Find the degree of the extension of $\sqrt{3+2\sqrt{2}}$ over $\mathbb{Q}$.

Question

I have done the following work:

Let $\alpha = \sqrt{3+2\sqrt{2}}$. Square both sides, so $\alpha^2 = 3+2\sqrt{2}$.

Then $(\alpha^2 - 3)^2 = (2\sqrt{2})^2$. Simplify, we can get $\alpha^4-6\alpha^2+1=0$. This means that $\alpha$ is a root of the monic polynomial $p(x)=x^4-6x^2+1 \in \mathbb{Q}[x]$. Then this means $\alpha$ is algebraic over $\mathbb{Q}$.

Now, if $p(x)$ is irreducible over $\mathbb{Q}$, then the degree of the extension field $\mathbb{Q}(\sqrt{3+2\sqrt{2}})$ is equal to the degree of $p(x)$, which is 4. However, $p(x)$ is reducible over $\mathbb{Q}$ since \begin{align*} p(x)&=x^4-6x^2+1\\ &=(x^4-2x^2+1)-4x^2\\ &= (x^2-1)^2-4x^2\\ &=(x^2-2x-1)(x^2+2x-1) \end{align*} I saw another post saying that this means the degree of the extension field is 2. I don't understand this at all. Could someone help me with this? Thanks a lot.

Answer 1

Of course, the degree is not 1, because $\alpha$ is irrational. It is at most two, because it's a root of either $x^2-2x-1$ or $x^2+2x-1$, so the basis over $\mathbb Q$ will be $(1, \alpha)$, since $(1, \alpha, \alpha^2)$ is linearly dependent (because appropriate polynomial has a root at $\alpha$). Therefore, the degree is exactly two.

Answer 2

Because $$\sqrt{3+2\sqrt2}=\sqrt{1+2\sqrt2+2}=\sqrt{(1+\sqrt2)^2}=1+\sqrt2,$$ which says that the degree is equal to $2$.