Degree of extension $\mathbb{Q}(\sqrt{3+2\sqrt{2}})$ over $\mathbb{Q}$

Question

I have found the minimal polynomial $x^4-6x^2+1$ so degree should be $4$. But it is given $2$ in the book. What am I doing wrong?

Answer 1

$x^4-6x^2+1 = (x^2 - 2 x - 1) (x^2 + 2 x - 1)$ is not a minimal polynomial over $\mathbf{Q}$ as it is reducible. This shows you btw that $\sqrt{3+2\sqrt{2}}$ is a root of a degree two polynimial, giving the idea that it must be a square. From this, you can easily conclude.

Answer 2

Observe that $$3+2\sqrt{2}=1+2\sqrt{2}+2=1+2\sqrt{2}+(\sqrt{2})^2=(1+\sqrt{2})^2.$$ Hence, $$\sqrt{3+2\sqrt{2}}=1+\sqrt{2}$$ and the irreducible polynomial over $\mathbb{Q}$ $$f(x)=x^2-2x-1$$ has $1+\sqrt{2}$ as a root, we deduce that $[\mathbb{Q}(\sqrt{3+2\sqrt{2}}):\mathbb{Q}]\leq 2$. However, since $1+\sqrt{2}\notin \mathbb{Q}$, we have that the last inequality is indeed an equality.