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Let $L/K$ be abelian. There is a natural way to define the Artin reciprocity map on the ideles using the notion of an admissible cycle. I don't want to go into the details of what that is right now, but essentially what I'm having trouble with is the following. In order to show that the Artin map is well defined on the ideles $\mathbb I_K$, I must show:

Suppose $x \in K^{\ast}$ is a local norm at each ramified place. Also suppose that for any real place $v$ of $K$ (and corresponding embedding $ \sigma: K \rightarrow \mathbb{R}$) which has a complex place $w$ lying over it, we have $\sigma(x) > 0$. Then $$\prod\limits_{v } (\mathfrak p_v, L/K)^{ord_v(x)}$$ is the identity element of $Gal(L/K)$, where $(\mathfrak p, L/K)$ is the Frobenius element and $v$ runs through all the unramified places.

One way to do this would be to show that $x$ is a global norm, or at least a local norm at each place $v$ where $ord_v(x) \neq 0$. But I don't know if this is true. Any hints?

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    $\begingroup$ Notationally I am finding this quite confusing. By $(p_v,L/K)^{ord_v(x)}$ is this supposed to represent local norm residue symbol at $v$? $\endgroup$
    – Jack Yoon
    Feb 10 '15 at 0:38
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    $\begingroup$ Either way do you know 'product formula'? I feel that it should be useful for this question. $\endgroup$
    – Jack Yoon
    Feb 10 '15 at 0:46
  • $\begingroup$ Thank you for responding. I mean that $\mathfrak p_v$ is the prime ideal of $K$ corresponding to the place $v$, $(\mathfrak p_v, L/K)$ is the Frobenius element, and this automorphism is raised to the power of $\operatorname{ord}_v(x)$, which is the normalized valuation of $K$ (so that $\operatorname{ord}_v(\mathfrak p_v) = 1$). $\endgroup$
    – D_S
    Feb 10 '15 at 0:51
  • $\begingroup$ I have thought about the product formula, but I wasn't able to come up with a way it can be applied here. I don't know how the product formula comes together with Frobenius elements. I just know that $(\mathfrak p_v, L/K)$ has order $f(v)$ (the inertia). $\endgroup$
    – D_S
    Feb 10 '15 at 0:52
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This is actually pretty simple. The basic properties of the Artin symbol show that

$$(\mathfrak{N}^L_K(\mathfrak{U},L/K)=1$$

for any ideal $\mathfrak{U}$ not divisible by ramified primes since $(\mathfrak{a}, L/K)=(N^K_{k}(\mathfrak{a}), k)$ for $k\subseteq K\subseteq L$.

Now we note that the Artin map is surjective, and that the global cyclic norm index inequality gives, for a conductor $\mathfrak{c}$

$$[I(\mathfrak{c}):P_{\mathfrak{c}}\mathfrak{N}(\mathfrak{c})]\le [L:K]=[\operatorname{Gal}\left(L/K\right):1]$$

since the extension is Galois. But then immediately this implies that the kernel is exactly $P_{\mathfrak{c}}\mathfrak{N}(\mathfrak{c})$, which is exactly what you're trying to show. The inequality is important since showing this is the Artin kernel starts with showing that an admissible cycle exists, but if it does then the rest is easy.


Additional notes: $(1)$The local norm conditions ensure that the product of the $\mathfrak{p}^{v_{\mathfrak{p}}(\mathfrak{p})}$ turn into the norm of the ideal for the primes--multiplicativity means we need only consider $x$ which generate prime ideals, i.e. which localize to a non-unit at one place, since the local valuations are all that affect the local Artin symbols anyways, though the process requires sufficient typing that I don't want to regurgitate it here, since you have a definition on $I$, I assume they taught you the standard procedure, though comment if not.

$(2)$ Without knowing what an admissible cycle is, it's not possible to do this: the fact that the Artin kernel is exactly things which are congruent to 1 $\mod^* \mathfrak{c}$, this is at the heart of the proof of Artin reciprocity, and involves a lot of auxiliary work. If you want anything with greater detail, I recommend Lang's Algebraic Number Theory which is quite accessible, especially the chapter on the Artin symbol, I learned it there, and it explains the issues in a pretty straightforward way.

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  • $\begingroup$ Thanks for the response. I do know what an admissible cycle is, and now I see that by definition, a conductor $f$ would have the property that the Artin map would be trivial on $K_c \hookrightarrow P_c$, which is what I want. However, are you saying that it is NOT true in general that for an arbitrary admissible cycle $c$, the Artin map is trivial on $K_c$? $\endgroup$
    – D_S
    Feb 10 '15 at 3:08
  • $\begingroup$ I guess to clarify my question: Serge Lang defines the conductor to be an admissible cycle $c$ for which $P_c$ is contained in the kernel of the Artin map (as it is defined on ideals relatively prime to the discriminant). Is such a cycle ONLY divisible by ramified primes? $\endgroup$
    – D_S
    Feb 10 '15 at 3:15
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    $\begingroup$ @D_S any admissible cycle can work, but you just have to tweak a bit if you want to define the artin symbol on $I(\mathfrak{c})$ rather than $I(\mathfrak{f})$. Since you seem to have a copy of Lang, I recommend you look at Theorem 7 in Chapter VII, where he shows that $P_{\mathfrak{f}}\mathfrak{N}(\mathfrak{f})=P_{\mathfrak{c}}\mathfrak{N} $ $(\mathfrak{c})$. So that there's no problem at all for arbitrary admissible cycles, even if the cycle is not minimal. $\endgroup$ Feb 10 '15 at 3:28
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    $\begingroup$ @D_S in particular, it's especially direct, since the proof of that theorem exactly spells out how the isomorphism goes, so the transfer from $I(\mathfrak{f})/P_{\mathfrak{f}}\mathfrak{N}(\mathfrak{f})$ to $I(\mathfrak{c})/P_{\mathfrak{c}}\mathfrak{N}(\mathfrak{c})$ is exactly going through the isomorphism from the theorem. $\endgroup$ Feb 10 '15 at 3:32

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