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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.

  1. Does this construction generalise when binary coproducts don't exist?
  2. What about when the required free object doesn't exist?
  3. 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?
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What is a pointed ring or a pointed group? –  Theo Buehler Dec 26 '10 at 10:07
    
It seems to me that the accepted usage of "pointed object" in a category $\mathscr{C}$ (with a terminal object $1$) is the category $1/\mathscr{C}$ of objects under the terminal object, see ncatlab.org/nlab/show/pointed+object But this does not yield anything interesting in the category of rings or groups. –  Theo Buehler Dec 26 '10 at 10:08
    
Hmmm. That's one way of viewing it, I suppose. But I was thinking of a pointed object as an object with a distinguished element which is preserved under morphisms. So it's the category $F(1) / \mathcal{C}$, I suppose. –  Zhen Lin Dec 26 '10 at 11:33
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You're right. Observe that for any object $A \in \mathscr{C}$ 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. –  Theo Buehler Dec 26 '10 at 12:27
    
There is another commonly-used meaning of pointed ring: a ring without (necessarily having) an identity. Then the category Ring_* of these pointed rings is pointed, in the sense that it has zero morphisms. In fact the category of pointed rings in this sense is equivalent to Ring/Z (rather than Z/Ring which is the category of pointed rings in your sense). Since rings (especially commutative ones) are often thought of as being dual to spaces, this fits quite well with the idea of the category Top_* of pointed spaces being 1/Top. –  Steve Lack Dec 28 '10 at 0:04

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In your examples, there is an evident underlying functor $U: C \to Set$ which allows you to talk about points as elements of the underlying set. So a pointed $C$-object would be an object of the comma category $1 \downarrow U$. There is an evident projection $\pi: 1 \downarrow U \to C$. A left adjoint to this functor would be the process of adjoining an indeterminate.

A category $C$ equipped with an "underlying" functor $U$ to $Set$ is often called a concrete category, especially when $U$ is faithful (some authors drop this assumption). So concrete categories admit a notion of pointed object, and a notion of adjoining an indeterminate if the associated $\pi$ has a left adjoint $(-)[x]$. A priori this makes sense without the need to mention binary coproducts in $C$ or the free object on one element, although it seems to me instances of that would be somewhat artificial. (For example, if $C$ has an initial object $0$, then $0[x] = 1 \to Uc$ would exhibit $c$ as the free object on one element, precisely because $0[x]$ is initial in $1 \downarrow U$.)

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Ah. That seems to be a much more compact way to talk about it, and more amenable to generalisation. Thanks! –  Zhen Lin Dec 26 '10 at 14:16

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