Is $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}):\mathbb{Q}$ a simple extension?
My conjecture is no, simply because I failed to come up with an $\alpha$ such that $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})=\mathbb{Q}(\alpha)$; but then I struggle to deduce a contradiction by assuming $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})=\mathbb{Q}(\alpha)$ for some $\alpha\in\mathbb{C}$.
Proof: Suppose $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})=\mathbb{Q}(\alpha)$ for some $\alpha\in\mathbb{C}$. It is obvious that $\alpha$ cannot be transcendental, hence $\mathbb{Q}(\alpha):\mathbb{Q}$ is a simple algebraic extension. Then $\mathbb{Q}[t]/<m>$ is isomorphic to $\mathbb{Q}(\alpha)$, where $m(t)$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$. Also, if $deg(m)=n$, then $\{1, \alpha, ..., \alpha^{n-1}\}$ is a basis for $\mathbb{Q}(\alpha)$. Screeching halt.
Any hint would be appreciated-many thanks in advance! No solutions please.