TITLE: The ring of integers in $\mathbb{Q}[\zeta]$ is $\mathbb{Z}[\zeta]$
QUESTION [5 upvotes]: 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

REPLY [4 votes]: 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 $\mathbb Z+(1-\zeta)\mathcal O_K$.