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Why $R$-Mod is a small category? There is a way to recognize small categories? For example Grp (i.e. category of all groups) is large because every set can be equiped with a group structure.

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Why do you think this is true? –  Thomas Andrews Jun 24 '12 at 18:16
    
because there is a fully faithfull functor between every small abelian category and $R$-Mod. –  fair-coin tossing Jun 24 '12 at 18:19
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@Galoisfan: You can make fully faithful functors between discrete categories of different cardinalities, so the existence of a fully faithful functor between two categories does not tell you anything about their smallness or largeness in general. –  Zev Chonoles Jun 24 '12 at 18:21
    
Any category of algebraic structures will be large (because there is always at least one model, and there is a proper class of sets of any given cardinality), and generally will not be equivalent to a small category. –  Zhen Lin Jun 24 '12 at 19:24

1 Answer 1

For any ring $R$, the category $R$-Mod isn't a small category: for any set $S$ one can form the $R$-module $R^{S}=\{f:S\to R\}$, and $R^S\neq R^T$ for any two distinct sets $S\neq T$ (though of course they may be isomorphic), so there are "at least as many $R$-modules as sets", and so the collection of all $R$-modules is a proper class.

In my experience with categories so far, I've never come across a situation where it wasn't clear from the outset (i.e., using what we already know about whatever the objects of our category are) whether a category was small, whether it was large, or whether it made no difference to the discussion at hand.

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Ok, but if $R$ is commutative? –  fair-coin tossing Jun 24 '12 at 18:13
    
@Galoisfan it doesn't matter if $R$ is commutative or not, –  Thomas Andrews Jun 24 '12 at 18:18
    
@Galoisfan: I was talking about commutative rings $R$ (besides, for non-commuative rings $R$, we must specify left $R$-module, right $R$-module, or $R$-bimodule), but the argument works just as well for non-commutative rings too. –  Zev Chonoles Jun 24 '12 at 18:23

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