# How to find a basis for $\mathbb{Q}(\sqrt{2+\sqrt{3}})$ over $\mathbb{Q}$

basis for $\mathbb{Q}(\sqrt{2+\sqrt{3}})$ over $\mathbb{Q}$

The minimum polynomial that I found is $x^4+4x^2+1$. Since that is degree $4$ the basis should also have four elements. However, I do not know which elements that would be. The two that I can think of are $1$ and $\sqrt{2+\sqrt{3}}$. Should $(1+\sqrt{2+\sqrt{3}})$ also be in the basis? Then, maybe, $((\sqrt{2+\sqrt{3}})+2+\sqrt{3})$ is also in the basis? Could someone help?

• Let $\alpha=\sqrt{2+\sqrt{3}}$, then the basis is $1,\alpha,\alpha^2,\alpha^3$ May 1 '18 at 18:07
• $1,x,x^2,x^3$ must be linearly independent and generate the whole field.
– user551819
May 1 '18 at 18:07
• Solve the biquadratic equation $x^4+4x^2+1=0$ by setting $y=x^2$ and $y^2+4y+1=0$. May 1 '18 at 18:08
• If you find a basis $\{ \alpha_1, \alpha_2 \}$ for $\mathbb{Q}(\sqrt{3})$ over $\mathbb{Q}$, then find a basis $\{ \beta_1, \beta_2 \}$ for $\mathbb{Q}(\sqrt{2+\sqrt{3}})$ over $\mathbb{Q}(\sqrt{3})$, then $\{ \alpha_i \beta_j \}$ will be a basis for $\mathbb{Q}(\sqrt{2+\sqrt{3}})$ over $\mathbb{Q}$. May 1 '18 at 18:10

Let's assume that $\alpha \in K$ is algebraic over $\mathbb{Q}$; that is, there is a polynomial (the minimal polynomial) $p_\alpha(x):=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, where $a_i \in \mathbb{Q}$, so that $\alpha$ is a root of $p_\alpha(x)$, i.e. $p_\alpha(\alpha)=0$. Then the degree of the extension $K/\mathbb{Q}$ is the degree of $p_\alpha(x)$. Moreover, $\{1=\alpha^0,\alpha,\alpha^2,\ldots,\alpha^{n-1}\}$ is a basis for $K$ as a $\mathbb{Q}$-vector space.
In your problem, you have $\alpha=\sqrt{2+\sqrt{3}}$. You found the minimal polynomial to be $x^4+4x^2+1$ (note that it should be $x^4-4x^2+1$ so there is a sign error somewhere). Just be sure to prove that this polynomial is irreducible. Then by the comments above, a basis for $K$ over $\mathbb{Q}$ is $\{1,\sqrt{2+\sqrt{3}}, \sqrt{2+\sqrt{3}}^{\,2},\sqrt{2+\sqrt{3}}^{\,3}\}$.