## Show that a given a is irreducible in QQ(a)

Let $\theta$ be a root of $p(x) = x^3 + 3x + 7$. Show that $\theta$ is irreducible in $\mathbb{Q}(\theta)$.

First we will verify that $p(x)$ is irreducible over $\mathbb{Q}$. To that end, since $p$ has degree 3, it suffices to show that it has no linear factor. Via the rational root theorem, it then suffices to note that $p(7) = 371$ while $p(-7) = -357$. So $p(x)$ is the minimal polynomial for $\theta$ over $\mathbb{Q}$.

Note that $p(x)$ is the field polynomial for $\theta$ over $\mathbb{Q}(\theta)$. Thus $N(\theta) = 7$ over this field; since the norm of $\theta$ is a rational prime, $\theta$ is irreducible as an algebraic integer in $\mathbb{Q}(\theta)$.