Let $A = (\dotsc \twoheadrightarrow A_2 \twoheadrightarrow A_1 \twoheadrightarrow A_0)$ be a (commutative) pro-ring with surjective transition maps. Consider the category $\mathcal{M} := \varprojlim_i \,\mathsf{Mod}(A_i)$: Objects are families of right $A_i$-modules $M=(M_i)$ together with isomorphisms $M_{i+1} \otimes_{A_{i+1}} A_i \cong M_i$. We let $\widehat{M} := \varprojlim_i M_i$. For each $j$, there is a natural epimorphism of $A_j$-modules $$\alpha_j : \widehat{M} \otimes_{\widehat{A}} A_j \to M_j,~ (m_i)_i \otimes a \mapsto m_j \cdot a.$$ Question. Is $\alpha_j$ an isomorphism?

Of course we may assume $j=0$. The answer is yes when $A_i = R/p^i$ for some commutative ring $R$ and some element $p \in R$. The proof for this requires some calculations and doesn't generalize to the case of arbitrary $A_i$.

Geometrically speaking, the question aims at understanding quasi-coherent sheaves on affine ind-schemes. Any literature about this is also appreciated.

  • $\begingroup$ You do not say what the morphisms in $\mathcal{M}$ are. In fact you may need to make precise whether the objects' data includes the given isomorphisms (in case that matters.) $\endgroup$ – Tim Porter Sep 8 '14 at 11:43
  • $\begingroup$ Sure, the isomorphisms belong to the data. A morphism $M \to N$ is a family of morphisms $M_i \to N_i$ compatible with the isomorphisms. Actually all this follows from the definition $\mathcal{M} := \varprojlim_i \, \mathsf{Mod}(A_i)$. $\endgroup$ – Martin Brandenburg Sep 8 '14 at 15:15
  • $\begingroup$ A part from the geometric motivation, why do you introduce the category $\mathcal{M}$? I feel you are asking whether you can recover the $j$-th piece of a projective system from its projective limit, under thw assumption that transition maps at level of ring are onto. Is that right? $\endgroup$ – Filippo Alberto Edoardo Sep 10 '14 at 10:05
  • $\begingroup$ @Martin: Actually, I do not see why you say that the $\alpha_j$'s are surjective. The counterexample I had in mind to $\alpha_h$ being an isomorphism comes from Iwasawa theory, but in general in that case one has neither surjectivity nor injectivity. Jence I wonder if it fits - therefore, to start with, I am trying (without success) to understand why in your setting should surjectivity be clear. $\endgroup$ – Filippo Alberto Edoardo Sep 10 '14 at 15:07
  • $\begingroup$ The maps $M_{j+1} \to M_i$ are surjective. It follows easily that each projection $\widehat{M} \to M_j$ is surjective (construct inverse images recursively). Hence, also $\alpha_j$ is surjective. $\endgroup$ – Martin Brandenburg Sep 10 '14 at 15:26

Since this just got bumped ...

I think the answer https://math.stackexchange.com/a/938076/88262, to a special case of this question that Martin asked on Math.SE, settles this question by giving a counterexample.

In summary, there is a counterexample where $A_i=k[x_1,\dots,x_i]$ modulo the ideal generated by polynomials of degree two, each $M_i$ is the direct sum of copies of $A_i$ indexed by the natural numbers, and $j=0$.

  • $\begingroup$ Oh yes, thank you for reminding me that this was answered. $\endgroup$ – Martin Brandenburg Jan 8 '15 at 10:25

The kernel $\widehat{K_j}$ of $\widehat M\to M_j$ consists of sequences of the form $(\cdots\mapsto m_{j+2}\mapsto m_{j+1}\mapsto0\mapsto0\mapsto\cdots\mapsto0)$ whereas the kernel $\widehat{I_j}$ of $\widehat A\to A_j$ consists of sequences of the form $(\cdots\mapsto a_{j+2}\mapsto a_{j+1}\mapsto0\mapsto0\mapsto\cdots\mapsto0)$. Thus $$\widehat{K_j}=\varprojlim_iK_{ji}\textrm{, }\widehat{I_j}=\varprojlim_iI_{ji}$$ with $K_{ji}=\mathrm{Ker}(M_{j+i}\to M_i)$ and $I_{ji}=\mathrm{Ker}(A_{j+i}\to A_i)$.

Now the given isomorphisms allow to identify $M_j$ with $M_{j+1}/(M_{j+1}\mathrm{Ker}(A_{j+1}\to A_j))$; more generally they provide identification of $K_{ji}$ with $M_{j+i}I_{ji}$ for all $i$. It follows that $\widehat{K_j}=\widehat M\widehat{I_j}$.

This in turn implies $M_j=\widehat M/\widehat M\widehat{I_j}$. On the other hand $\widehat M\otimes_{\widehat A}A_j$ is also $\widehat M/\widehat M\widehat{I_j}$ since $A_j=\widehat A/\widehat{I_j}$.

  • $\begingroup$ Thank you. But $\widehat{K_j} = \widehat{I_j} \widehat{M}$ is unclear to me (and clearly this is a direct reformulation of the claim). How do you prove $\subseteq$? $\endgroup$ – Martin Brandenburg Sep 10 '14 at 0:47
  • $\begingroup$ @MartinBrandenburg I've tried to be more accurate. Please have a look - I still have the feeling it is not entirely rigorous... $\endgroup$ – მამუკა ჯიბლაძე Sep 10 '14 at 6:39
  • $\begingroup$ From $K_{ji} = M_{j+i} I_{ij}$ for all $i$ we only get $\widehat{K_j} = \varprojlim_i \, (M_{j+i} \cdot I_{ji})$, but why does this equal $\varprojlim_i \, M_{j+i} \cdot \varprojlim_i \, I_{ji} = \widehat{M} \widehat{I_j}$? Again, $\subseteq$ is unclear. $\endgroup$ – Martin Brandenburg Sep 10 '14 at 8:11

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