# 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 '19 at 11:03
• is the basis correct? i wrote this in an exam I want to make sure @lhf
– user515599
Feb 6 '19 at 6:44

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 $$K/F$$.