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I just started learning about (global) class field theory and I am having some problems with the setup.

Suppose $\mathfrak{m}=\mathfrak{m}_f \mathfrak{m}_\infty$ be a modulus of an abelian extension $L/K$ of number fields. We define $$ K_\mathfrak{m}:=\{a/b~|~a,b\in\mathcal{O}_K, \text{ and }ab \text{ is prime to }\mathfrak{m}_f\} $$ and $$ K_{\mathfrak{m},1}:=\{x\in K_{\mathfrak{m}}~|~x\equiv 1\bmod{\mathfrak{m}}\}. $$ Here, by $x\equiv 1\bmod \mathfrak{m}$, one means two things: (i) $x\equiv 1 \bmod {\mathfrak{m}_f}$ and (ii) $\sigma(x)>0$, for all real embeddings $\sigma$ dividing $\mathfrak{m}_\infty$. I want to understand what is the purpose of (ii). I cannot find any book which explains that part. Also I would like to know if there is way to find $\sigma$'s with such property. Like, for example, if $K=\mathbb{Q}$ and $L=\mathbb{Q}(\zeta_n)$, then one can take $\mathfrak{m}=n\infty$, where $\infty$ is the unique real infinite place. I do not understand which one of the embeddings $\sigma_k:\zeta\to \zeta^k$ ($\gcd(k,n)=1$), the $\infty$ is supposed to be.

KCd
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ShyamalSayak
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  • After considering the various alternative interpretations, I suspect CFT here refers to "class field theory". I'd use the verbose version since in its present form the Question makes only that one mention of CFT. – hardmath Aug 14 '23 at 04:21
  • I edited the way you typed congruences, in particular changing your use of \pmod to \bmod. The spacing with \pmod looked terrible but the spacing with \bmod always looks fantastic, so I advise you always to use \bmod. :) – KCd Aug 14 '23 at 04:50
  • @KCd Thank you Sir for the suggestion. – ShyamalSayak Aug 14 '23 at 04:59
  • One way to workaround the extra \pmod spacing in \displaystyle is to wrap the \pmod in a \textstyle. Using \bmod is incorrect @KCd – Bill Dubuque Aug 14 '23 at 05:09
  • @BillDubuque in my experience \bmod always looks great, so I don't see in what sense it is incorrect. Well, there was just one time I saw a weird spacing with \bmod, which occurred when I was using "mod" outside the context of a congruence condition (namely when writing about a congruence class like $2 \bmod 5$ instead of something like the relation $x \equiv y \bmod 5$). Normally even in that situation with a congruence class \bmod still looks fine, but one time it did not. I understand the "b" in \bmod refers to "binary" as in "the binary congruence relation denoted mod". – KCd Aug 14 '23 at 05:16
  • @KcD bmod is for a binary mod operation, not congruence relations. See here and here for further discussion. Iirc fixes for this problem are discussed on the LaTeX SE site. – Bill Dubuque Aug 14 '23 at 05:20
  • @BillDubuque okay, I did not know that was the intention behind the command \bmod. Nevertheless, I think \bmod produces very nice-looking output in congruence relations and I never felt parentheses had to be used in a systematic way around "mod $m$" in congruence relations when the modulus $m$ is a single character or other short expression, so I've never used \pmod. – KCd Aug 14 '23 at 05:27
  • @KCd How would you write $,x \equiv x\bmod m\pmod{!m},$ using your nonstandard style? $\ \ $ – Bill Dubuque Aug 14 '23 at 05:33
  • @BillDubuque heh-heh, I'd never do anything like that, since I think that kind of double use of "mod" (where the two uses of mod have different meanings) is clunky, at least in mathematics. To express the idea you asked about, I would write $x \equiv a \bmod m$ where $0 \leq a \leq m-1$. – KCd Aug 14 '23 at 05:36
  • @KcD So you can't notate mod operations, which makes it cumbersome to do calculations, e.g. you can't use the powerful mod distributive law: $,cn\equiv 0,\Rightarrow, ca\equiv c(a\bmod n) \ \pmod{! m},,$ which provides a slick operational way to do CRT calculations. And, yes, we do often calculate in number theory. – Bill Dubuque Aug 14 '23 at 05:45
  • Over many years of teaching number theory I have never been unable to express what I want to say when teaching CRT or any other topic related to modular arithmetic (and students do learn in the course how to use CRT in examples). One can always create a temporary notation if necessary, e.g., when stripping off the last base $10$ digit in the decimal expansion of a positive integer $n$, like going from $1234$ to $123$, instead of writing the new number as $(n - (n \bmod 10))/10$ I can just say what I want to do in words and call the new number $n'$, e.g., if $n = 1234$ then $n' = 123$. – KCd Aug 14 '23 at 08:07
  • @KCd Of course you can still express it, but it is more cumbersome. The benefit of having normal form reduction operations is that they enable efficient straight-line calculations, e.g. see here (integer) and here (polynomial) for simple instances of their use in slick CRT / interpolation calculations. – Bill Dubuque Aug 14 '23 at 18:18

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You say you can't find any book which explains the need for condition (ii). Just try reading accounts of class field theory while ignoring that condition and you will eventually find things stop being correct.

Consider the need for positive primes in quadratic reciprocity, e.g., when $(p)$ is a nonzero prime ideal in the integers, $(p)$ splits in $\mathbf Q(i)$ if and only if $p \equiv 1 \bmod 4$ when $p$ is positive. It would be wrong to say a nonzero prime ideal $(p)$ splits completely in $\mathbf Q(i)$ if and only if $p \equiv 1 \bmod 4$ while imposing no sign condition on the generator $p$ because $(-3)$ does not split completely in $\mathbf Q(i)$ and $-3 \equiv 1 \bmod 4$. So $(p)$ splits completely in $\mathbf Q(i)$ if and only if $p \equiv 1 \bmod 4\infty$.

More generally, consider the rule for prime ideals in $\mathbf Z$ to split completely in cyclotomic fields. Each cyclotomic field is $\mathbf Q(\zeta_m)$ for a unique positive integer $m$ such that $m \not\equiv 2 \bmod 4$. Using such $m$, a nonzero prime ideal $(p)$ splits completely in $\mathbf Q(\zeta_m)$ if and only if $p$ is positive and $p \equiv 1 \bmod m$, not just when $p \equiv 1 \bmod m$ without a sign condition. So $(p)$ splits completely in $\mathbf Q(\zeta_m)$ if and only if $p \equiv 1 \bmod m\infty$.

When $m$ is a positive integer, the ray class group with modulus $(m)$ is $\mathbf Q(\zeta_m + \zeta_m^{-1})$, and it is not true that every finite abelian extension of $\mathbf Q$ is contained in such a field: we only get finite abelian extensions inside $\mathbf R$ that way. You'd never find imaginary quadratic fields or cyclotomic fields $\mathbf Q(\zeta_m)$ with $m \geq 3$ this way. The rule for a nonzero prime ideal $(p)$ in $\mathbf Z$ to split completely in $\mathbf Q(\zeta_m + \zeta_m^{-1})$ is that $p \equiv \pm 1 \bmod m$, which does not depend on whether $p$ is positive or negative, so we can say the nonzero prime ideals $(p)$ splitting completely in $\mathbf Q(\zeta_m + \zeta_m^{-1})$ are those with some generator $p$ where $p \equiv 1 \bmod (m)$. The ray class field mod $(m)\infty$ is $\mathbf Q(\zeta_m)$, and such fields contain all finite abelian extensions of $\mathbf Q$. You can't express that idea in the language of multiplicative congruence conditions if you don't allow yourself to use positivity conditions on generators of prime ideals.

KCd
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  • What is the $\infty$ in the cyclotomic extension, i.e., for which k, $\sigma_k$ is $\infty$? And also are you saying that the definition of $K_{\mathfrak{m}}$ is related to the condition of primes splitting completely? – ShyamalSayak Aug 14 '23 at 05:05
  • I don't understand your question: a cyclotomic field $\mathbf Q(\zeta_m)$ with $m \geq 3$ has no real embeddings. But the base field $\mathbf Q$ has a real embedding and that is what the $\infty$ refers to: the fields $\mathbf Q(\zeta_m)$ are ray class fields over the rational field $\mathbf Q$. Reread what the real embeddings are related to. – KCd Aug 14 '23 at 05:07
  • You are absolutely right. I misunderstood where we were considering real embeddings from. I thank you for your wonderful explanation above. – ShyamalSayak Aug 14 '23 at 14:02