TITLE: How to show polynomial is irreducible in extension field of rationals?
QUESTION [5 upvotes]: How would one show this? is there any test?
say $x^3 - 2x - 5$ over $\mathbb{Q}[\sqrt{2}]$
This is just a made-up example, but I'm interested in the general process and how it differs from showing irreducible over $\mathbb{Q}$.

REPLY [3 votes]: In this case it's easy. Suppose $\theta$ is a root of $x^3-2x-5$. Observe that $x^3-2x-5$ has no rational roots (say by the rational root test) so is irreducible over $\mathbb Q$. Thus $\mathbb Q[\theta]$ has degree $3$. If $\theta\in \mathbb Q[\sqrt{2}]$ then we would have $\mathbb Q[\theta]\subseteq \mathbb Q[\sqrt2]$, thus $[\mathbb Q[\theta]:\mathbb Q]|[\mathbb Q[\sqrt2]:\mathbb Q]$, i.e. $3|2$, a contradiction.