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I'm studying the Algebraic Number Theory Notes of Robert B. Ash. I really like his notes, but I don't understand a suggestion he gives at page 7 of chapter 8 (Factoring of prime ideals in Galois Extensions).

He's using all the theory of that chapter to discover new properties of cyclotomic fields. So pick $\zeta$ a primitive $m^{th}$ root of unity and let $L=\mathbb Q(\zeta)$, $ A=\mathbb Z$ and $K=\mathbb Q$.

Consider $p$ rational prime that does not divide $m$. Say $B$ the integral closure of $A$ in $L$ and that $(p)$ factors in $B$ as $Q_1....Q_g$. We know that the relative degree $f$ is the same for all $Q_i$.

He wants to find the Frobenius automorphism $\sigma$ explicitly. We know that $\sigma$ has the property that $\sigma(x)\equiv x^p\pmod {Q_i}$ for all $i$ and for all $x\in B$. From this, why do we deduce that $\sigma(\zeta)=\zeta^p$?

Richard
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  • I edited your question to reflect some necessary conditions, eg. $(m,p)=1$ otherwise you will have ramification at $p$ which of course defeats the point of the Frobenius. – Adam Hughes Nov 29 '16 at 18:59
  • Ah, so you did. Good catch. I made some other minor (but necessary) changes as well, so I'll just leave things as-is. – Adam Hughes Nov 29 '16 at 20:48

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All finite extensions of $\Bbb F_p$ are separable (in fact Galois), so all polynomials are separable--note some authors demand separable to mean "distinct roots" I use the "irreducible factors have distinct roots" definition, if you are using the former, then clearly this is not always true as $x^{pk}-1$ has repeated roots modulo $p$.

Now, the point of the suggestion is that $\zeta\mapsto\zeta^p$ generates the Galois group of $x^m-1$ over $\Bbb F_p$ because it generates all finite extension Galois groups over $\Bbb F_p$. We know that it permutes the roots of $x^m-1$ in the big field as well, so it lifts to an element of $\text{Gal}(L/\Bbb Q)$. What Ash is referencing is his theorem 8.1.8 which says that the map from the decomposition group is surjective in the case that the residue field extension is separable on page 3 of his notes. This is used to life the mod $Q$ Frobenius to the Frobenius element in the big Galois group.

Adam Hughes
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  • @Richard you don't need the system of congruences per se, you need to look at what the Frobenius is, not which primes it's trivial on or which sub-extension we're looking at. The key factor here (at least as-stated) is to determine the Frobenius explicitly, and the point of the Frobenius is that it has the effect noted on the residue field. What we have done is verify that it does exactly that, so we need go no further. – Adam Hughes Nov 29 '16 at 20:36
  • @Richard certainly if $\zeta\mapsto\zeta^p$ in the big field, it does so in the residue field as well, since we have the usual surjection from the Decomposition group of a given prime onto the Galois group of the residue field extension. Then by definition this is a Frobenius element. Professor Ash uses the separability to show that the map $\zeta\mapsto\zeta^p$ is indeed an automorphism. Why do you think it should be related to systems of congruences? Those are usually used to determine a sub-extension or establish which elements have a trivial action, not to show the lifting property. – Adam Hughes Nov 29 '16 at 20:45
  • @Richard I went and took a look at Ash's notes to try and find a way to explain it using his specific results, and I have rephrased my answer to reference that. Take a look. – Adam Hughes Nov 29 '16 at 20:54
  • @Richard That first thing you said in your last post doesn't make sense, why should you use a system of congruences just because they're talking about separability, those two ideas are not logically linked. As for your second point, see my previous comment. – Adam Hughes Nov 29 '16 at 20:55
  • @Richard you should clarify what you're aware of. Earlier in the notes professor Ash notes that there is a surjective map from the decomposition group--and since this is the abelian case from the full Galois group--onto the residue field Galois group. Clearly the field automorphism $\zeta\mapsto\zeta^p$ maps onto the same map in the residue field, or is that what you cannot see? By definition an element of the big Galois group which hits the mod $p$ Frobenius is the Frobenius in the big field, Ash's notes even define it in this way. – Adam Hughes Nov 29 '16 at 21:22
  • @Richard perfect! Then I'll vote to close this. – Adam Hughes Nov 29 '16 at 21:50