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!

  • $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.