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.
Small categories
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1By the way, [the statement that every set can be equipped with a group structure is equivalent to the axiom of choice](http://math.stackexchange.com/questions/105433/does-every-set-have-a-group-structure). – 2012-06-24
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0Why do you think this is true? – 2012-06-24
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0because there is a fully faithfull functor between every small abelian category and $R$-Mod. – 2012-06-24
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1@Galoisfan: You can make fully faithful functors between [discrete categories](http://en.wikipedia.org/wiki/Discrete_category) 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. – 2012-06-24
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0Any 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. – 2012-06-24
2 Answers
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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0Ok, but if $R$ is commutative? – 2012-06-24
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0@Galoisfan it doesn't matter if $R$ is commutative or not, – 2012-06-24
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0@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. – 2012-06-24
One does not need to invoke the axiom of choice to show that the category of groups is not small. One can merely observe that there is no largest cardinality of a group. (In particular, for any cardinal number, there is a group of larger cardinality.) It is irrelevant whether there is a group of every cardinality.