# Basic question about power series and complete group algebras

This is a pretty basic question, but I suspect it might be too exotic for math.stackexchange.

Let $\mathbb{Z}_p$ be the $p$-adic integers. For free pro-$p$ group $F_r$ of rank $r$, we can consider the complete group algebra

$$\mathbb{Z}_p[[F_r]] := \varprojlim_U \mathbb{Z}_p[F_r/U]$$ as $U$ ranges over all open subgroups of $F_r$. Let $x_1,\ldots,x_r$ are generators for $F_r$, which we identify with their images in $\mathbb{Z}_p[[F_r]]$. We may consider the ring of formal power series $\mathbb{Z}_p[[u_1,\ldots,u_r]]$ in the noncommuting variables $u_1,\ldots,u_r$.

According to various sources (Ribes-Zalesski 5.9.1, Ihara - On Galois Representations Arising from Towers...), by sending $u_i\mapsto (x_i-1)$, apparently we get an isomorphism $$\mathbb{Z}_p[[u_1,\ldots,u_r]]_\text{nc}\stackrel{\sim}{\longrightarrow} \mathbb{Z}_p[[F_r]]$$ I'm trying to work out the simplest case, where $r = 1$, where I think of $F_1$ as $\varprojlim_n \mu_{p^n}$ with generator $x = (\zeta_p,\zeta_{p^2},\zeta_{p^3},\ldots)$ and I don't understand how the image of, say, $$a_0 + a_1u + a_2u^2 + a_3u^3 + \cdots$$ is well defined, where $u = u_1$. Even if each $a_i = 1$, the image of $1+u+u^2 + \cdots$ in the group ring $\mathbb{Z}_p[\mu_p]$ should be $$1 + (\zeta_p-1) + (\zeta_p-1)^2 + (\zeta_p-1)^3 + \cdots$$ which would seem to have constant term $1-1+1-1+1-1+1-\cdots$''.

Okay, so this is probably linked to the remark that the augmentation ideal is topologically nilpotent, which I don't see. Is it still generated by $f-1$, $f\in F_1$? (perhaps $x-1,x^2-1,x^3-1,\ldots$ would do?). Even if it's some weird topological thing I'm missing, surely that shouldn't come into play at the "finite level" $\mathbb{Z}_p[\mu_p]$?

Not sure what the best tags are for this...

• No doubt someone will spell it out for you, but if you want to figure it out yourself then look at any book or resource containing some basic Iwasawa theory (for example Washington's "cyclotomic fields") to see the construction explicitly when r=1. The point is that Z_p[[T]]/((1+T)^{p^n}-1) is isomorphic to Z_p[Z/p^nZ]. – eric Nov 18 '15 at 20:08
• @eric Perfect I'll take a look! Thanks! – Will Chen Nov 18 '15 at 20:33
• The comment above, of @eric, says it all, but to your specific question, at the very first level, if $\zeta$ is your generator of the cyclic group of order $p$, then $(\zeta-1)^p$ is already divisible by $p$, so the powers of $\zeta-1$ go to zero. – Lubin Nov 18 '15 at 20:40
• @oxeimon When you calculate the constant term by $1−1+1−1+1−1+1−\ldots ''$ you forget that $\zeta_p^p=1$ and so the constant term of $(\zeta_p-1)^p$ is not $(-1)^p$. – Andrei Jaikin Nov 19 '15 at 11:36