split in cyclotomic field $K=Q(\zeta_n)$ a cyclotomic extension:
$p$ splits completely in $K$ if and only if $p\equiv 1\ (mod\ n)$
I don't know how i could prove, I search a kind of cyclotomic reciprocity law 
Many thanks
 A: By a result of Dedekind—see this Keith Conrad handout for details—the decomposition of $p$ in $\mathbf Z[\zeta_n]$ is determined by the factorization of the $n$-th cyclotomic polynomial $\Phi_n(X)$ modulo $p$. Let's assume that $p \equiv 1 \bmod n$. Since $X^n - 1$ has derivative $nX^{n - 1}$ and $p\nmid n$, we know that the reduction of $\Phi_n$ is separable and hence $p$ does not ramify. We also see that reduction maps the $n$-th roots of unity in $\mathbf Z[\zeta_n]$ bijectively onto those in $\mathbf Z[\zeta_n]/\mathfrak p$ for any $\mathfrak p$ lying above $p$.
So it remains to show that the reduction $\bar\Phi_n$ splits completely. But $n$ divides the order of the cyclic group $\mathbf F_p^* = (\mathbf Z/p\mathbf Z)^*$ and hence the residue field of $p$ already contains the $n$-th roots of unity, which generate $\mathbf Z[\zeta_n]/\mathfrak p$ over $\mathbf F_p$. The converse seems to use the same ideas—I'll try to spell that out later.
There is indeed a more general law for how primes split in cyclotomic extensions. Off the top of my head this is explained in Birch's article in Cassels and Frölich, but maybe I can find something online.
