# 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

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Might I refer you to the book: Algebraic Number Theory by Jürgen Neukirch? It is a complete and reasonably arranged book. (amazon.com/…) – awllower Feb 7 '12 at 13:16

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.
Sorry, but I am still confused: to show that the reduction of $\Phi$ splits completely, we are told that all the roots of that polynomial are elements of the residue field of $p$, by means of the fact that $p$ is congruent to 1 modulo $n$. But why? Thanks in any case for sharing. – awllower Feb 7 '12 at 13:09
@awllower $\mathbf F_p^*$ is a cyclic group. A cyclic group has a (unique!) subgroup of order $m$ for each $m$ dividing the order of the group. In our case, there is a subgroup of order $n$, and this consists of the $n$-th roots of unity. – Dylan Moreland Feb 7 '12 at 13:17
And those elements in this subgroup will be the $n$-th roots of unity? I cannot figure out this step, thanks for the elaboration. – awllower Feb 7 '12 at 13:19
Well, for one inclusion: certainly each $x$ in that subgroup will satisfy $x^n = 1$, because $n$ is the order of the subgroup. For the reverse there are a few ways to see it. For one, the $n$-th roots of $1$ are the roots of the polynomial $x^n - 1$, which can have at most $n$ roots. – Dylan Moreland Feb 7 '12 at 13:20