The subgroup generated by a set is found by taking the intersection of all subgroups containing that set

The field of fractions Q(D) of an integral domain D is the "smallest field containing D", informally we can think Q(D) as the intersection of all fields containing D, whatever that means

In the first example it actually makes sense to take intersection, but "the intersection of all fields containing D", as in the usual set theoretic intersection, is meaningless. However there's still this idea of "intersection". What made me think of a possible way to make this meaningful was the proof of Freyd’s adjoint theorem; a solution set is found in the comma category and product taken over this set, and then apply a suitable equalizer. Notice how products in a preorder is the greatest lower bound, so in the power set of a set X, a product of a collection of subsets is just their intersection.

If we take a group G, and let P(G) be the preorder category of subsets ordered by inclusion, and S(G) be the preorder category of subgroups ordered by inclusion, then it’s clear the forgetful functor S(G)->P(G) has a left adjoint which assigns each subset X of G the subgroup generated by X. This is due to the fact that the comma category over X has all products (intersection of subgroups is a subgroup.

Now I was reading about injective envelops, and how it’s the “skinnest” injective module for which a given module may be embedded. Instead of the usual proof, could we proceed by: Let M be a R-module, Q be the category consisting of pairs (I,f:M->I) for I an injective module f an injection, and arrows injections that commute. This is not empty because R-mod has enough injectives.If Q has a solution set and this solution set has a product then clearly this product is weakly initial and thus is an injective envelop of M.

I’m having a bit of trouble expressing myself and so I hope you get my idea. My question can be summarized by: is there a way of expressing “the intersection of all X containing Y” like in my examples without resorting to set theoretic intersection?

canbe viewed through set-theoretic intersection: just put your domain D in a very large field K, and then intersect allsubfieldsof K containing D. Of course, I doubt it's the right way of thinking about Q(D), as it would cause more problems than it's worth (e.g. 2 different K's should give you isomorphic Q's; probably easy to prove but clumsy). But it ispossible. $\endgroup$ – Thierry Zell Sep 12 '10 at 14:42