# what's the conductor of a ray class field?

Let $K$ be a number field. The theorems of class field theory tell us that given any modulus $\mathfrak{m}$ for $K$, there is a unique Abelian extension $K_{\mathfrak{m}}$ such that the kernel of the Artin map of $K_{\mathfrak{m}}/K$ with respect to $\mathfrak{m}$ is precisely the subgroup of principal fractional ideals congruent to $1 \pmod{\mathfrak{m}}$. This is the Ray class field.

Moreover, we know that the conductor of $K_{\mathfrak{m}}/K$ divides $\mathfrak{m}$. Is it equal in general? If not, what are counterexamples?

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Dear Tony, Have you tried the case $K = \mathbb Q$, and various (small!) choices of $\mathfrak m$? Regards, – Matt E May 15 '12 at 20:04
No, the conductor of $K_{\mathfrak{m}}$ can properly divide $\mathfrak{m}$. Hint: this already happens for $K = \mathbb{Q}$. The basic culprit behind this is the fact that $\mathbb{Q}(\zeta_{2n}) = \mathbb{Q}(\zeta_n)$ when $n$ is odd. – Pete L. Clark May 15 '12 at 20:05
Oops, silly me. So I see that it already fails for some cyclotomic fields (namely, when adjoining a 2* (2n-1) root of unity)? – Tony May 15 '12 at 20:08