Finding $p(x)$ such that $\mathbb{Q}(\sqrt{1+\sqrt{5}})$ is ring isomorphic to $\mathbb{Q[x]}/\langle p(x)\rangle$ I am trying to find a polynomial $p(x)$ in $\mathbb{Q}[x]$ such that $\mathbb{Q}(\sqrt{1+\sqrt{5}})$ is ring isomorphic to $\mathbb{Q[x]}/\langle p(x)\rangle$.
This is what I tried to do:
Consider the polynomial $p(x)=x^4-2x^2-4$. Solving the quadratic equation we get the roots $\sqrt{1+\sqrt{5}},-\sqrt{1+\sqrt{5}}, \sqrt{1-\sqrt{5}} \text{ and} -\sqrt{1-\sqrt{5}}$. We will try to show that $p(x)$ is irreducible over $\mathbb{Q}$. suppose $p(x)=r(x)q(x)$. If $p(X)$ is not irreducible, none of $r$ and $q$ has degree $0$. None can have degree $1$ since none of the roots is in $\mathbb{Q}$. So we are left with the case when each has degree 2. 
Suppose $r(x)=x^2+ax+b$. Let $r(x)=(x-a_1)(x-a_2)$ with $a_1,a_2$ in the set of roots of $p(x)$. We can do some casework to see that $a_1+a_2=-a \in \mathbb{Q}$ iff $(a_1,a_2)=(\sqrt{1+\sqrt{5}},-\sqrt{1+\sqrt{5}})$ or $(\sqrt{1-\sqrt{5}},-\sqrt{1-\sqrt{5}})$. This implies $a_1a_2=b \not \in \mathbb{Q}$, a contradiction to the fact that $r(x)\in Q[x]$. Hence $p(x)$ is irreducible and since $\sqrt{1+\sqrt{5}}$ is a zero of $p(x)$, we have the conclusion that $\mathbb{Q}(\sqrt{1+\sqrt{5}})$ is ring isomorphic to $\mathbb{Q[x]}/\langle p(x)\rangle$.
Question 1: Is this proof correct?
Question 2: Could anyone suggest an alternate proof of $\mathbb{Q}(\sqrt{1+\sqrt{5}})$ is ring isomorphic to $\mathbb{Q[x]}/\langle p(x)\rangle$. Thanks.
 A: Observe that $\sqrt{5} \in \mathbb{Q}(\sqrt{1+\sqrt{5}})$ so $\mathbb{Q}(\sqrt{5})$ is an intermediate field of the extension $\mathbb{Q}(\sqrt{1+\sqrt{5}}) / \mathbb{Q}$. It's easy to see that $\sqrt{1+\sqrt{5}}$ is not in $\mathbb{Q}(\sqrt{5})$ and therefore $[\mathbb{Q}(\sqrt{1+\sqrt{5}}) : \mathbb{Q}(\sqrt{5})] = 2$. Because $[\mathbb{Q}(\sqrt{5}) : \mathbb{Q}] = 2$, you have  $[\mathbb{Q}(\sqrt{1+\sqrt{5}}) : \mathbb{Q}] = 4$. The result follows because $\sqrt{1+\sqrt{5}}$ is a root of $p(x) \in \mathbb{Q}[x]$.
A: A result is that if $ f \in K[x] $ is irreducible and $ f(\alpha) = 0 $, then we have an isomorphism $ K[x]/(f) \cong K(\alpha) $. Therefore, it suffices to find an irreducible polynomial of which $ \alpha = \sqrt{1 + \sqrt{5}} $ is a root. We can see that $ (\alpha^2 - 1)^2 - 5 = 0 $, so $ \alpha $ is a root of $ X^4 - 2X^2 - 4 $. This polynomial has no roots in the finite field $ \mathbb{F}_3 $, so if it reduces in $ \mathbb{F}_3[X] $ it must split into quadratic factors, and therefore have a root in $ \mathbb{F}_9 $. Let $ w \in \mathbb{F}_9^{\times} $ be such a root. Then, we have
$$ w^8 - 1 = (w^4 - 1)(w^4 + 1) = (2w^2 + 3)(2w^2 + 5) = w^4 + w^2 = 3w^2 + 4 = 1 \neq 0 $$
which is impossible for an element of the multiplicative group $ \mathbb{F}_9^{\times} $. We conclude that $ X^4 - 2X^2 - 4 $ is irreducible in $ \mathbb{F}_3[X] $, and therefore in $ \mathbb{Q}[X] $. We may then pick $ p(X) = X^4 - 2X^2 - 4 $.
Your proof is fine, I felt that proving irreducibility this way instead of doing casework was faster. In any case, it is an alternative solution.
