p-adic Eisenstein series I'm trying to understand the basic properties of the p-adic Eisenstein series.
Let $p$ be a prime number.
Define the group $X = 
\begin{cases} 
\mathbb{Z}_p\times \mathbb{Z}/(p-1)\mathbb{Z}  & \mbox{if }p \neq2 \\
\mathbb{Z}_2 & \mbox{if } p=2
\end{cases}$ 
where $\mathbb{Z}_p$ is the ring of $p$-adic integers.
If $k\in X$ and $d$ is a positive integer then is it true that $d^{k-1}\in \mathbb{Z}_p$? If so, why?
Thank you for your help.

Like your previous question, there's a slight philosophical issue: the question should not be "is $d^{k-1} \in \mathbb{Z}_p$", but "when and how is $d^{k-1}$ defined"? It's far from obvious what the definition should be, but once you know what the conventional definition is, the fact that it gives you something in $\mathbb{Z}_p$ whenever it is defined is totally obvious :-)
So we have to do something to define $d^{x}$ for $x \in X$, and it will only work if $d \in \mathbb{Z}_p^\times$. The usual definition is as follows. Let me assume $p \ne 2$ -- you can work out the necessary modifications for $p = 2$ yourself.
Suppose $x = (x_1, x_2)$ where $x_1 \in \mathbb{Z}_p$ and $x_2 \in \mathbb{Z} / (p-1)\mathbb{Z}$. Write $d = \langle d \rangle \tau(d)$, where $\tau(d)$ is a $(p-1)$-st root of unity and $\langle d \rangle$ is in $1 + p\mathbb{Z}_p$ (this can always be done, in a unique way, for any $d \in \mathbb{Z}_p^\times$). Then we define
$$ d^x = \langle d \rangle^{x_1} \tau(d)^{x_2} $$
which is well-defined (using your previous question) and lies in $\mathbb{Z}_p^\times$. It's easy to check that this agrees with the "natural" definition of $d^x$ when $x \in \mathbb{Z}$ (which lives inside $X$ in the obvious way). In fact $X$ is exactly the set of group homomorphisms $\mathbb{Z}_p^\times \to \mathbb{Z}_p^\times$.
If $k \in X$ we can now define $d^{k-1} = d^k / d$, where $d^k$ is defined as above.
There's no sensible definition of $d^x$ for $x \in X$ and $d \in p\mathbb{Z}$, which is why the definition of the coefficient of $q^n$ in the $p$-adic Eisenstein series involves a sum over only those divisors of $n$ coprime to $p$.
