# Linearly disjoint field extensions

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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+1. Nice question. – Pierre-Yves Gaillard Aug 14 '11 at 16:11

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 $\mathbb{Q}(\sqrt 2, \sqrt 3)$.
I don't manage to show that f(t) is irreducible over $\mathbb{Q}(\sqrt{2},\sqrt{3})$. Of course if it was not irreducible, then it should split into two polynomials of degree 2, but i don't find the contradiction ... Nevertheless the result is true since $\mathbb{Q}(\sqrt{2},\sqrt{3})\cap \mathbb{Q}(\zeta_5)=\mathbb{Q}$. – Louis La Brocante Aug 14 '11 at 15:24
@Rolando: The coefficients of any factor of $f(t)$ (over the complex numbers) are either non-real or in $\mathbf{Q(\sqrt5)}\setminus\mathbf{Q}$. Does that help? – Jyrki Lahtonen Aug 14 '11 at 15:49