\section{Completed rings}
Let $\widehat{R}(G)$ be the completion of $R(G)$ with respect to the $I$-adic topology, with $I$ the augmentation ideal. Since this topology coincides with the $F$-topology and the $\Gamma$-topology, the generators of the graded rings $\rr^*(G)$ and $R^*(G)$ are topological generators for $\widehat{R}(G)$.
\begin{Remark}
The restriction and induction maps naturally extend to $\widehat{R}(G)$ and satisfy the same restriciton-corestriction formulas as in the not complete case.
\end{Remark}
\subsection{\texorpdfstring{$p$}{p}-groups}
\paragraph*{Notation.} Fix a prime $p$ and assume $G$ is a $p$-group of exponent $e$. Then the characters of $G$ all have values in $\ZZ[\mu]$, with $\mu$ a primitive $p^e$-th root of unity. We want to look at the $p$-adic completion of $\ZZ[\mu]$. The $p$-adic valuation on $\ZZ[\mu]$ is well-defined: any two embeddings $\QQ(\mu) \rightarrow \overline{\QQ}_p$ are Galois-conjugated and thus have the same image, so we can pick some primitive $p^e$-th root of unity in $\overline{\QQ}_p$ and call it $\mu$ again. Then the image of $\QQ(\mu)$ in $\overline{\QQ}_p$ is $\QQ_p(\mu)$. There is a unique valuation on $\QQ_p(\mu)$ extending that on $\QQ_p$, and since $\ZZ[\mu]$ is the ring of integers of $\QQ(\mu)$, there is a unique $p$-adic valuation on $\ZZ[\mu]$ extending that on $\ZZ$. The completed ring $\ZZ_p[\mu]$ is the ring of integers of $\QQ_p(\mu)$. Let $K = \QQ_p[\mu]$, and let $A = \ZZ_p[\mu]$.\\
Back to representation rings: we consider the evaluation morphisms $R(G) \rightarrow \ZZ[\mu]$, which we see as having values in $A$. This makes $R(G)$ into a subring of the ring of functions $\mathcal{F}(G,A)$.
\begin{Proposition}
The topological closure of $R(G)$ in $\mathcal{F}(G,A)$ is $R(G)\otimes \ZZ_p$.
\end{Proposition}
\begin{proof}
The elements of a basis for $R(G)$ (that is, the irreducible characters of $G$) are still linearly independent in $\mathcal{F}(G,K)$, thus they are linearly independent over $A$ and over $\ZZ_p$. Thus $R(G)\otimes\ZZ_p$ is a subring of $\mathcal{F}(G,A)$. In fact, since $\ZZ_p$ is closed in $A$, so is $R(G)\otimes \ZZ_p$ in $\mathcal{F}(G,A)$; thus $R(G) \otimes \ZZ_p$ is the closure of $R(G)$ in $\mathcal{F}(G,A)$.
\end{proof}
The evaluation morphisms $\phi: R(G) \rightarrow A$ are continuous with respect to the $I$-adic topology on $R(G)$ and the $p$-adic topology on $A$, so each $\phi$ can be extended to $\hat{\phi}:\widehat{R}(G) \rightarrow A$. Define
\[
i: \widehat{R}(G) \rightarrow \mathcal{F}(G,A)
\]
as $i(\chi) = (g \mapsto \hat{\phi}_g(\chi))$.
\begin{Proposition}
The map $i$ is injective.
\end{Proposition}
\begin{proof}
We want to show that if $\hat{\phi}_g(\chi) = 0$ for all $g \in G$ then $\chi = 0$, or in other words, if $(x_n)_n$ is a sequence of characters in $R(G)$ such that $(x_n)$ converges and $\lim x_n(g) = 0$ for all $g \in G$, then $x_n$ converges to the zero character. \\
\begin{itemize}
\item First assume $G$ is cyclic of order $q = p^r$ and fix a generator $\gamma$ of $G$. Recall that:
\[
R(G) \cong \frac{\ZZ[\rho]}{(\rho^q-1)}
\]
where $\rho(\gamma) = \mu$, and $\Gamma^M = I^M = (1-\rho)^M$. Note that $R(G)$ is also generated as a ring by $X := \rho_1$. \\
Let $(x_n)_n \in R(G)^{\NN}$ be a sequence of characters and suppose that $(x_n(g))_n$ tends to zero $p$-adically, for all $g \in G$. First note that since $(x_n(1))$ tends to zero, we must have $x_n \in I$ for $n$ large enough. Fix $M\in\NN$ and suppose that there is an $N$ such that $x_n \in I^M$ for all $n>N$; I claim that for $n$ large enough, $x_n \in I^{M+1} = X^{M+1}$. Recall that
\[
R^*(G) \cong \frac{\ZZ[\overline{X}]}{(q\overline{X})}
\]
where $\overline{X}$ denotes the class of $X$ in $R^*(G)$. In particular, we have $\Gamma^M / \Gamma^{M+1} = \ZZ / p^r\ZZ$. Write:
\[
x_n = \left( a_{n,0} + a_{n,1}X + \cdots + a_{n,m}X^m \right) X^M = a_{n,0}X^M + f(X)X^{M+1}
\]
for some polynomial $f$. Then for all $g \in G$,
\[
v_p(\phi_g(x_n)) \geq \min\left\{ v_p\left(\phi_g\left(a_{n,0}X^M\right)\right), v_p\left(\phi_g\left(f(X)X^{M+1}\right)\right) \right\}.
\]
Now, we know that, for $n$ large enough and for all $g \in G$, $v_p(\phi_g(x_n)) \geq v_p(\phi_g(X^M)) + r$, which is possible only if $a_{n,0} \equiv 0 \ (\Mod p^r)$. Since $\Gamma^M/\Gamma^{M+1} \cong \ZZ/p^n\ZZ$, this implies that $x_n \in \Gamma^{M+1}$, which proves the claim.
\item Consider the map:
\[
\psi: \widehat{R}(G) \rightarrow \bigoplus\widehat{R}(C)
\]
where the direct sum is over all cyclic subgroups of $G$, and $\psi$ is induced by restriction of characters. If $i(\chi) = 0$ for some $\chi \in \widehat{R}(G)$, in particular $i(\res^{G}_{C}(\chi)) = 0$ for all cyclic subgroups $C \leq G$. Since $i$ is injective on cyclic groups, this implies that $\psi(\chi) = 0$. \\
To see that $\psi$ is injective, let $\chi \in \widehat{R}(G)$ satisfy $\psi(\chi) = 0$, that is, $\res^{G}_{C}(\chi) = 0$ for all cyclic subgroups $C$. A theorem of Artin (\cite[13.1, Th. 30]{serre}) states that every rational character of $G$ is a linear combination with rational coefficients of permutation characters defined by cyclic subgroups of $G$. Thus we can write:
\[
\mathbb{1} = \sum_{C \leq G \text{ cycl.}}\lambda_C\QQ[G/C],
\]
with each $\lambda_C \in \QQ$. Let $m\in \NN$ such that $m\cdot\lambda_C \in \ZZ$ for all $C$. Now, for any cyclic subgroups $C\leq G$, we have $\res^{G}_{C}(\chi) = 0$, and in particular $\lambda_C\cdot\res^{G}_{C}(\chi) = 0$. Thus
\[
m\cdot\chi = m\cdot\mathbb{1}\cdot\chi = \sum_{C \leq G \text{ cyc.}}m\cdot\lambda_C\QQ[G/C]\cdot\chi = \sum_{C \leq G \text{ cycl.}}m\cdot\lambda_C\ind_{C}^{G}(\res^{G}_{C}\chi) = 0
\]
and thus $m\cdot\chi = 0$ in $\widehat{R}(G)$. Since $I$-adic completion is exact (see \cite[Th. 10.12]{atiyah-macdonald}), there is no nontrivial torsion in $\widehat{R}(G)$, and thus $\chi = 0$.
\end{itemize}
\end{proof}
Thus $\widehat{R}(G)$ can be viewed as a subring of $\mathcal{F}(G,A)$. Since $R(G) = \ZZ\oplus I$, we have $\widehat{R}(G) = \ZZ\oplus \hat{I}$, with $\hat{I} = \lim_{n\geq 1} I^n/I^{n+1}$. Since $I^n/I^{n+1}$ is finite for all $n \geq 1$, the limit $\hat{I}$ is compact. It contains $I$ as a dense subset, so it is the closure $I$, that is, $\hat{I} = I\otimes \ZZ_p$. Thus $\widehat{R}(G)$ is a subring of $\ZZ_p\oplus\hat{I} = R(G)\otimes \ZZ_p$.\\
In summary, the generators of $\rr^*(G)$ or $R^*(G)$ give us generators of the completed ring $\widehat{R}(G)$, itself a dense subring of the completion $R(G)\otimes \ZZ_p$ of $R(G)$ in $\mathcal{F}(G,A)$. So generators of the graded rings are (topological) generators of $R(G) \otimes \ZZ_p$.
\begin{Remark}
When $G$ is a $p$-group, there is no prime-to-$p$ torsion in $R^*(G)$. Thus the kernel of the map:
\[
\Gamma^n(G)\otimes \ZZ_p \longrightarrow \frac{\Gamma^n(G)}{\Gamma^{n+1}(G)} \otimes \ZZ_p
\]
is exactly $\Gamma^{n+1}(G)\otimes\ZZ_p$. This implies
\[
R^*(G)\otimes \ZZ_p \cong \frac{\Gamma^{n}(G)\otimes \ZZ_p}{\Gamma^{n+1}(G) \otimes \ZZ_p}.
\]
Or in other words, tensoring with $\ZZ_p$ commutes with taking graded rings. This is also true for saturated rings.
\end{Remark}
\subsection{General finite groups}
Let $G$ be any finite group, and fix a prime $p$. Let $H_p = Syl_p(G)$ and $\widehat{R}_p(G) = \widehat{R}(G) \otimes \ZZ_p$. The restriction and induction maps are both continuous, so they extend to maps
\[
\res: \ \widehat{R}_p(G) \leftrightarrows \widehat{R}_p(H_p) \ : \ \ind
\]
\begin{Proposition}
The map $\res: \widehat{R}_p(G) \longrightarrow \widehat{R}_p(H_p)$ is injective.
\end{Proposition}
\begin{proof}
The restriction-corestriction formula is still valid on $\widehat{R}_p(G)$:
\[
\ind_{H}^{G}\res^{G}_{H_p}(x) = \CC[G/H_p]x
\]
I claim that $\CC[G/H_p] \in \widehat{R}_p(G)\x$. Indeed:
\[
\CC[G/H_p] = \varepsilon(\CC[G/H_p]) + \left( \CC[G/H_p] - \varepsilon(\CC[G/H_p]) \right) = \varepsilon(\CC[G/H_p])(1-u)
\]
where $u\in I$. Note that $\left( \varepsilon(\CC[G/H_p]), p\right) = 1$, so $\varepsilon(\CC[G/H_p])$ is invertible in $\widehat{R}_p(G)$. Moreover, since $u \in I$, the sum $1+u+u^2+\cdots$ converges and $(1-u)$ is invertible with inverse
\[
(1-u)^{-1} = 1+u+u^2+u^3+\cdots.
\]
Thus $\CC[G/H_p] \in \widehat{R}_p(G)\x$. This means that $\ind_{H_p}^{G}\res^{G}_{H_p}$ is injective, and thus $\res$ is.
\end{proof}
Note that $\widehat{R}_p(-)$ is again a Mackey functor \todo{(check!)}, so that:
\[
\widehat{R}_p(G) \cong (\widehat{R}_p(H_p))^{N_G(H_p)}
\]
