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

Solution 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.

Solution 2:

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

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

Solution 3:

One may also note that $\mathbb{Q}(\sqrt{2})\subseteq \mathbb{Q}(\sqrt{3+2\sqrt{2}})$ and $x^2-3-2\sqrt{2}\in \mathbb{Q}(\sqrt{2})[x]$ kills$\sqrt{3+2\sqrt{2}}$. So determining the degree of $\mathbb{Q}(\sqrt{3+2\sqrt{2}})$ over $\mathbb{Q}(\sqrt{2})$ (sufficient to answer the original question by multiplicativity of degree) amounts to figuring out if $$ x^2-3-2\sqrt{2} $$ is irreducible in $\mathbb{Q}(\sqrt{2})$, which is easy since this is asking if it has a root in $\mathbb{Q}(\sqrt{2})$. Indeed, a solution to the system of equations derived from $$ (a+b\sqrt{2})^2-3-2\sqrt{2} $$ yields $a=b=1$ as a solution. Thus $\mathbb{Q}(\sqrt{2})=\mathbb{Q}(\sqrt{3+2\sqrt{2}})$ and the field extension is quadratic.