I'm curious, is a model for a theory of groups (in the sense of Lawvere's algebraic theories) in the category of rings a group ring? Similarly, is a model for a theory of rings in the category of groups a group ring?
Model of a group in category of rings
3 Answers
To address the question in a comment, there isn't a variety of group rings. In any variety of universal algebra, the underlying set of a product of two algebras is the product of the underlying sets. However, letting $\mu_n$ be the cyclic group of $n$ elements and $\omega$ be a primitive cube root of unity, we have:
- $ Z[\mu_2] \times Z[\mu_1] \cong (Z \times Z) \times Z $
- $ Z[\mu_3] \cong Z \times Z\left[\omega\right] $
$ Z[\mu_3]$ is the only group ring whose additive group is $Z^3$, so these two rings would have to be isomorphic, but they are not. Thus, the product of group rings is not given by the product of the underlying sets.
From some perspective, there are two obstacles here. The first is that the theory of group rings is naturally two sorted: one sort for the group and one for the ring. The other is that the group ring is supposed to be free in the appropriate sense.
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0Hrm. What am I missing then? You have two sorts, you add in the group axioms, you add in the ring axioms, and then you add in the axiom that says the group is a subobject of the ring. Oh! I guess what's missing is that the group ring is *free*, which is (?) a geometric axiom, not an essentially algebraic one. I've fixed my answer. I think I had in my mind a more general theory that merely needed the groups to be a subgroup of the multiplicative group of the ring. – 2012-07-21
There are no nonzero group objects in the category of rings. The problem is that the identity is supposed to be a morphism $e : 1 \to R$ where $1$ is the terminal object, but in $\text{Rng}$ the terminal object is the zero ring, and there are no morphisms from the zero ring to any nonzero ring.
(A group object in the opposite of the category of commutative rings, on the other hand, is a group scheme. And if you want to think about group rings categorically, the way to do it is to consider the left adjoint to the forgetful functor $\text{Rng} \to \text{Grp}$ sending a ring to its group of units.)
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0@Matt: The terminal group is the zero group, so both $0$ and $1$ have to be the identity element of the group, so $0=1$. – 2012-07-21
While the notion of group object internal to the category of rings (without $1$) yields a ring with zero multiplication, and a group object internal to the category of groups yields an abelian group, it is an important starting point that groupoids internal to the same categories are very much non trivial objects. The subject is studied in general in for example
Internal Categories and Groupoids in Congruence Modular Varieties Authors: Janelidze G.; Pedicchio M.C., Journal of Algebra, 193 (1997) 552-570.
Groupoids internal to groups, or groups internal to groupoids, were shown to be equivalent to crossed modules in
R. Brown and C.B. Spencer, ``$\cal G$-groupoids, crossed modules and the fundamental groupoid of a topological group'', Proc. Kon. Ned. Akad. v. Wet. 7 (1976) 296-302.
Since crossed modules occur in homotopy theory, for second relative homotopy groups, all this was motivation for seeking and applying higher homotopy groupoids, and so going from abelian homotopy groups to more complicated nonabelian algebraic structures to model homotopy theory.
Groupoids internal to groupoids are called double groupoids, and are even somewhat mysterious, though many special cases have been studied.