understanding $\mathrm{Spec}\mathbf{Z}[\mu_{p-1},\frac{1}{(p-1)p}] \cap \mathbf{Z}_p$ What's $\mathrm{Spec}\mathbf{Z}[\mu_{p-1},\frac{1}{(p-1)p}] \cap \mathbf{Z}_p$, where $\mu_{p-1}$ denotes the $(p-1)$-th roots of unity and $\mathbf{Z}_p$ the $p$-adic integers?
It should be $\mathbf{Z}[\mu_{p-1}]$ without (the places above the primes dividing $p-1$ and all but one places above $p$).
 A: To answer your question, you first have to choose a map
$\mathbb Q[\mu_{p-1}] \subset \mathbb Q_p.$
To this end, note that $p \equiv 1 \bmod p-1,$ hence $p$ splits completely
in $\mathbb Q[\mu_{p-1}]$, and so choosing such an embedding amounts to
choosing one prime $\mathfrak p$ above $p$.   The embedding is then characterized by the fact that the maximal ideal of $\mathbb Z_p$ pulls back 
to $\mathfrak p$, while all the other prime ideals dividing $p$ generate the unit ideal of $\mathbb Z_p$.
In order to keep things concrete, it will be easier to think about actual elements (or, if you like, pricipal ideals) rather than prime ideals that are not necessarily principal.
To this end, choose a power $h$
such that $\mathfrak p^h$ is principal, say equal to $\pi$.  We may then write
$p^h = \pi \pi',$ where $\pi'$ is prime to $\mathfrak p$. (The ideal generated
by $\pi'$ is the product of the $h$th powers of all the primes other than
$\mathfrak p$ which divide $p$.)
Then we see that $$\mathbb Z[\mu_{p-1},\frac{1}{(p-1)p}] = \mathbb Z[\mu_{p-1},\frac{1}{(p-1)^h\pi\pi'}].$$
The elements $p-1$ and $\pi'$ are invertible in $\mathbb Z_p$, while 
$\pi$ is not (it generates the $h$th power of the maximal ideal of $\mathbb Z_p$).    Thus we see that
$$\mathbb Z[\mu_{p-1},\frac{1}{(p-1)p}] \cap \mathbb Z_p
= \mathbb Z[\mu_{p-1},\frac{1}{(p-1)\pi'}],$$
and so its Spec coincides with that of $\mathbb Z[\mu_{p-1}]$ with
the prime above $p-1$ and all the primes dividing $\pi'$ (i.e. all but one
of the primes over $p$) removed (confirming your claim).
