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 )

• I thought that I had answered your second question (about $R_1,R_2$) in the affirmative. – Mohan Dec 21 '18 at 21:12
• @Mohan Thank you, could you give some ideas for the proof? I have thought about those questions since long ago.. – sawdada Dec 21 '18 at 21:19

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 $$A\to B$$.

• Thank you, the idea of using Isom functor is very good! Though it has some subtleties due to non-representability in general. In the last paragraph, as we need to find a $k$-point, maybe we need to assume $k$ is algebraically closed. – sawdada Dec 21 '18 at 22:49
• It is not true that morphisms of algebraic groups are closed (think of a projection $\mathbb{G}_a\times\mathbb{G}_a\to\mathbb{G}_a$). What is true, and suffices for the argument, is that it has closed image. More precisely, $f:G\to H$ factors uniquely as $G\to G/\mathrm{ker}\,f\to H$ where the first (resp. second) map is faithfully flat (resp. a closed immersion). – Laurent Moret-Bailly Dec 22 '18 at 7:46
• @LaurentMoret-Bailly You are right. I meant (and only needed) that the image is closed. Thank you. – Mohan Dec 22 '18 at 16:00
• It seems to me that this argument only addresses the equicharacteristic case. – R. van Dobben de Bruyn Dec 22 '18 at 16:55
• @R.vanDobbendeBruyn. Presumably you can reduce the mixed characteristic case to the equicharacteristic case by applying the Greenberg functors to the group schemes $G_n$. – Jason Starr Dec 23 '18 at 10:41