# Degree and basis of field extension $\mathbb{Q}[\sqrt{2+\sqrt{5}}]$

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?

• – lhf Feb 3 at 11:03
• is the basis correct? i wrote this in an exam I want to make sure @lhf – badatmath Feb 6 at 6:44