# The “skinnest” object

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?

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I want to point out that your second example can be viewed through set-theoretic intersection: just put your domain D in a very large field K, and then intersect all subfields of 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 is possible. – Thierry Zell Sep 12 '10 at 14:42
And I want to point out that in some examples (injective hull of a module, algebraic closure of a field) the skinniest widget (to borrow a term from the answer below) is unique up to isomorphism without being unique up to unique isomorphism. – Tom Goodwillie Sep 13 '10 at 3:08
English language comment: there is no such word as "skinnest". From the mathematical context, I am guessing you mean "skinniest" rather than "having the most skin"? – Pete L. Clark Sep 13 '10 at 10:14

By the adjoint functor theorem, as you say, the construction can be done in two stages, given the solution-set condition and enough limits. First, take a product of the solution set to get a weakly inital widget $W$.
Then take the intersection of all the sub-widgets of $W$; and this gives the initial widget you want. In the widest generality, this is the “intersection” in the categorical sense of being a limit of various subobjects of a fixed object, i.e. a limit in $\mathrm{Sub}(W)$. But in most common examples, e.g. in any algebraic category over $\mathbf{Sets}$, this'll be intersection in the normal set-theoretic sense (since the forgetful functor down to $\mathbf{Sets}$ preserves/reflects limits).