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I read online (in https://mathoverflow.net/questions/184347/intersection-of-a-ring-class-field-of-a-quadratic-field-k-with-the-cyclotomic-ex) that an (apparently not so trivial) exercise in class field theory shows that the roots of unity contained in ring class fields (of any conductor) on an imaginary quadratic field $K$ are always contained in $K(\mu_{12})$, where $\mu_{12}$ is the group of 12-th roots of unity.

Unfortunately, I can't find any proof or reference of this fact: can someone help me?

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    $\begingroup$ apparently it is $\mu_{24}$ not $\mu_{12}$, see the end of mathoverflow.net/a/195041/84768 $\endgroup$ Commented Mar 8, 2022 at 0:51
  • $\begingroup$ Thankyou very much for your comments and your answer. It is not 100% clear to me how to relate different abelian extensions using Artin's map, but I will search in literature. You're right, it seems that it is $\mu_{24}$. But a natural question arises: in the mathoverflow answer he finds $\zeta_8$ when $K=\mathbb{Q}(i)$. Maybe there is some issue with the fact that $\mathbb{Q}(i)$ contains units different from $\pm 1$? So my question can now be: if $K$ has only $\{\pm1\}$ as units (i.e. $K\ne \mathbb{Q}(i),\mathbb{Q}(\zeta_3)$), can we restrict to $\zeta_{12}$? $\endgroup$ Commented Mar 8, 2022 at 9:38
  • $\begingroup$ The negative result that I'm using is clear "if $(.,L/K)$ doesn't factor through $(.,E/K)$ then $L\not\subset E$", the positive result that "if $L/K,E/K$ are abelian and $(.,L/K)$ factors through $(.,E/K)$ then $L \subset E$" follows from that we can deduce $[E:K]$ and $[EL:K]$ from the surjectivity of the Artin maps and that $(.,L/K)\times (.,E/K)$ is the Artin map of $EL/K$ (embedding $Gal(EL/K)$ into $Gal(L/K)\times Gal(E/K)$) $\endgroup$ Commented Mar 8, 2022 at 10:47
  • $\begingroup$ Ok, I was just wondering about the domain of the Artin maps, because (from what I know) the domain depends on the primes ramifying in the extension. But maybe in our case we can say that if $n\nmid d$ then of course $\zeta_n\notin K_O$, as $K_O/K$ is ramified only at primes dividing $d$. Hence we can reason just for $n|d$, so we can fix the modulus $d$ and reason with both artin maps defined in the class field of ideals coprime with d. And in here I understand what you mean by "factoring maps", since both artin maps have the same domain. $\endgroup$ Commented Mar 8, 2022 at 11:18
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    $\begingroup$ Well anyway we can restrict the Artin map to the unramified primes/product of primes, I didn't clarify this but there are infinitely many possible coprime $a^2\ne 1\bmod n$ in my answer so all but finitely many must be a product of unramified primes. The Artin map is surjective when restricted to (the group of fractional ideals generated by) "all but finitey many (unramified) primes". $\endgroup$ Commented Mar 8, 2022 at 11:21

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Not an expert of those things so please check that I'm not confused with something, and I can only show that the roots of unity of $K_O$ are contained in $\langle\zeta_{24}\rangle$.

We have an order $O=\Bbb{Z}+dO_K$.

The Artin map of the ring class field $K_O/K$ is $$(I,K_O/K)= [I\cap O]\ \in Cl(O)$$

For $a\in \Bbb{Z},\gcd(a,d)=1$ then $aO_K\cap O=aO$.

Proof: take $b\in \Bbb{Z},ab=1\bmod n$, if $av=c+dw\in aO_K\cap O$ with $c\in \Bbb{Z},v,w\in O_K$ then $v=abv=bc\in (O_K/dO_K)^\times$ and hence $v\in O$.

Fix $n\in \Bbb{Z},n\nmid 24$. The Artin map of $K(\zeta_n)/K$ is $$(I,K(\zeta_n)/K)=N(I)\in \Bbb{Z}/n\Bbb{Z}^\times$$

For some $a\in \Bbb{Z},\gcd(a,dn)=1$ we have $a^2 \ne 1\bmod n$ so that $$(aO_K,K(\zeta_n)/K)=N(aO_K)=a^2\ne 1\in \Bbb{Z}/n\Bbb{Z}^\times$$ whereas

$$(aO_K,K_O/K)=[aO_K\cap O]=[aO]=1\in Cl(O)$$

whence $\zeta_n\not \in K_O$.

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