It's well-known that over $\Bbb Q$ the factorization into irreducible cyclotomic polynomials$$x^n-1=\prod_{d|n}\Phi_d(x)$$and over $\Bbb F_{p^m}$ it's well-known as well from basic Galois theory that $\Phi_d(x)$ factors into $\displaystyle\frac{\phi(d)}{k}$ factors of degree $k$, where $k$ is the order of $p^m$ in $(\Bbb Z/d\Bbb Z)^\times$.
However, little do I known about the factorization in an arbitrary field of characteristic $0$ other than $\Bbb Q$, or an infinite field of characteristic $p$. How does the factorization look like in those cases?
The answer I am looking for does not need to involve the explicit form of the irreducible factors, but rather how the degree of $n$ is distributed for the degree of the irreducible factors, just like the description of the factorization mentioned over $\Bbb F_{p^m}$.
As a particular case I am concerned about, when is $\Phi_d$ irreducible over a field $F$?
Best Answer
For any field $K$ and integer $n$, the Galois group $\mathrm{Gal}(K(\zeta_n)/K)$ can be canonically identified with a subgroup of the unit group $U_n:=(\mathbb{Z}/n\mathbb{Z})^\times$ (by identifying an automorphism $\sigma$ with the (coset of the) integer $k$ such that $\sigma(\zeta_n)=(\zeta_n)^k$). The factorization of $x^n-1$ over $K$ is exactly determined by this subgroup, which we will call $G$:
First of all, since $x^n-1=\prod_{d\mid n}\Phi_d(x)$ in any field, it is enough to concentrate on the $\Phi_d$. It is perhaps useful to first have a look at the case $d=n$. Since any root of $\Phi_n$ generates the whole Galois extension, this polynomial factors into several irreducibles of the same degree each. This degree is nothing but the order of $G$, so that the factorization of $\Phi_n$ over $K$ is into $\frac{\varphi(n)}{|G|}$ polynomials of degree $|G|$ each.
Now, for each $d\mid n$, the story is really the same but for the intermediate extension $K(\zeta_d)$. It is therefore enough to determine how the Galois group of this smaller extension relates to $G$. By a little computation, we see that the quotient map $\mathrm{Gal}(K(\zeta_n)/K)\to \mathrm{Gal}(K(\zeta_d)/K)$ is compatible with the natural map $$\pi_d:U_n\to U_d, k+n\mathbb{Z}\mapsto k+d\mathbb{Z}.$$ Hence, by the same reasoning as above, $\Phi_d$ must factor into $\frac{\varphi(d)}{|\pi_d(G)|}$ polynomials of degree $|\pi_d(G)|$ each, which gives the full factorization of $x^n-1$.
I should also mention that every subgroup of $U_n$ does in fact arise as such a Galois group (in characteristic $0$) - this is clear from the Galois correspondence, since $\Phi_n$ is irreducible over $\mathbb{Q}$ and as such gives the full unit group $U_n$ as its Galois group.
I don't really have a good answer for your final question though - it seems hard to state the irreducibility of $\Phi_n$ in any other useful way for any specific $n$. If you are interested in irreducibility of all $\Phi_n$ at once, you might employ Kronecker-Weber: All these polynomials are irreducible over a field $K$ if and only if $K$ has characteristic $0$ and any finite normal subfield of $K$ (finite over $\mathbb{Q}$, that is) has a perfect Galois group (which is to say that $K$ has no nontrivial abelian subextensions). There are quite a few different flavours of such fields - for instance, any purely transcendental extension of $\mathbb{Q}$ will do, as will any finite algebraic extension with a simple nonabelian Galois group (say). Still another example would be the field obtained by adjoining a single cube root of every prime. All in all, these seem so different in spirit that it seems hard to give a useful criterion that encapsulates them all.