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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.

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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 $\mathbb{Q}(\sqrt 2, \sqrt 3)$.

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    Ok, i managed to do it, although in a really messy way. $T$hank you.2011-08-14