Existence of isomorphism mod every power of the maximal ideal

Question

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 )

Answer 1

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 $A\to B$.