Forgive me if this is a well-known observation/result, but I'm quite new to graduate-level algebra and I was wondering if there are generalisations of the constructions I describe below.

It's straightforward to see that the functor "adjoin an indeterminate to a ring" is the unit of the adjunction $F : \mathbf{Ring} \to \mathbf{Ring}_* : U$, where $U: \mathbf{Ring}_* \to \mathbf{Ring}$ is the forgetful functor from the category of pointed rings to the category of rings.

Along similar lines, the functor "disjoint-union an element to a set" is the unit of the adjunction $F : \mathbf{Set} \to \mathbf{Set}_* : U$, and if I'm not mistaken, the functor "coproduct with $\mathbb{Z}$" is the unit of the adjunction $F : \mathbf{Grp} \to \mathbf{Grp}_* : U$. Indeed, in general, it seems that if a category has a notion of "free object on one generator", a notion of "pointed" objects, and binary coproducts, then the forgetful functor from the category of pointed objects has a left adjoint, and the unit of the adjunction is the functor which takes objects to their coproduct with the free object on one generator.

- Does this construction generalise when binary coproducts don't exist?
- What about when the required free object doesn't exist?
- Is there a (even) more general way to describe this operation of "adjoining an indeterminate", e.g. when there's no notion of pointed objects?

anyobject $A \in \mathscr{C}$ you can form the comma category $A/\mathscr{C}$ of objects under $A$. If the forgetful functor $U: A/\mathscr{C} \to \mathscr{C}$ has a left adjoint $F$ then the identity $\hom_{A / \mathscr{C}}{(F(X), (A \to Y))} = \hom_{\mathscr{C}}{(X,Y)}$ shows that $F(X)$ must be given by the coproduct with $A$. Conversely, if all coproducts with A exist then this forgetful functor has a left adjoint. $\endgroup$ – Theo Buehler Dec 26 '10 at 12:27