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I have following question about a remark in J. Neukirch's Algebraic Number Theory around page 397.

The context: We consider ideal theoretic formulation of global class field theory of a number field $K$.
The statement is that to every modulus $\mathfrak{m} = \prod_{\mathfrak{p} \nmid \infty}\mathfrak{p}^{n_p}$ (in modern terms a fractional ideal not dividing infinite places) we can associate generalized congruence subgroups $C^{\mathfrak{m}}_K$ wrt $\mathfrak{m}$ of "usual" class group $C_K$ of $K$. Then there exist a class field $K^{\mathfrak{m}}$ with respect modulus $\mathfrak{m}$ such that the Galois group of $K^{\mathfrak{m}}/K$ is isomorphic to $C_K/C^{\mathfrak{m}}_K$. This $K^{\mathfrak{m}}$ is called the ray class field with respect $\mathfrak{m}$. Note, that's not a 1-to-1 correspondence, but it inverts inclusions, namely
$\mathfrak{m}' \subset \mathfrak{m} $ implies $K^{\mathfrak{m}} \subset K^{\mathfrak{m}'}$, but in general not the converse.

Let $L/K$ be a finite abelian extension. The conductor $\mathfrak{f}$ of $L/K$ in Neukirch's book is defined to be the gcd of all modules $\mathfrak{m}$ such that $L \subset K^{\mathfrak{m}}$.

This "defect" leads to an interesting remark on page 397 after Definition (6.4):

By definition $K^{\mathfrak{f}}/K$ is therefore the smallest ray class field containing $L/K$. But it is not true in general that $\mathfrak{m}$ is the conductor of $K^{\mathfrak{m}}/K$.

Question: Does there exist an explicit relation between $\mathfrak{m}$ and the conduction $\mathfrak{f}_m$ of $K^{\mathfrak{m}}/K$? Clearly, $\mathfrak{f}_m$ divides $\mathfrak{m}$ as fractional ideal.

But is it possible to give to the "defect factor" between $\mathfrak{f}_m$ and $\mathfrak{m}$ a "geometric meaning"? Does it hide some "deep" arithmetic information or is it more of less a historical appendage from times before the interpretation of class field theory with topological methods via adelic formalism a la Chevalley?

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I don't know what "a historical appedage from times before the interpretation of class field theory with topological methods via adelic formalism a la Chevalley" means, but if $\mathfrak{m}$ is a prime ideal and the global unit group $\mathcal{O}^{\times}_K$ surjects onto the units of the residue field $(\mathcal{O}_K/\mathfrak{m})^{\times}$ then the ray class field of $K^{\mathfrak{m}}$ is unramified at $\mathfrak{m}$, and so has conductor $1$ not conductor $\mathfrak{m}$. If the image of the unit group is not surjective, then $K^{\mathfrak{m}}$ has conductor $\mathfrak{m}$. I don't think there is much more to say. The general case is along similar lines. If you fix a real quadratic field then the first case should happen for infinitely many $\mathfrak{m}$ under GRH

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  • $\begingroup$ If $\mathbb{m}$ is prime (= finite "place"), then $\mathbb{m}$ is the only(!) finite place where $K^{\mathfrak{m}}/K$ ramifies, and surjection from $\mathcal{O}_K^{\times}$ to $(\mathcal{O}_K/\mathfrak{m})^{\times}$ is always trivially fullfilled, so I not understand what you mean by that $K^{\mathfrak{m}}$ is unramified in $\mathfrak{m}$. $\endgroup$
    – user267839
    Commented Feb 29 at 0:46
  • $\begingroup$ Your comment reveals some glaring misconceptions, and suggests that you are trying to study CFT without having ever computed a genuine example. This is a mistake. $\endgroup$
    – user523984
    Commented Feb 29 at 1:20
  • $\begingroup$ @user267839 Surjectivity of the ring homomorphism $\mathcal O_K\to\mathcal O_K/\mathfrak m$ does not imply surjectivity of the group homomorphism $\mathcal O_K^\times\to(\mathcal O_K/\mathfrak m)^\times$. Try $K=\mathbf Q$: how often is $\mathbf Z^\times \to (\mathbf Z/m\mathbf Z)^\times$ surjective? Even if the unit group is infinite, such surjectivity is certainly not always true. Try $K=\mathbf Q(\sqrt{2})$: you want $(\mathbf Z[\sqrt{2}]/\mathfrak p)^\times = \pm\langle 1+\sqrt{2} \bmod \mathfrak p\rangle$ for prime ideals $\mathfrak p$, which is a nontrivial constraint on $\mathfrak p$. $\endgroup$
    – KConrad
    Commented Feb 29 at 1:44
  • $\begingroup$ I wrote more about my last example in an answer to the MO question mathoverflow.net/questions/31495/…. $\endgroup$
    – KConrad
    Commented Feb 29 at 1:48
  • $\begingroup$ @user523984: I see, sorry, I overlooked the "unit group" of $O_K^{\times}$ part in your answer. Then clearly the surjectivity is a nontrivial (...even expectably rather rarely fullfillable) condition as KConrad's example demonstrates. Nevertheless, could you maybe sketch roughly the idea why if we assume that this surjectivity considition in your answer is satisfied, then $K^{\mathfrak{m}}$ must be unramified over $\mathfrak{m}$? $\endgroup$
    – user267839
    Commented Feb 29 at 11:07

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