I am working on a proof in the lecture of Milne "Proposition 6.2 b)" but there is a step I don't get:

We have an inclusion $\mathbb{Z}\hookrightarrow \mathcal{O}_K$ that induces the following isomorphism $\mathbb{Z}/(p)\to \mathcal{O}_K/(1-\zeta)$.
This means $\mathcal{O}_K=\mathbb{Z}+(1-\zeta)\mathcal{O}_K$.
I guess this should be easy to see, but where do we get this equation from? The rest of the proof is clear.

I'd be happy if someone could help me with this equation!
All the best, Luca

  • $\begingroup$ Clearly if $\Bbb Z\to{\cal O}_K/(1-\zeta)$ is onto then every element of ${\cal O}_K$ is in $\Bbb Z$ up to a multiple of $(1-\zeta)$! $\endgroup$ – anon Oct 29 '13 at 15:50

If you agree that $\mathbb Z/(p)\to\mathcal O_K/(1-\zeta)$ is an isomorphism, then you must agree that $\mathcal O_K=\{0,1,\cdots,p-1\}+(1-\zeta)\mathcal O_K$. This last is contained in $\mathbb Z+(1-\zeta)\mathcal O_K$.

  • $\begingroup$ The equation ${\cal O}_K=\Bbb Z+(1-\zeta){\cal O}_K$ is not actually in the notes (or I didn't see it), which indicates that this is Luca's reimagining of the isomorphism. Thus, Luca is essentially asking why the isomorphism is true, which isn't trivial. $\endgroup$ – anon Oct 29 '13 at 15:45
  • 1
    $\begingroup$ It is in the notes: $\mathcal{O}_K=\mathbb{Z}+\pi \mathcal{O}_K$, where $\pi=1-\zeta$. And Milne proved before that $|\mathcal{O}_k/(\pi):\mathbb{Z}/(p)|=1$. $\endgroup$ – Dietrich Burde Oct 29 '13 at 15:46
  • $\begingroup$ Sorry, I did not read the next page. [In my defense, it is a bit strange someone would be able to read the rest of the proof but not understand this bit.] $\endgroup$ – anon Oct 29 '13 at 15:48
  • $\begingroup$ Thanks a lot. We confused ourselves working on this proof ;) $\endgroup$ – Luca Oct 29 '13 at 18:00

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