Timeline for Roots of unity inside ring class fields of an imaginary quadratic field $K$ always contained in $K(\mu_{12})$?
Current License: CC BY-SA 4.0
9 events
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| Mar 10, 2022 at 10:31 | comment | added | Fraz | Ok thankyou. By the way, do you know where I can find some justifications for the construction made in mathoverflow.net/questions/184347/… ? I tryed to search in Cox's book, but I can't find such precise statements | |
| Mar 8, 2022 at 11:21 | comment | added | reuns | 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". | |
| Mar 8, 2022 at 11:18 | comment | added | Fraz | 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. | |
| Mar 8, 2022 at 10:47 | comment | added | reuns | 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 $Gal(L/K)\times Gal(E/K)$) | |
| Mar 8, 2022 at 9:38 | comment | added | Fraz | 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 $\zeta_{12}$? | |
| Mar 8, 2022 at 9:23 | vote | accept | Fraz | ||
| Mar 8, 2022 at 0:51 | comment | added | reuns | apparently it is $\mu_{24}$ not $\mu_{12}$, see the end of mathoverflow.net/a/195041/84768 | |
| Mar 7, 2022 at 14:14 | answer | added | reuns | timeline score: 3 | |
| Mar 7, 2022 at 12:05 | history | asked | Fraz | CC BY-SA 4.0 |