(2 - 4\alpha^2 + \alpha^4) = 0$$

Now obviously $\sqrt{2} = \alpha^2 - 2 \in \mathbb{Q}(\alpha)$ and so $\pm\frac{\sqrt{2}}{\alpha} \in \mathbb{Q}(\alpha)$. Thus the roots are $\pm \alpha, \pm \frac{\sqrt{2}}{\alpha}$ (make sure to prove that in fact the 4 distinct roots) and as they are all in $\mathbb{Q}(\alpha)$ we have proven the claim.

Answer 2

As noted by @Stefan4024, the polynomial $f(x) = x^4-4x^2+2$ is irreducible so it is the minimal polynomial of $\alpha$.

Notice that

$$f(x) = (x^4-4x^2+4)- 2 = (x^2-2)^2- 2$$

so the roots are $\sqrt{2+\sqrt2}, -\sqrt{2+\sqrt2}, \sqrt{2-\sqrt2},-\sqrt{2-\sqrt2}$.

Now clearly $-\sqrt{2+\sqrt2} = -\alpha \in \mathbb{Q}(\alpha)$.

Also notice that $\sqrt{2} = \alpha^2-2 \in \mathbb{Q}(\alpha)$ so $$\sqrt{2-\sqrt2} = \frac{\sqrt{2}}{\sqrt{2+\sqrt{2}}} = \frac{\alpha^2-2}{\alpha} \in \mathbb{Q}(\alpha)$$

and then also $-\sqrt{2-\sqrt2} \in \mathbb{Q}(\alpha)$.

We conclude that $\mathbb{Q}(\alpha)$ is the splitting field of $f(x)$ over $\mathbb{Q}$.