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Is there a general procedure for showing that the polynomial $x^p - n$ is irreducible over the cyclotomic field $Q(\zeta_p)$? ($\zeta_p$ a primitive pth root of unity, and $n \in \mathbb{N}$. Maybe $n$ can also be prime.)

I have run into a few exercises today where such a technique would be useful.

In particular, the tricky case is seems to be when $p > 3$. (Otherwise we are guaranteed a root in $Q[\zeta_p]$ if $x^p - n$ is not irreducible there, from which we get all roots, so the polynomial splits. But then comparing degrees shows that this is impossible, since now $Q[\sqrt[p][n]]$ (degree $p$) embeds in $Q[\zeta_p]$ (degree $p - 1$).)

Edit: The shorter answer that I submitted below appears to be correct.

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(My other answer is unnecessarily complicated and therefore probably wrong, but I will let it continue to exist as a gentle reminder to myself of other useful techniques - in particular Gauss's lemma, and Eisenstein's criterion coupled with a well chosen automorphism.)

The following is easy to show: if $F | L$ and $E | L$ have relatively prime degrees $n$ and $m$, then $FE | Q$ has degree $nm$. (Corollary 22 in Dummit and Foote, the chapter on field theory.)

We can apply this to the stated problem in the following way:

$Q(\zeta_p, \sqrt[p]{n})$ has as subfields $Q(\zeta_p)$ and $Q(\sqrt[p]{n})$. The former has degree $p-1$ and the latter degree $p$. Therefore, $Q(\zeta_p, \sqrt[p]{n})$ has degree $p(p-1)$. Considering $Q(\zeta_p, \sqrt[p]{n})$ as an extension of $Q(\zeta_p)$, we see that the minimal polynomial of $\sqrt[p]{n}$ has degree $p$, and thus coincides with $x^p - n$. Therefore, $x^p - n$ is irreducible over $Q(\zeta_p)$.

In fact, this shows that $x^q - n$ is irreducible over $Q[\zeta_p]$ when $p - 1$ and $q$ are relatively prime.

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This answer is complicated and probably wrong. The other answer I submitted makes sense.

I put together a partial solution that (maybe) works only in the case when $Z[\zeta_p]$ is a UFD and the exponent of $x$ a power of a prime in $Z[\zeta_p]$. I would appreciate it if someone could check my work, since I only vaguely know stuff about algebraic number theory.

New question: When is $(q)$ a prime ideal in $Z[\zeta_p]$?

Proposition: For $p$ a prime so that $Z[\zeta_p]$ is a UFD, and a $q$ a prime so that $(q)$ is prime in $Z[\zeta_p]$, then given $e, m \in \mathbb{N}$, with $m$ subject to the restriction implied by the lemma below, the polynomial $x^{q^e} + m$ is irreducible in $Q[\zeta_p]$.

Proof(?): For factorization in the field of fractions $K$ is equivalent to factorization in the ring of integers $R$ (Gauss's Lemma) when the polynomial is primitive in $R$, we reduce this to irreducibility in $Z[\zeta_p]$. Since $x \to x+ n $ is an automorphism of $Z[\zeta_p][X]$, $x^{q^e} + m$ irreducible is equivalent to $(x+n)^{q^e} + m$ irreducible. We expand the latter polynomial, and see that we can apply Eisenstein's criterion if we can choose $n$ so that $q | n^{q^e} + m$ but $q^2 \not | n^{q^e} + m$. (For $q^e$ choose $k$ is divisible by $q$ for $k = 1, \ldots q^e - 1$.) Therefore, $x^{q^e} + m$ is irreducible over $Q[\zeta_p]$.

So we only need to show that we can always find $n$ making this requirement on $n + m$ true.

Lemma: Let $q^e$ be fixed. Given an $m \in \mathbb{N}$ (subject to some restrictions?) and $p$ prime, there is an $n \in \mathbb{N}$ so that $p | n^{q^e} + m$ but $p^2 \not | n^{q^e} + m$.

Question: How does this restrict $m$?

(Since the class number of $Z[\zeta_p]$ is 1 for $p < 23$, this would cover many cases that come up in practice - that is, in homework problems.)

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