Exercise 9


Theorem: For any ring , has no subobject classifier.

Proof: Assume it did have a subobject classifier . Note that the terminal object in is the zero module, which is also initial, so we have , and so there is exactly one choice for the map . Since is a sub-object classifier, we have for any monic , a unique classifying map such that is the pullback of along . But, considering as the unique map from the initial object, this is precisely the statement that is the kernel of . More compactly, this means for any submodule of , must be the kernel of its classifying map . Now consider that the zero module is a subobject of any module – thus, for any module there must be a map that has trivial kernel – i.e., an injection from into . Then, in particular, this is true for the -module ( has all small limits). But, the underlying set of this module has cardinality , a contradiction.