How do I show that $\sqrt{5}\notin \mathbb{Q}(\sqrt{2},\sqrt{3})$? How do I show that $\sqrt{5}\notin \mathbb{Q}(\sqrt{2},\sqrt{3})$?
Since $X^2-5$ is the minimal polynomial of $\sqrt{5}$ over $\mathbb{Q}$ and its degree is not relatively prime to $[\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}]$, it cannot be shown that $X^2-5$ is irreducible in $\mathbb{Q}(\sqrt{2},\sqrt{3})[X]$ in this way.
How do I show this?
 A: If you know a little Galois theory, then one way of going about this is to start by listing the intermediate fields between $\mathbb{Q}$ and $\mathbb{Q}[\sqrt{2}, \sqrt{3}]$, of which there are only three: $\mathbb{Q}[\sqrt{2}]$, $\mathbb{Q}[\sqrt{3}]$, and $\mathbb{Q}[\sqrt{6}]$.  (Galois theory guarantees this list is exhaustive.)  
If $\sqrt{5} \in \mathbb{Q}[\sqrt{2}, \sqrt{3}]$, then we necessarily have $\mathbb{Q}[\sqrt{5}] \subsetneq \mathbb{Q}[\sqrt{2}, \sqrt{3}] \implies \mathbb{Q}[\sqrt{5}] \cong \mathbb{Q}[\sqrt{a}]$ for $a = 2, 3, \text{ or } 6$, and a simple argument using the properties of isomorphisms shows that this is not possible.  
A: Every nonempty subset of $\,S = \{\sqrt2,\sqrt3,\sqrt 5\}\,$ has product $\not\in \Bbb Q\,$ thus   $\,[\Bbb Q(\sqrt2,\sqrt3,\sqrt 5):\Bbb Q] = 8\,$ by this answer. Said informally, multiplicative independence implies linear independence (the reason for such becomes clearer when one studies Galois theory of Kummer extensions).
One can also give a direct elementary proof using the simple Lemma in the linked answer.
