I am working through this problem but seem to have hit a roadblock, can anyone help?
Suppose that $[E:\mathbb{Q}] = 2$. Show that there is an integer $d$ such that $E = \mathbb{Q}(\sqrt{d})$ where $d$ is not divisible by the square of any prime.
Suppose $[E:\mathbb{Q}] = 2$. Let $\alpha\in E\setminus\mathbb{Q}$, there is also a minimal polynomial of $\deg \alpha\geq 2$ but cannot be more than 2 since $[E:\mathbb{Q}] = 2 = [E:\mathbb{Q}(\alpha)]$ would cause a contradiction with $[E : \mathbb{Q}(\alpha)]\ge 1$
Therefore $[E : \mathbb{Q}(\alpha)] = 1$. $[\mathbb{Q}(\alpha) : \mathbb{Q}] = 2$ and $[E : \mathbb{Q}(\alpha)]\Rightarrow E = \mathbb{Q}(\alpha)$.
Let the minimal polynomial of $\alpha$ be equal to $f(x) = x^2 + ax + b - \dfrac12(-a + (\sqrt{a})^2 – 4b)$. Keep in mind that $\frac12(-a + (\sqrt{a})^2 – 4b)=\frac12(-a + (\sqrt{a})^2 – 4b)$.
$a^2 – 4b$ is not a perfect square in $\mathbb{Q}$ because $[\mathbb{Q}(\alpha) : \mathbb{Q}]=1$ is a contradiction.
As such $(\sqrt{a})^2 – 4b \in E\setminus\mathbb{Q}$ and $\mathbb{Q}(\alpha) = \mathbb{Q}((\sqrt{a})^2 – 4b)$.