Determine the degree of the extension $\mathbb{Q}(\sqrt{3 + 2\sqrt{2}})/\mathbb Q$.

Question

The following is from a set of exercises and solutions.

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

The solution says that the degree is $2$ since $\mathbb{Q}(\sqrt{2}) = \mathbb{Q}(\sqrt{3 + 2\sqrt{2}})$. I understand that the LHS is a subset of the RHS since $$ \sqrt{2} = \frac{(\sqrt{3 + 2\sqrt{2}})^2 - 3}{2}. $$

How can the RHS be a subset of the LHS? In other words, how can $\sqrt{3 + 2\sqrt{2}}$ be in $\mathbb{Q}(\sqrt{2})$?

The author of the solution mentioned that $(1 + \sqrt{2})^2 = 3 + 2\sqrt{2}$ but I do not see how this helps.

Answer 1

This is Exercise 10 of $\S$13.2 of Dummit and Foote. The exercise immediately preceding it is the following:

Let $F$ be a field of characteristic $\neq 2$. Let $a,b$ be elements of the field $F$ with $b$ not a square in $F$. Prove that a necessary and sufficient condition for $\sqrt{a + \sqrt{b}} = \sqrt{m} + \sqrt{n}$ for some $m$ and $n$ in $F$ is that $a^2 - b$ is a square in $F$. Use this to determine when the field $\mathbb{Q}(\sqrt{a + \sqrt{b}})$ ($a,b, \in \mathbb{Q}$) is biquadratic over $\mathbb{Q}$.

Taking $a = 3$ and $b=8$, we find that $a^2 - b = 9 - 8 = 1$ is indeed a square, so the extension is biquadratic. Furthermore, one can actually determine $m$ and $n$ from $a$ and $b$ (I can say more about this if you like): $$ m = \frac{a + \sqrt{a^2 - b}}{2} \qquad n = \frac{a - \sqrt{a^2 - b}}{2} \, . $$ Thus for $a = 3$ and $b = 8$, we have $m = 2$ and $n = 1$, as claimed in the solution.

Answer 2

Since $\left(1+\sqrt2\right)^2=3+2\sqrt2$, we have $\sqrt{3+2\sqrt2}=1+\sqrt2$.

Therefore, $\mathbb{Q}\!\left(\sqrt{3+2\sqrt2}\right)=\mathbb{Q}\!\left(1+\sqrt2\right)=\mathbb{Q}\!\left(\sqrt2\right)$

Answer 3

$\alpha=\sqrt{3+2\sqrt{2}}\implies \alpha^4-6\alpha^2+1=0$.

So,$x^4-6x^2+1$ annihilates $\alpha$.Now if we could show that this polynomial is irreducible over $\mathbb Q$ then we are done.Now we use mod $3$ irreducibility test as in $\mathbb Z_3$ this polynomial reduces to $x^4+1=0$ which is irreducible over $\mathbb Z_3$ and hence irreducible over $\mathbb Z$ also.Hence it is irreducible over $\mathbb Q$ also by Gauss's lemma.