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 )

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    $\begingroup$ I thought that I had answered your second question (about $R_1,R_2$) in the affirmative. $\endgroup$
    – Mohan
    Dec 21, 2018 at 21:12
  • $\begingroup$ @Mohan Thank you, could you give some ideas for the proof? I have thought about those questions since long ago.. $\endgroup$
    – sawdada
    Dec 21, 2018 at 21:19

1 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 $A\to B$.

  • $\begingroup$ 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. $\endgroup$
    – sawdada
    Dec 21, 2018 at 22:49
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    $\begingroup$ 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). $\endgroup$ Dec 22, 2018 at 7:46
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    $\begingroup$ @LaurentMoret-Bailly You are right. I meant (and only needed) that the image is closed. Thank you. $\endgroup$
    – Mohan
    Dec 22, 2018 at 16:00
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    $\begingroup$ It seems to me that this argument only addresses the equicharacteristic case. $\endgroup$ Dec 22, 2018 at 16:55
  • $\begingroup$ @R.vanDobbendeBruyn. Presumably you can reduce the mixed characteristic case to the equicharacteristic case by applying the Greenberg functors to the group schemes $G_n$. $\endgroup$ Dec 23, 2018 at 10:41

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