Prove $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}$ is a simple extension My problem is "Prove $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5}):\mathbb{Q}$ is a simple extension"
I met a similar problem "Prove $\mathbb{Q}(\sqrt{2},\sqrt{3}):\mathbb{Q}$ is a simple extension" and I showed that it is equal to Prove $\mathbb{Q}(\sqrt{2}+\sqrt{3}):\mathbb{Q}$. I let $u=\sqrt{2}+\sqrt{3}$ and deduce that $u-\sqrt{2}=\sqrt{3})$, then square it and prove $\sqrt{2} \in Q(\sqrt{2}+\sqrt{3})$. However, the same technique cannot be applied in the first problem since we have 3 square root.
I actually found a proof for the first problem, by letting $u=\sqrt{2}+\sqrt{3}+\sqrt{5}$ and calculating $u,u^2,..., u^5$ and prove $\sqrt{2},\sqrt{3}\in Q(u)$. Though it's not so nice. Looking for a simpler answer
 A: Consider the set of field homomorphism of $\mathbb{Q}(\sqrt2,\sqrt3,\sqrt5)$ to $ \bar{\mathbb{Q}}$, the algebraic closure of $\mathbb{Q}$, and call it $\operatorname{Hom}$.
For $k \in \{2,3,5\}$ and $f \in \operatorname{Hom}$, we can send $\sqrt{k}$ only to $\sqrt{k}$ or to $-\sqrt{k}$, as $f(\sqrt{k})$ must be a root of $x^2-k$. Combining these possibilities, we obtain $2^3=8$ different elments in $\operatorname{Hom}$. Consider the element $\alpha = \sqrt{2}+\sqrt{3}+\sqrt{5}$.
Note that for two different elements $f,g \in \operatorname{Hom}$, we have $f(\alpha) \neq g(\alpha)$, for example $\sqrt2+\sqrt3+\sqrt5 \neq \sqrt2+ \sqrt3-\sqrt5$. Now consider $P$, the minimal polynomial of $\alpha$ over $\mathbb{Q}$.
For any $f \in \operatorname{Hom}$ $f(\alpha)$ is a root of $P$. Thus $P$ has at least $8$ distinct roots in $\bar{\mathbb{Q}}$, thus $\deg(P) \geq 8$.
But $\deg(P)=[\mathbb{Q}(\alpha):\mathbb{Q}] \leq [\mathbb{Q}(\sqrt2,\sqrt3,\sqrt5):\mathbb{Q}] = 8$, so $\deg(P) = [\mathbb{Q}(\alpha):\mathbb{Q}] = 8$, so $[\mathbb{Q}(\sqrt2,\sqrt3,\sqrt5):\mathbb{Q}(\alpha)] = 1$, i.e. $\mathbb{Q}(\alpha)=\mathbb{Q}(\sqrt2,\sqrt3,\sqrt5)$.
