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$K=Q(\zeta_n)$ 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

1 Answer 1

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.

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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
    
Ah! I see the point. I was apt to regard that finite field as something unnatural, whereby viewing it as different from our prime field. Sorry for that. Now is this proof of one inclusion complete and great. :) –  awllower Feb 7 '12 at 13:23

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