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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?

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