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).

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