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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?

YCor
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user267839
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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

user523984
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  • 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 $\mathfrak{m}$. – user267839 Feb 29 '24 at 00:46
  • 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. – user523984 Feb 29 '24 at 01:20
  • @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 $\mathfrak p$. – KConrad Feb 29 '24 at 01:44
  • I wrote more about my last example in an answer to the MO question https://mathoverflow.net/questions/31495/when-does-a-ring-surjection-imply-a-surjection-of-the-group-of-units. – KConrad Feb 29 '24 at 01:48
  • @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 $\mathfrak{m}$? – user267839 Feb 29 '24 at 11:07
  • Step 1 is to understand the definition of the ray class group. Preferably in simple classical Language. Step 2 is to compute several examples, say for conductors $\mathfrak{p}$ in $K = \mathbf{Q}(\sqrt{2})$ as in the comment @KConrad. Either you will think about these computations and my answer will be clear, or you will fail to compute then and then you need to think about more basic questions. – user523984 Feb 29 '24 at 14:45
  • To avoid misunderstandings, in the following I will refer to definitions/notions from these notes: https://math.mit.edu/classes/18.785/2017fa/LectureNotes22.pdf , which which works with definition of ray class group, which I would call "classical" (...in contrast to adelic formulation), is that what you mean by "classical language." If yes, lat's deal firstly with case of the the modulus $\mathfrak{m}=\mathfrak{p}$ a single finite prime as in your answer. Then I assume that this condition you mentioned in your answer that the ramification behavior of the ray call field $K^{\mathfrak{m}}/K$ – user267839 Mar 01 '24 at 00:54
  • in $\mathfrak{m}$ is completely depending on if the units in$\mathcal{O}^{\times}_K$ surject onto the units of $(\mathcal{O}_K/\mathfrak{m})^{\times}$ is captured by condition posed in 3rd dotted part in the linked notes, ie those on the subgroup $K^{\mathfrak{m},1} \subset K^{\times}$ using the notions from there, right? So if such $\alpha$ with $v_m(\alpha-1)>1$ this implies in context of your criterion that the units of $O_K$ surject onto unity of residue field wrt to $\mathfrak{m}$; that's how your surjectivity criterion is related to the descriptions in the linked notes, right? – user267839 Mar 01 '24 at 01:07
  • If what I wrote so far make sense, let's come briefly to general case, ie when $\mathfrak{m}$ is any fractional ideal. When you write "The general case is along similar lines", then you just mean that it is consistent with local-to-global principle; in other words can analyzed at every prime $\mathfrak{p}$ dividing $\mathfrak{m}$ as in "key" case $\mathfrak{m} = \mathfrak{p}$ above. If I misunderstand that point, could you correct me what you then else mean by that "The general case is along similar lines"? – user267839 Mar 01 '24 at 01:14