How to find a primitive element of $\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5})$ over $\mathbb{Q}$?
I think that $[\mathbb{Q}(\sqrt{2}, \sqrt{3}, \sqrt{5}):\mathbb{Q}] = 8$, but not really sure how to prove it. But even if so, how to find the right element? Thanks.
The tower of extensions $$ \Bbb Q\subset\Bbb Q(\sqrt2)\subset\Bbb Q(\sqrt2,\sqrt3)\subset\Bbb Q(\sqrt2,\sqrt3,\sqrt5) $$ is proper at every step, so since each step is a quadratic extension the total degree is 8. A $\Bbb Q$-basis is $\{1,\sqrt2,\sqrt3,\sqrt5,\sqrt6,\sqrt{10},\sqrt{15},\sqrt{30}\}$. The 8 $\Bbb Q$-automorphism are those determined by any choice of signs in $$ \sqrt m\mapsto\pm\sqrt m\qquad m\in\{2,3,5\}. $$ The proof of the theorem of the primitive element says that an element whose conjugates are all different is primitive. Can you find one?