Existence of isomorphism mod every power of the maximal ideal This problem is a continuation of Hensel lemma and rational points in complete noetherian local ring.
Let $A$ be a complete noetherian local ring and $\mathfrak{m}$ be its maximal ideal. 
Assume $S_1$ and $S_2$ are two finitely generated $A$-algebras (no flatness assumption), and $S_1/\mathfrak{m}^n \cong S_2/\mathfrak{m}^n$ as $A$-algebras for every $n$. If we assume $S_i$ is the inverse limit of $S_i/\mathfrak{m}^n$ respectively, then do we have $S_1 \cong S_2$ as $A$-algebras?
Again the problem is about compatibility. If one try to use the Hom functor, there is a subtlety of representability. The problem can also be stated for finite type schemes.
Moreover, I have one similar question: if $R_1$, $R_2$ are two complete noetherian local rings and $R_1/\mathfrak{m_1}^n \cong R_1/\mathfrak{m_2}^n$, then do we have $R_1 \cong R_2$ as rings?
The similar question for finitely generated modules is true as $Isom$ satisfies Mittag-Leffler condition (an endomorphism of finitely generated module is an isomorphism iff it's surjective, so this can be checked mod $\mathfrak{m}$, and Hom mod ${\mathfrak{m}}^n$ are modules of finite lengths )
 A: Here is a lovely argument shown to me by Madhav Nori.
Let $A,B$ be complete local rings, quotients of a power series ring in
finitely many variables over a field $k$. If $A/\mathfrak{M}_A^n\cong
B/\mathfrak{M}_B^n$ for all $n$ then $A\cong B$ where as usual
$\mathfrak{M}$ denotes the maximal ideal.
Proof: Let us denote by $A_n=A/\mathfrak{M}_A^n$ and
similarly $B_n$. Let $G_n=\mathrm{Aut}\, A_n$. Then $G_n$ is an
algebraic group over $k$. Let $I_n=\mathrm{Isom}(A_n,B_n)$, the set of
$k$-algebra isomorphisms, which is naturally a variety over $k$. In our
situation, $G_n\cong I_n$ as algebraic varieties by using any of the
isomorphisms in $I_n$, which is non-empty by assumption. In other
words, if $\phi\in I_n$, then we have a morphism $G_n\to I_n$ given by
$g\mapsto \phi\circ g^{-1}$ which one easily checks is an
isomorphism. (In standard language, $I_n$ is a principal homogeneous
space over $G_n$.)
Basic fact from Algebraic groups: If $f:G\to H$ is a group morphism of
algebraic groups, $f(G)$ is closed.
We have natural restriction maps $r_{m,n}:I_m\to I_n$ for $m>n$. This
induces by using any element of $I_m$ a  group morphism from
$G_m\to G_n$ and hence, we see that $r_{m,n}(I_m)$ is closed in $I_n$. Let
$$K_n=\cap_{m>n} r_{m,n}(I_m).$$
Then $K_n$ is a closed subvariety of $I_n$ and it is non-empty. One
can easily check that $r_{m,n}(K_m)= K_n$. Thus we have a surjective
projective system $\{K_n\}$. Since these are all non-empty, it has an
element in the projective limit. That is, there exists $\phi_n\in K_n$
so that $r_{m,n}(\phi_m)=\phi_n$ for all $m>n$. These $\phi$'s give a
compatible collection of isomorphisms from $A_n\to B_n$ and thus we
get an isomorphism in the projective limit $A\to B$.
