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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 $K(\mu_{24})$.

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

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

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 $K(\mu_n)/K$ is $$(I,K(\mu_n)/K)=N(I)\in \Bbb{Z}/n\Bbb{Z}^\times$$

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

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

whence $\mu_n\not \in K_O$.

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 $\zeta_n\not \in K_O$.

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 $K(\mu_{24})$.

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

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

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 $K(\mu_n)/K$ 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 $$(aO_K,K(\zeta_n)/K)=N(a)=a^2\ne 1\in \Bbb{Z}/n\Bbb{Z}^\times$$ whereas

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

whence $\mu_n\not \in K_O$.

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 $K(\mu_{24})$.

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

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

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 $K(\mu_n)/K$ is $$(I,K(\mu_n)/K)=N(I)\in \Bbb{Z}/n\Bbb{Z}^\times$$

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

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

whence $\mu_n\not \in K_O$.

Not an expert of those things so please check that I'm not confused with something.

$O=\Bbb{Z}+dO_K$,

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

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 $K(\mu_n)/K$ 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 $$(aO_K,K(\zeta_n)/K)=N(a)=a^2\ne 1\in \Bbb{Z}/n\Bbb{Z}^\times$$ whereas

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

whence $\zeta_n\not \in K_O$.

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 $K(\mu_{24})$.

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

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

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 $K(\mu_n)/K$ 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 $$(aO_K,K(\zeta_n)/K)=N(a)=a^2\ne 1\in \Bbb{Z}/n\Bbb{Z}^\times$$ whereas

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

whence $\mu_n\not \in K_O$.