Cyclotomic Integers? Let $K=\mathbb{Q}(\zeta_{p^\infty})$ be the extension of $\mathbb{Q}$ obtained by adjoining all $p-$power roots of unity. 
My question is : how to show that the the ring of integes of $K,$ $O_K$ is not a Dedekind domain ?
Thanks.
 A: Hint: It's not Noetherian. Try and show that above some prime of $\mathbb{Z}$ there are infinitely many primes. Check this by arguing that you can find extensions of arbitrarily large degree where the prime splits. 
EDIT: Of course, you knew that it had to be non-Noetherianess since normality and dimension $1$ are true for any integral closure of $\mathbb{Z}$ in a field extension of $\mathbb{Q}$.
A: At this point, I think there's been enough discussion in the comments to turn it into a proper answer.
As Alex noted, our goal is really to prove that $\mathcal O_K$ is not Noetherian. To do so, we will prove that some rational prime $\ell$ "factors into infinitely many factors" in $\mathcal O_K$.
We will do this for $p=\ell$. Recall that 
$$(p)=(1-\zeta_p)^{p-1}=(1-\zeta_{p^2})^{p^2-p}=\ldots=(1-\zeta_{p^n})^{\varphi(p^n)}=\ldots$$
Then we have an increasing chain 
$$(p)\subset(1-\zeta_p)\subset(1-\zeta_{p^2})\subset\ldots\subset(1-\zeta_{p^n})\subset\ldots$$
and thus the ring is not Noetherian. One can think of this as a failure of unique factorization into prime ideals, which characterizes Dedekind rings. In this case, the uniqueness is not the problem, but rather the existence of a factorization in the first place--we can keep on factoring and factoring $(p)$ but will never get all the way down to prime ideals.
It turns out that the only possible choice for $\ell$ was $p$: Suppose $\ell\neq p$ and let $n$ be the greatest integer so that $\zeta_{p^n}\in\mathbb Q_\ell$. Then $[\mathbb Q_\ell(\zeta_{p^{n+m}}):\mathbb Q_\ell]=\frac{\varphi(p^{n+m})}{\varphi(p^n)}=[\mathbb Q(\zeta_{p^{n+m}}):\mathbb Q(\zeta_{p^n})]$, so primes lying above $\ell$ in $\mathbb Q(\zeta_{p^n})$ remain prime in larger subextensions.
