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I'm working on a proof to show that $\mathbb{Q}(\zeta)$ has class number one, where $\zeta$ is a primitive 20-th root of unity. Now in this exercise there is a hint:

"Show that all prime ideals above any of the primes 2 [...] are principal. [...] Consider the quadratic subfields of $\mathbb{Q}(\zeta)$. The prime $2$ may be treated via $\mathbb{Q}(i)$ [...]"

Now I know in a Galois extension how to gain certain information from the inertia/decomposition group, however I don't know how I can connect the theory of class groups to intermediate fields. I must be missing some key connection, because obviously there must be some way to use that e.g. $\mathbb{Q}(i)$ has class number 1 in order to conclude that a prime ideal above 2 is principal.

Can someone give me a hint/explanation on the connection or point me to some resources? I'm not looking for a solution to the exercise above (it's rather meant to illustrate the general question).

Thanks a lot in advance!

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  • $\begingroup$ I'm not sure about general theory, but for the prime $2$ above I guess they want you to show it ramifies in $\mathbb{Q}(i)$ but is inert in $\mathbb{Q}(\zeta_5)$ so taking the compositum will tell you it must be principal in $\mathbb{Q}(\zeta_{20})$. $\endgroup$ – Matt B Jan 10 '17 at 13:51
  • $\begingroup$ Okay, thanks. Can you provide some details why you are able to draw this conclusion from the inertness / ramification in those two fields? $\endgroup$ – johnnycrab Jan 10 '17 at 14:26
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The idea I'm using is that if a prime $\mathfrak{p}$ of $F$ is totally inert in the extension $K/F$, then $\mathfrak{P}=\mathfrak{p}\mathcal{O}_K$ is also prime. Therefore $\mathfrak{P}$ is principal if $\mathfrak{p}$ is (with the same generator).

Note that $\mathbb{Q}(i)/\mathbb{Q}$ is totally ramified at $2$ which in particular means that the residue degree is $1$. On the other hand, we have another subfield $\mathbb{Q}(\zeta_5)$ of $\mathbb{Q}(\zeta_{20})$ (where $\zeta_n$ is a primitive $n$th root of unity), which is totally inert at $2$ with residue degree $4$.

Since $[\mathbb{Q}(\zeta_{20}) : \mathbb{Q}]=8$, using properties of the residue and ramification degrees, we can show that the extension $\mathbb{Q}(\zeta_{20})/\mathbb{Q}(i)$ is totally inert at $2$ (by which I mean the residue degree equals the degree of the extension). All that's left to do in this case is to show that the ramified prime above $2$ in $\mathbb{Q}(i)$ is principal (it has generator $1+i$), so we are done.

Note that if the prime happened to split anywhere then we would have to be more careful.

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  • $\begingroup$ Very interesting! Before I upvote: As far as I have seen, $\mathbb{Q}(\zeta_5)$ is actually the inertia field of (2) in $\mathbb{Q}(\zeta_{20})$. Which would imply that 2 totally ramifies. Am I wrong? $\endgroup$ – johnnycrab Jan 10 '17 at 15:55
  • $\begingroup$ @johnnycrab You're right; that was a typo so I'll edit. $\endgroup$ – Matt B Jan 10 '17 at 16:07
  • $\begingroup$ Very nice answer, thanks for the effort! $\endgroup$ – johnnycrab Jan 10 '17 at 16:08
  • $\begingroup$ You could of course do it that way round instead of course though; I've shown $2$ is still prime (hence principal) in $\mathbb{Q}(\zeta_5)$ and then you're left with the same computation with the ramified part. $\endgroup$ – Matt B Jan 10 '17 at 16:10
  • $\begingroup$ Yes, I thought so as well. I just got confused by the typo. $\endgroup$ – johnnycrab Jan 10 '17 at 16:12

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