I want to find the degree and basis of the field extension $\mathbb{Q}(\sqrt{2+\sqrt{5}})$.
let $\alpha=\sqrt{2+\sqrt{5}}$. $$\alpha^2=2+\sqrt{5},\quad \alpha^4-4\alpha^2-1=0.$$
So possible minimal polynomial $f(X)=X^4-X^2-1$. $f$ is irreducible over $\mathbb{Q}$ since its roots are all non rationals and so its possible factors are of degree $2$ not $1$. Looking at the roots shows irreducibility and hence $f$ is the minimal polynomial and degree of extension is $4$.
My problem is with the basis. Since $\alpha^2=2+\sqrt{5}$, then $\sqrt{5}$ is in $\mathbb{Q}(\alpha)$. So $\mathbb{Q}(\alpha)$ has degree $4$ over $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{5})$ has degree $2$ over $\mathbb{Q}$ using Tower Law gives degree $\mathbb{Q}(\alpha)$ over $\mathbb{Q}\sqrt{5}$ is $2$.
So a basis for $\mathbb{Q}(\alpha)$ is the product of the bases. $$B=\{1,\sqrt{5},\alpha,\alpha \sqrt{5}\}$$
Is the basis and method correct?
This is correct.
Let $L/K/F$ be a tower of field extensions.
Let $\{ k_i\}_i$ be a basis for $K/F$.
Let $\{ l_j \}_j$ be a basis for $L/K$.
Then the product $\{ k_i \cdot l_j \}_{i,j}$ is a basis for $L/F$.
This is reasonable as it has the correct size by tower law.
We can express every element $l \in L$ in terms of this basis. First expand $l = \sum_{j} l_j c_j$ with each $c_j \in K$, then expand each $c_j = \sum_{i} k_i d_{j,i}$ and we have $l = \sum_{j,i} (l_j k_i) d_{j,i}$ with each $d_{j,i} \in F$.
Secondly it is a linearly independent basis, suppose $0 = \sum_{j,i} (l_j k_i) d_{j,i}$ then you can group the terms as a sum of $K$ coefficients in the basis of $L/K$ so it will be zero if each coefficient is zero. Now you can apply linear independence of $K/F$.