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I am looking for a counterexample to the following:

Let $(A_i)_{i = 1}^{\infty}$ be an infinite family of rings with unity. Let $\text{Mod}A$ denote the category of right $A$ modules for a ring $A$. Let $\mathcal{C} = \text{Mod}A_1 \times \text{Mod}A_1 \times \cdots$ be the infinite Cartesian product of categories defined in the simplest way. Let $\mathcal{D} = \text{Mod} (A_1 \times A_2 \cdots)$. Show $\mathcal C$ is not equivalent to $\mathcal{D}$ in general.

I am unable to find a counterexample. I think the simplest counterexamples should work, say by taking $A_i = \mathbb{Z}_2$ but I do not know how to show the resulting categories to be non-equivalent.

Any hints are welcome.

Subham Jaiswal
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    huh... really? It seems like the objects correspond, at least. Must be something else subtle, if so. – rschwieb Aug 16 '23 at 18:26
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    Any such nonequivalence will come down to the difference in some infinity silliness. The objects in $\mathcal{D}$ are only able to have finitely many summation terms acting on any module element at any given time (rings only like finite sums even if the elements are infinite lists) while the objects in $\mathcal{C}$ can have infinitely many things acting independently of each other at the same time. This is of course not an answer, but it may help in your search. – Geoff Aug 16 '23 at 18:34
  • @rschwieb, I agree. I now see that there are natural choice of functors in both directions which are inverses in the case of finite products. – Subham Jaiswal Aug 17 '23 at 03:46
  • @Geoff (Nonrhetorical) What do you mean "The objects in $\mathcal D$ are only able to have finitely many summation terms acting on any module element at any given time"? I don't follow the "finitely many summation terms" part. The natural action of $\prod A_i$ on $\prod M_i$ has infinitely many ring elements acting on infinitely many module elements, with no summation or finiteness in sight. – rschwieb Aug 17 '23 at 11:21

1 Answers1

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$D$ will generally have more simple objects.

In $\text{Mod}(A)$ for $A$ a commutative ring the simple objects are the simple modules, namely the quotients $A/m$ where $m$ is a maximal ideal. You can check that the simple objects in $C$ are the simple objects in each $\text{Mod}(A_i)$ individually (and zeroes in the other coordinates), so if each $A_i$ is commutative these are quotients $A_i/m$ by the maximal ideals of each $A_i$ individually. However, the simple objects in $D$ are of the form $(\prod_i A_i)/m$ where $m$ is a maximal ideal of $\prod_i A_i$, and generally there are more of these than maximal ideals of the $A_i$ individually.

Specifically, if each $A_i$ is a field (so $A_i = \mathbb{F}_2$ works fine as a counterexample) then you can show that the maximal ideals of $\prod_i A_i$ correspond to ultrafilters on the index set $I$. These come in two types: the principal ultrafilters correspond to each $A_i$ individually, but if $I$ is infinite then there are non-principal ultrafilters and these correspond to exotic maximal ideals $m$ where the quotient $(\prod_i A_i)/m$ is an ultraproduct. You can learn more about this sort of thing here.

If $I$ is countable there are uncountably many non-principal ultrafilters (this is necessary or else $\text{Spec} \prod A_i$ can't be compact). So in this case $C$ has countably many simple objects but $D$ has uncountably many simple objects.

Edit: For a more geometric perspective on this argument, again restricting to the commutative case, $C$ is the category of quasicoherent sheaves on the coproduct $\bigsqcup_i \text{Spec } A_i$ taken in the category of schemes, whereas $D$ is the category of quasicoherent sheaves on $\text{Spec } \prod_i A_i$, which is the same coproduct but taken in the category of affine schemes. This second scheme is much stranger than the first scheme and contains many new points coming from ultrafilters; it is a kind of compactification of the first scheme, and indeed if each $A_i$ is a field the underlying topological space is exactly the Stone-Cech compactification $\beta I$ of the index set $I$ (with the discrete topology). The simple objects being counted above correspond to closed points of these schemes. What we learn from this, among other things, is that the inclusion of affine schemes into schemes does not preserve infinite coproducts.

I think this issue can be "fixed" by regarding $\prod A_i$ not as an abstract ring but as a topological ring with the product topology, or maybe some more sophisticated object like a pro-ring or a condensed ring or something, but I haven't checked the details.

As an alternative argument, Gabriel's theorem characterizes categories of modules over rings precisely as the cocomplete abelian categories with a compact projective generator, so we can simply check that $C$ fails to satisfy this condition (if the index set $I$ is infinite and the $A_i$ are nonzero). It is clearly a cocomplete abelian category so we need to check for a compact projective generator. And it's not hard to see that

  1. a compact projective object in $C$ must have finite support (by splitting an object up into a direct sum component-by-component then homming into it), but
  2. a generator in $C$ must have full support (or else it can't detect objects whose support is disjoint from it),

so no object can be simultaneously compact projective and a generator.

Qiaochu Yuan
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    It's not obvious from the way these arguments are written but roughly speaking the fundamental issue is that it's not possible to write an element of an infinite product $\prod_i A_i$ as an infinite sum of elements $a_i \in A_i$ the way it is for a finite product. Adding a topology or some other structure can "fix" this. – Qiaochu Yuan Aug 16 '23 at 20:01
  • +1 Gotta love a good counting argument! – rschwieb Aug 17 '23 at 11:36