What are the prime ideals in rings of cyclotomic integers? Is a good characterization of Spec $\mathbb{Z}[\zeta_n]$ known? Same question for its unit group.
 A: The extension $\mathbb{Q}(\zeta_n)|\mathbb{Q}$ is abelian of group $(\mathbb{Z}/n\mathbb{Z})^\times$ so class field theory tells you everything about the prime ideals in $\mathbb{Z}[\zeta_n]$, the ring of integers of $\mathbb{Q}(\zeta_n)$.
You should try to do the cases $n=3,4$ by hand.
As for the group $\mathbb{Z}[\zeta_n]^\times$, an explicit subgroup of "cyclotomic units" can be constructed which has finite index.  
Any book on Cyclotomic Fields (Lang, Washington) should help.  For a start, you can look up  Chapter VI of Fröhlich-Taylor.
A: Let me summarise what Hilbert says in his Zahlbericht about the behaviour of rational primes in the cyclotomic field $\mathbb{Q}(\zeta)$, where $\zeta$ is a primitive $l$-th root of $1$ and $l$ is an odd prime.  You can read the original at the Göttingen site or a French translation at the Grenoble site.
Satz 117. The ideal $\mathfrak{l}=(1-\zeta)\mathbb{Z}[\zeta]$ is prime of residual degree $1$, and $l\mathbb{Z}[\zeta]=\mathfrak{l}^{l-1}$.
Satz 118. The discriminant of the field $\;\mathbb{Q}(\zeta)$ is $(-1)^{(l-1)/2}l^{l-2}$.
Satz 119. If $p\neq l$ is a rational prime,  $f>0$ is the smallest exponent such that $p^f\equiv1\pmod l$, and $e$ is defined by $ef=l-1$, then
$$
p\mathbb{Z}[\zeta]=\mathfrak{p}_1\ldots\mathfrak{p}_e,
$$
where the $\mathfrak{p}_i$ are distinct prime ideals of residual degree $f$.
These results go back to Kummer (1847).  All this was much before anyone dreamt of Class Field Theory.
A: Theorem:  Let $\alpha$ be an algebraic integer such that $\mathbb{Z}[\alpha]$ is integrally closed, and let its minimal polynomial be $f(x)$.  Let $p$ be a prime, and let 
$\displaystyle f(x) \equiv \prod_{i=1}^{k} f_i(x)^{e_i} \bmod p$
in $\mathbb{F}_p[x]$.  Then the prime ideals lying above $p$ in $\mathbb{Z}[\alpha]$ are precisely the maximal ideals $(p, f_i(\alpha))$, and the product of these ideals (with the multipicities $e_i$) is $(p)$.  (Theorem 8.1.3.)
In this particular case we have $f(x) = \Phi_n(x)$.  When $(p, n) = 1$, its factorization in $\mathbb{F}_p[x]$ is determined by the action of the Frobenius map on the elements of order $n$ in the multiplicative group of $\overline{ \mathbb{F}_p }$, which is in turn determined by the minimal $f$ such that $p^f - 1 \equiv 0 \bmod n$ as described in Chandan's answer.  (This $f$ is the size of every orbit, hence the degree of every irreducible factor.)  When $p | n$ write $n = p^k m$ where $(m, p) = 1$, hence $x^n - 1 \equiv (x^m - 1)^{p^k} \bmod p$.  Then I believe that $\Phi_n(x) \equiv \Phi_m(x)^{p^k - p^{k-1}} \bmod p$ and you can repeat the above, but you'd have to check with a real number theorist on that.  (Edit:  Indeed, it's true over $\mathbb{Z}$ that $\Phi_n(x) = \frac{ \Phi_m(x^{p^k}) }{ \Phi_m(x^{p^{k-1}}) }$.)
