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I am referring to Proposition 1.6 of CHAPTER 5 in Milne's notes on Class Field Theory, it claims that given a modulus $\mathfrak m$ of $K$, and its ray class group $\mathcal C_{\mathfrak m}$,

Every class in $\mathcal C_{\mathfrak m}$ is represented by an integral ideal $\mathfrak a$, and two integral ideals $\mathfrak a$ and $\mathfrak b$ represent the same class in $\mathcal C_\mathfrak m$ if and only if there exit nonzero $a,b \in \mathcal O_K$ such that $$a \equiv b \equiv 1 \mod{\mathfrak m_0},$$ and $a$ and $b$ have the same sign for every real prime dividing $\mathfrak m$.

My question is this:

For a proof, taking an element $\mathfrak a$ of $I^{S(\mathfrak m)}$ and writing it as $\mathfrak b \mathfrak c^{-1}$ with $\mathfrak b$ and $\mathfrak c$ integral ideals in $I^{S(\mathfrak m)}$, he chooses a nonzero $c \in \mathfrak c \cap K^S$, (which exists buy CRT) and notes that $c \mathfrak a$ is integral. What guarantees that $c \in K_{\mathfrak m,1}$, since $\mathcal C_{\mathfrak m}$ is defined to be $I^{S(\mathfrak m)}/K_{\mathfrak m,1}$? From the construction suggested, it only follows that the prime factoisation of $c$ doesn't involve any finite primes of $S(\mathfrak m)$. Indeed, we can apply CRT after requiring that $c$ be $1 \equiv \wp^{\mathfrak m(\wp)}$ for all finite primes $\wp$ dividing $\mathfrak m$, but it still doesn't guarantee that $c$ is totally positive.

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  • $\begingroup$ ad 2: multiply the congruence by the inverse residue class of a. $\endgroup$ – franz lemmermeyer Aug 5 '17 at 10:53
  • $\begingroup$ Thanks! I realised it soon afterwards. So I deleted that part of the question in an edit. I still have no clue regarding the total positivity of $c$. $\endgroup$ – Sameer Kulkarni Aug 5 '17 at 11:27
  • $\begingroup$ I'm not sure I understand your problem, but why can't you choose $c$ totally positive? Every nonzero residue class contains totally positive elements by the strong approximation theorem (CRT). $\endgroup$ – franz lemmermeyer Aug 5 '17 at 12:03
  • $\begingroup$ Ah I see. That answers it. I didn't think of the approximation theorem. Thanks!! $\endgroup$ – Sameer Kulkarni Aug 5 '17 at 12:16

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