# When does an irreducible polynomial stay irreducible in a cyclotomic extension?

Suppose that $P(x)\in\mathbb{Q}[x]$ is irreducible over $\mathbb{Q}$, and let $K$ be the $n$-th cyclotomic field. Is there a simple criterion to tell if $P$ remains irreducible over $K$? (Preferably a necessary and sufficient condition, unlike Eisenstein's.)

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A polynomial $P(x) \in \mathbb{Q}[x]$ is irreducible iff it is a prime element in the ring $\mathbb{Q}[x]$ iff $\mathbb{Q}[x]/(P(x))$ is a field, say $L$. To say that the polynomial remains irreducible in an extension $K/\mathbb{Q}$ is to say that $\mathbb{Q}[x]/(P(x)) \otimes_{\mathbb{Q}} K$ is a field, i.e., that the fields $K$ and $L$ are linearly disjoint over $\mathbb{Q}$.
In your case, your extension $K/\mathbb{Q}$ is Galois, so you're in luck: if $K,L/\mathbb{Q}$ are two finite degree field extensions at least one of which is Galois, then linear disjointness is equivalent to $K \cap L = \mathbb{Q}$.
So there's your necessary and sufficient condition: you want $L \cap \mathbb{Q}(\zeta_N) = \mathbb{Q}$. This is a condition that a mathematical software package will be able to check for given $P$ and $N$. I'm not a computational number theorist, but I would have to imagine that a computer will have a much easier time checking this than, say, factoring $P$ over $\mathbb{Q}(\zeta_N)$.
Also, the intersection of $K$ and $L$ will depend on the embedding of each, but presumably whether it's $\mathbb{Q}$ won't care? – John Conecker May 10 '12 at 23:49
@jdc: a priori the definition of linear disjointness depends on embedding into some big field, yes, but one can show that e.g. for finite extensions the notion is independent of the choice of embedding. See for instance $\S 12.2$ of math.uga.edu/~pete/FieldTheory.pdf. – Pete L. Clark May 11 '12 at 0:17
And to wrap things up nicely, this other Math Stack Exchange question discusses a sufficient condition for when the intersection is $\mathbb{Q}$. – John Conecker May 11 '12 at 20:20