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For a global function field $K$, What is the difference between the ray class group modulo $\mathfrak{m}$ (as defined in say, JS Milne's CFT notes) and the $S$-class group (as defined in say, Rosen's book "Number Theory in Function Fields") for $S$ being equal to the support of modulus $\mathfrak{m}$ ?

the ray class group is $I(\mathfrak{m})/P(\mathfrak{m})$, where $I(\mathfrak{m})$ is the free abelian group generated by primes which do not divide $\mathfrak{m}$ (sounds like the group of $S$-divisors). And $P(\mathfrak{m})$ is the set of principle divisors of $\{a \in K^* : v_p(x -1 ) \geq \mathfrak{m}_p \}$ (I am ignoring real places because $K$ is a function field). This looks a lot like the principle $S$ divisors to me.

Rosen states that the S-class group is finite, whereas the notes from Pete Clark at http://math.uga.edu/~pete/8410Chapter8.pdf suggests that the ray class group mod $\mathfrak{m}$ is only finite if the constant field of $K$ is algebraically closed. I note that Clark does not require that $\mathfrak{m}_p > 0$, could this be the cause of the discrepancy, or am I otherwise confused ?

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  • $\begingroup$ ray class groups are extensions of class groups, S-class groups are quotients. $\endgroup$ – franz lemmermeyer May 24 '17 at 5:42
  • $\begingroup$ that is easy enough to see in the number field case, as the the ray modulo $\mathfrak{m}$ is generated by real primes, which would never occur in the S-divisors. But it is not clear to me how this follows in the function field case. And in any case doesn't tell me where my proof is wrong. I am guessing that $P(\mathfrak{m}) \subsetneq \text{div}_S(K^*)$ ? $\endgroup$ – mebassett May 24 '17 at 14:03

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