Proof of the general reciprocity law (Algebraic Number Theory)

Question

I'm studying Neukirch's Algebraic number theory, p. 300~301, (6.3) Theorem.

We assume knowledge of abstract galois theory, from his book p.275~

I now trying to understand the underlined statement.

Q.1) What is the fixed field $M$ of a p-Sylow subgroup? Does it means that the index of $G_M$, where $G_M$ is a $p$-Sylow subgroup of $G$(c.f. his book p.275. $G$ is maybe a given profinte group.) But in the proof, he writes that "$M|K$ need not be Galois", and "but we may use the left part of the diagram, where $r_{L|M}$ is surjective." From his writing, it seems to possible that we may choose $M$ such that $M|K$ is a subextension of $L|K$. Is it possible?

Q.2) What is the $p$-Sylow subgroup $S_p$ of $A_K/N_{L|K}A_L$(Definition is given in his book 277)? If $A_K/N_{L|K}A_L$ is finite, then since $A_K/N_{L|K}A_L$ is abelian so if $p$ is divides the order of $A_K/N_{L|K}A_L$, then we may prove that there is only one $p$-Sylow subgroup of $A_K/N_{L|K}A_L$. But

Q.2-1) Is $A_K/N_{L|K}A_L$ a finite?

Q.2-2) Does $p$ divides the order of $A_K/N_{L|K}A_L$?

Q.3) Why $([M : K], p) =1 $?

Any one helps? Thanks for reading.

For 1): if $M/K$ is a subextension of a finite Galois extension $L/K$, then $L/M$ is Galois and its Galois group identifies to a subgroup of the Galois group of $L/K$. In this case, I think that $M$ is chosen such that this subgroup is a $p$-Sylow of $Gal(L/K)$. — Aphelli, Nov 15 '21 at 11:38
For 3), $[M:K]$ is the index of $Gal(L/M)$ in $Gal(L/K)$. As the former is a $p$-Sylow of the latter, the index cannot be divisible by $p$. — Aphelli, Nov 15 '21 at 11:42
For 2): we don’t all have the book at hand, so it’ll be easier if you include definitions for $A_K$. I think it’s abelian by any reasonable definition, and it’s probably defined so that the quotient is finite. There’s an easy definition for the $p$-Sylow of a finite abelian group $B$: it’s the set of elements such that their order is a power of $p$. — Aphelli, Nov 15 '21 at 11:45
I uploaded an image about definition of $A_K$. As p.276, since $A$ is abelian, $A_K$ is also abelian. Then the quotient $A_K/N_{L/K}A_L$ (p.277) is finite group? — Plantation, Nov 15 '21 at 12:57
And I understand the 3), if $M$ is sellected such that $G(L|M)$ is a $p$-Sylow of $G(L|K)$. Then can we really find such subextension $M/K$? — Plantation, Nov 15 '21 at 13:00
For 3): that’s just the usual Galois correspondence between subextensions of a Galois extension and subgroups of the Galois group. For 2): no idea in general. But if $S_p$ is defined instead as “the set of elements whose order is a power of $p$”, then the image by a group homomorphism of a $p$-Sylow is always contained in the “$p$-Sylow” of the range. — Aphelli, Nov 15 '21 at 15:43
Yes. Thank you. Following your comment, similarly, let $S_p$ be the maximal $p$-subgroup of $A_K/N_{L|K}A_L$. Here $p$-group means that every its element has order a power of $p$. (Such $S_p$ may exists by using Zorn's Lemma). But in this case, how can we show that "That this holds true for all p amounts to saying that $r_{L|K}$ is surjective" and "$[M:K] :S_p \to S_p$ ($[M:K]$'th power map) is surjective?" (in p.301) — Plantation, Nov 16 '21 at 03:22
First, to show that $r_{L|K}$ is surjective, if $A_K/N_{L|K}A_L$ is finite, then I used the fact that a finite group can be generated by the union of all Sylow subgroups( c.f. https://math.stackexchange.com/questions/487488/finite-group-generated-by-the-union-of-its-sylow-p-i-subgroups ) This also works for infinite caset of $A_K/N_{L|K}A_L$ ? — Plantation, Nov 16 '21 at 03:26
Second, to show that $[M:K]$ is surjective, for finite case of $A_K/N_{L|K}A_L$ I used that "https://math.stackexchange.com/questions/304667/powering-map-is-surjective-when-the-power-is-relatively-prime-to-the-order-of-th". But how can we show for $A_K/N{L|K}A_L$ not necessarily finite? Similar mechanism works for the infinite case of $A_K/ N_{L|K}A_L$? — Plantation, Nov 16 '21 at 03:29
Uhm..Yes. For the surjectivity of $[M:K]$, I somewhat understand. And for the first question Q.1), I tried to find such subextension and stucked with some technical(?) problem and uplodead related question (https://math.stackexchange.com/questions/4307408/intermediate-subgroup-between-closed-subgroups-of-profinite-group-is-closed) Can you refer to this question and give some comment? — Plantation, Nov 16 '21 at 04:54
Note that $X=A_K/N_{L/K}A_L$ is $[L:K]$-torsion. So the subgroup of all elements whose order is a power of $p$ is well defined (and the largest $p$-group in $X$) and $X$ is a direct sum of such subgroups (by Chinese remainder theorem). Then: if “this” holds true for all $p$ (whatever “this” precisely is), then the $p$-Sylow of the group is in the image of $r_{L/K}$. As $X$ is the direct sum of its $p$-Sylows, this means $r_{L/K}$ is onto. The final thing is a fact in a $p$-group: if $n$ is coprime to $p$, then $x \longmapsto x^n$ is bijective. — Aphelli, Nov 16 '21 at 08:44
Some confusion. First, $[L:K]$-torsion means that for every element $a\in X$, $a^{[L:K]} =1$? Now I can't understand the well definedness of the $p$-group well. Second, you wrote that "the $p$-Sylow of the grop is in the image of $r_{L/K}$". But is $N_{{M_{p}}|K}$(I write $M_{p}$ for emphasizing that $M$ depends on $p$) correct, instead of $r_{L/K}$? — Plantation, Nov 16 '21 at 10:55
Third, you mean the 'direct sum of such subgroups' as internal direct sum of its Sylows? Then $X$ is generated by the union of the Sylows and using the surjectivity of $r_{L|M_{p}}$ and the commutative diagrams involving $N_{M_{p}|K}$ and $r_{L|M_{p}}$, we may show that $r_{L|K}$ is surjective. (As I commented for finite case of $X$ above). But is really $X$ a direct sum of such group? Why? What version of Chinese remainder theorem you mean? How can we apply the theorem? — Plantation, Nov 16 '21 at 10:55
Uhm.. as your comment, by the commutativity of the left diagram involving $N_{M_{p}|K}$ and $r_{L|M_{p}}$, "If $S_p \subseteq im(N_{M_{p}/K})$ for all $p$, then $S_p \subseteq im(r_{L|K})$ for all $p$." So why $X$ is the direct sum of $S_p$ 's.. — Plantation, Nov 16 '21 at 11:29

Proof of the general reciprocity law (Algebraic Number Theory)

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