Finding a basis for $K=\mathbb{Q}(\sqrt{5-2\sqrt{2}})$

Let $K=\mathbb{Q}(\sqrt{5-2\sqrt{2}})$. I need to Compute $[K:\mathbb{Q}]$ and find a basis for $K$ as a vector space over $\mathbb{Q}$. I also need to find a polynomial $p(x)\in \mathbb{Z}[x]$, such that $K=\mathbb{Q}[x]/(p(x))$.

Here is what I have so far :

Let $\alpha = \sqrt{5-2\sqrt{2}}$, $$\alpha^2=5-2\sqrt{2} \Rightarrow (5-\alpha^2)^2=8 \Rightarrow \alpha^4-10\alpha^2+17=0.$$ I can prove that $p(x) := x^4-10x^2+17$ is irreducible over $\mathbb{Q}$, then I can say that $p(x)$ is the minimal polynomial of $\alpha$. Therefore $$[\mathbb{Q}(\alpha):\mathbb{Q}]=4$$

But I am not sure how to find basis elements, The thing that comes to my mind is $$\mathcal{B}=\{1,\sqrt2,\sqrt{5-2\sqrt{2}},\sqrt2 \cdot \sqrt{5-2\sqrt{2}} \}$$ But how can I prove that the elements in $\mathcal{B}$ are $\mathbb{Q}$-linearly independent ?

• Very good approach, the basis would be $\{1,\alpha, \alpha^2,\alpha^3\}$. Nov 24 '14 at 11:39

The following is a standard argument:

Let $K = F(a)$ for some $a$ algebraic over $F$, and let $[K:F]=d$. Then the minimal polynomial $m(x) \in F[x]$ of $a$ has degree $d$ and $1,a,a^2, \dots, a^{d-1}$ are a basis of $F$ over $K$.

To see this, you can note that $1,a,a^2, \dots, a^{d-1}$ are $d$ elements which are linearly independent (otherwise you would have a polynomial of degree $d-1$ vanishing on $a$).

In your particular case, you have just that $1, \alpha, \alpha^2, \alpha^3$ is a basis of $K$ over $\mathbb{Q}$.

• This works, too. The basis proposed OP is in the spirit of constructing a basis for a tower of extensions as pairwise products of elements of the two bases. Nov 24 '14 at 11:44

Your suggested basis is correct. The general principle at work is that if $L/K/F$ is a tower of finite extensions of fields such that $\{\alpha_1,\alpha_2,\ldots, \alpha_n\}$ is a basis for $K/F$ and $\{\beta_1,\beta_2,\ldots,\beta_m\}$ is a basis for $L/K$, then $$\{\alpha_i\beta_j\mid 1\le i\le n, 1\le j\le m\}$$ is a basis for $L/F$.

In the present case $F=\Bbb{Q}$, $K=\Bbb{Q}(\alpha^2)=\Bbb{Q}(\sqrt2)$ and $L=K(\alpha)$. As you know that $p(x)$ is irreducible, then $[L:F]=\deg p(x)=4$, so we can deduce that $[L:K]=2$. Therefore $\{1,\alpha\}$ is a basis for $L/K$. And, of course, $\{1,\sqrt2\}$ is a basis for $K/F$.