The problem is from Matsumura 8.5.
Let $A$ be a Noetherian ring and $I$ a proper ideal of $A$. Consider the multiplicative set $S=\{1+a: a \in I \}$. Then $A_{S}$ is a Zariski ring with ideal of definition $IA_{S}$, and its completion coincides with the $I$-adic completion of $A$.
My question is, how can one show $\tilde{A}=\tilde{A_{S}}$, where the left hand side is $I$-adic completion of $A$ and the right hand side is $IA_{S}$-adic completion of $A_{S}$?
What I found is that, using Theorem 8.12, if we let $I=(a_{1},\cdots, a_{n})$, then $IA_{S}$ is also generated by $a_{1}/1,\cdots, a_{S}/1$, thus $$\tilde{A_{S}} \cong A_{S}[[X_{1},\cdots, X_{n}]]/(X_{1}-a_{1},\cdots, X_{n}-a_{n}),\quad \tilde{A} \cong A[[X_{1},\cdots, X_{n}]]/(X_{1}-a_{1},\cdots, X_{n}-a_{n})$$ However, I don't know how this two rings are isomorphic. Moreover, it seems that in general $A_{S}[[X_{1},\cdots, X_{n}]] \not\cong A[[X_{1},\cdots, X_{n}]]$, right? Do you have any idea for this statement?
To show that $A_S$ is a Zariski ring, we must verify $IA_S\subseteq \operatorname{rad}(A_S)$. This follows from the well-known characterisation (Matsumura, p.3) $$\operatorname{rad}(A_S)=\left\{x\in A_S\ \middle|\ 1+ax\text{ is a unit for all }a\in A_S\right\}\,.$$Indeed, every element $x\in IA_S$ is of the form $y/t$ for some $y\in I$ and some $t\in 1+I$, and every $a\in A_S$ can be written as $a=b/s$ for some $b\in A$ and $s\in 1+I$. Hence $1+ax=(st+yb)/(st)$, which is a unit because $st+yb\in 1+I$.
To show that the natural map $\widehat{A}=\lim_{n\geq 0}A/I^n\rightarrow \lim_{n\geq 0}A_S/I^nA_S=\widehat{A_S}$ is an isomorphism, it suffices to check that all $A/I^n\rightarrow \lim_{n\geq 0}A_S/I^nA_S$ are isomorphisms. Observe that $A_S/I^nA_S$ is the localisation of $A/I^n$ at the image of $S$ in it, i.e., the localisation $(A/I^n)_{S_n}$ at the multiplicative subset $S_n=1+I/I^n\subseteq A/I^n$. But $1+I/I^n$ already consists of units! Indeed, an inverse for $1+x$ with $x\in I$ is given by $\sum_{i=0}^{n-1}(-1)^ix^i$. Hence $(A/I^n)_{S_n}=(A/I^n)$ and we are done.