# 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 $p$.

• I have a few questions-- so would the natural embedding of $\mathbb{Z}$ into $X$ be given by $x\mapsto (x,0)$? Then by your answer to my subsequent question, would it be more accurate to say that $d^x$ being defined for all $x\in X$ implies that $d\in 1+p\mathbb{Z}_p$? (2) In the decomposition for $d$ should there be an additional factor equal to a power of the prime element $p$ generating the maximal ideal of $\mathbb{Z}_p$? (3) Is there an easy way of seeing that $X$ is the set of group homomorphisms $\mathbb{Z}^{\times}_p\rightarrow \mathbb{Z}^{\times}_p$? Commented Jul 6, 2013 at 14:04
• (4) As you say, for the $p$-adic Eisenstein series, the sum defining the coefficient of $q^n$ is taken over all divisors $d$ of $n$ coprime to $p$. Wouldn't this involve integers which are not congruent to 1 mod$(p)$, therefore we wouldn't be able to raise them to the power $k-1$? Thank you very much. Commented Jul 6, 2013 at 14:14
• Please ignore my second question- you're right of course since $d\in \mathbb{Z}_p- p\mathbb{Z}_p$ so $v_p(d)=0$. Commented Jul 6, 2013 at 16:02
• The natural embedding of $\mathbb{Z}$ into $X$ is given by $n \mapsto (n, n \bmod{p-1})$. Commented Jul 6, 2013 at 21:31
• Your questions (1) and (4) suggest you are labouring under a misunderstanding. The definition of $d^x$ for $x \in X$ makes sense when $d \in \mathbb{Z}_p^\times$, and this definition is different from the definition of $d^y$ for $y \in \mathbb{Z}_p$, which only makes sense when $d \in 1 + p\mathbb{Z}_p$. Commented Jul 7, 2013 at 8:00