I'm pretty sure that my conjecture $$ \operatorname{Hom}(\varprojlim_i R/\mathfrak{m}^i, A) = \operatorname{colim}_i \operatorname{Hom}(R/\mathfrak{m}^i, A), $$ is true. To prove it, just use the construction of the colimit as a disjoint union modulo equivalence. It is then quite easy to set up a bijection explicitly. Therefore pro-representability in Mazur's sense implies pro-representability in the sense of the nLab. I suspect that the converse does not hold, and I intend to look for a counterexample when I have time. Until then, I will leave the question up, because I think it's interesting, and it doesn't get discussed much in the literature.

Original question:

Let $\mathcal{C}$ be the category of complete, local, Noetherian rings with residue field $k$ (where $k$ is fixed and finite). Let $\mathcal{C}^0$ be the full subcategory of rings that are also Artinian.

Mazur (and everyone else studying deformation rings, it seems) defines a functor $F:\mathcal{C}^0 \to \textbf{Set}$ to be pro-representable if there is some $R\in \mathcal{C}$ such that $F(A) = \operatorname{Hom}(R, A)$ as a functor $A\mapsto \textbf{Set}$.

The nLab defines a functor to be pro-representable if it is a filtered colimit of representables. I would like to show that these notions are equivalent in the case of functors $\mathcal{C}^0\to \textbf{Set}$.

For $R \in \mathcal{C}$, we have $R = \varprojlim_i R/\mathfrak{m}^i$, where $\mathfrak{m}$ is the maximal ideal of $R$. Thus, I would conjecture naively that $$ \operatorname{Hom}(\varprojlim_i R/\mathfrak{m}^i, A) = \operatorname{colim}_i \operatorname{Hom}(R/\mathfrak{m}^i, A), $$ since the $R/\mathfrak{m}^i$ are in $\mathcal{C}^0$. However, I cannot see why this should be the case (and indeed I suspect that it is not), since the $\operatorname{Hom}$ functor does not play nicely with limits in the first argument. Using the universal properties, I am able to construct an injective map

$$ \operatorname{colim}_i \operatorname{Hom}(R/\mathfrak{m}^i, A) \to \operatorname{Hom}(\varprojlim_i R/\mathfrak{m}^i, A) $$ in the obvious way. However, I am not able to show that this map is surjective.

  • $\begingroup$ Ah, okay. I'll remove my comments to declutter this space a bit. $\endgroup$ – Jeroen van der Meer Feb 6 at 7:41

This isn't a full answer because I don't know enough about noetherian complete local rings, but it's too long for a comment. Essentially I try to isolate the key-points you would need to show that they're the same.

It's a very nice question. The point is that in many cases, pro-objects in a given category $C$ have a nice description in terms of a variant of $C$.

For instance, profinite groups are the same thing as pro-(finite groups). That is, topological groups that are topologically inverse limits of finite (discrete) groups are a model for pro-objects in the category of finite groups. So if you have a functor $\mathbf{finGrp\to Set}$, it is pro-representable in the nLab sense if and only if there is a profinite group $G$ such that it is isomorphic to $\hom(G,-)$. But there is work behind this proof, namely the equivalence I mentioned earlier.

This equivalence works for a lot of neat algebraic gadgets (although I have never thought about the generality in which it was true... that would be a very nice question too)

You can indeed always define a map $\mathrm{colim}_i\hom(R/m^i,-)\to \hom(\lim_i R/m^i,-)$ , and you can ask when is it an isomorphism ? Essentially, you're asking whether the objects of $C^0$ are cocompact in $C$.

As you point out, it's pretty easy to see that this is always injective, so the question is about surjectivity: does a map $R\to A$ factor through $R/m^i$ for some $i$ ? I don't know a lot about these rings, but iirc the map $R\to A$ has to be continuous for the adic topologies, and so because the maximal ideal in $A$ is nilpotent, some power of $m$ has to be sent to $0$, thus proving the claim. Maybe I'm mistaken here, but something like that has to be true for the claim to hold anyways (otherwise you could find a morphism where no power of $m$ gets mapped to $0$ and get a nonsurjectivity claim)

How about the converse ? Well if I'm not mistaken about the previous paragraph, and if the proof can be adapted to prove the more general statement that any map $\lim_i R_i \to A$ factors through some $R_i$, where all the $R_i$'s are in $C^0$ and $A\in C^0$, then it would follow that the canonical functor $Pro(C^0)\to C$ is fully faithful. Since clearly any object of $C$ is an inverse limit of objects in $C^0$, it would follow that it's an equivalence, in other words, that the two definitions of pro-representable agree, just like in the case of groups !

Ah, maybe I should mention : for the canonical functor to even exist, one would need $C$ to be closed under inverse limits which doesn't seem completely obvious at first glance, but is perhaps still true.

As a conclusion (of this incomplete answer), here are the points you should try to work out:

1- Does any map $R\to A$ in $C$ factor through some $R/m^i\to A$ if $A\in C^0$ ? (probably yes, by some continuity things)

2- Does $C$ have inverse limits, and if so, is $R\cong \lim_i R/m^i$ in $C$ ? (be careful: this claim is true in the category of commutative rings by definition of complete, but it's not necessarily true that limits in $C$ are computed on the underlying rings ! )

If the answer to those 2 is "yes", then by the first part of 2- you get a canonical functor $Pro(C^0)\to C$, which is fully faithful by 1-, and is therefore an equivalence by the second part of 2-.

COnversely, if you can disprove any of 1- or 2-, it follows that $Pro(C^0)\not\simeq C$, and so the two definitions don't agree. In any case, you will get an answer to your question by solving 1 and 2.

  • $\begingroup$ Thanks for your help! I'm pretty sure $(1)$ is true because the local ring homomorphism takes $m_R$ into the maximal ideal $m_A$ of $A$. Since $A$ is Artin local we have $m_A^n = 0$ for some $n$ (by Nakayama's Lemma), and so $m_R^n$ is in the kernel of the homomorphism. I currently suspect that $(2)$ is not true, because I have found some lemmas that gives sufficient conditions for the inverse limit to exist, which implies that it might not always be the case. $\endgroup$ – Qwertiops Feb 7 at 19:56
  • $\begingroup$ Ah you're right, that's what I was calling "continuity" but it's even easier than that it's just locality. Then that means you get a fully faithful functor $C\to Pro(C^0)$, which sends certain types of limits (namely it sends $R$ to $\lim_i R/m^i$ in the latter category), whose image contains $C^0$ but it might not contain all pro-objects : if you're right about 2, that would mean that the nLab definition is more general (although the Mazur-et al. definition is fully faithfully embedded in it) $\endgroup$ – Maxime Ramzi Feb 7 at 20:09

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