Proving irreducibiltity of $f(X)=X^3 -2 \in \mathbb{Q}[X]$ at $ \mathbb{Q}(\sqrt{2})$

Question

My try:

Based on some answers about irreducibilitty i tried to show that if $\mathbb{L}$ is the splitting field of $f$ then, $[\mathbb{L}: \mathbb{Q}(\sqrt{2})]=3$

By my calculations I get $\mathbb{L}=\mathbb{Q}(\sqrt{-3},\sqrt[3]{2})$, but i got to nowhere this way.

this way of proving make any sense?

Answer 1

Hint: A reducible polynomial of degree at most three has a root (over a field).

Answer 2

Eisenstein applies here, but you have to apply it with ingenuity. I assume that you know that $\sqrt2$ is a prime element of (the integers of) $\Bbb Q(\sqrt2\,)$. Now, I hope this sorites will convince you: \begin{align} X^3-2\text{ is irreducible over }\Bbb Q(\sqrt2\,) &\Longleftrightarrow X^3-\frac12\text{ is irreducible over }\Bbb Q(\sqrt2\,)\\ &\Longleftrightarrow\left(\frac X{\sqrt2}\right)^3-\frac12\text{ is irreducible over }\Bbb Q(\sqrt2\,)\\ &\Longleftrightarrow2\sqrt2\left[\left(\frac X{\sqrt2}\right)^3-\frac12\right]\text{ is irreducible over }\Bbb Q(\sqrt2\,)\\ &\Longleftrightarrow X^3-\sqrt2\text{ is irreducible over }\Bbb Q(\sqrt2\,)\,. \end{align}

And with this, you conclude that the splitting field $\Bbb L$ is of degree six over $\Bbb Q(\sqrt2\,)$.