TITLE: Linearly disjoint field extensions
QUESTION [5 upvotes]: Recall that if $k$ is a field, some field extensions $K_1/k$,..., $K_n/k$ are called linearly disjoint if the tensor product $K_1\otimes_k\cdots \otimes_k K_n$ is a field.
Let $\zeta_5$ be a pritive fifth root of $1$. I would like to show that the three field extensions $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$, $\mathbb{Q}(\sqrt{3})/\mathbb{Q}$ and $\mathbb{Q}(\zeta_5)/\mathbb{Q}$ are linearly disjoint over $\mathbb{Q}$.
The statement seems quite natural for me since the only non-trivial subextension of $\mathbb{Q}(\zeta_5)/\mathbb{Q}$ is $\mathbb{Q}(\sqrt{5})/\mathbb{Q}$ and since the field extensions $\mathbb{Q}(\sqrt{2})/\mathbb{Q}$ and $\mathbb{Q}(\sqrt{3})/\mathbb{Q}$ are clearly linearly disjoint. However i would like some help to prove this statement.

REPLY [9 votes]: Consider $f(t) = t^4 + t^3 + t^2 + t + 1$. It is clear that $\mathbb{Q}(\sqrt 2) \otimes_\mathbb{Q} \mathbb{Q}(\sqrt 3) = \mathbb{Q}(\sqrt 2, \sqrt 3)$. Now
$$
\mathbb{Q}(\sqrt 2) \otimes_\mathbb{Q} \mathbb{Q}(\sqrt 3) \otimes_\mathbb{Q} \mathbb{Q}(\zeta_5) = \mathbb{Q}(\sqrt 2, \sqrt 3) \otimes_\mathbb{Q} \mathbb{Q}[t]/(f(t)) = 
\mathbb{Q}(\sqrt 2, \sqrt 3)[t]/(f(t)).
$$
So it suffices to show that the polynomial $f(t)$ is irreducible over $\mathbb{Q}(\sqrt 2, \sqrt 3)$.