As George Janelidze pointed out to me, there was an error in what I wrote yesterday: the multiplicative monoid structure of C(X) determines X, and hence the ring structure of C(X), up to isomorphism (this is a 1949 result of A.N. Milgram), but it doesn't determine the additive structure uniquely, since one can take the standard addition and "conjugate" it by a multiplicative automorphism of R, in the same way that Steve points out for finite fields (e.g. one could define a new addition by f +' g = (f^3 + g^3)^{1/3}). Peter Johnstone On Wed, 5 Mar 2008, Steve Vickers wrote:
Dear Peter,
A special case is that of groups (with 0 adjoined) and fields. But even that is hard. In Paul Cohn's book on Skew Fields (p.145 of 1995 edition) he says "The problem of characterizing the multiplicative group of a field has not yet been solved even in the commutative case, but there are some results on the subgroups of fields." (The main result he cites, one of Amitsur's, says that a finite group G can be embedded as a subgroup in the multiplicative group of a [skew]field if and only if G is (i) cyclic, or (ii) a certain kind of metacyclic group, or (iii) a certain from of soluble goup with a quaternion subgroup, or (iv) the binary icosahedral group SL_2(F_5) of order 120.)
Even in finite fields F = Z/p, the additive structure is not determined on the nose by the multiplicative structure. If the group F* has a non-identity automorphism alpha then (extending alpha to take 0 to 0) a different addition for the same multiplication can be defined by x +' y = alpha^{-1}(alpha(x) + alpha(y)). This would be the original addition only of alpha preserves addition; but then since it preserves 0 and 1, and 1 is an additive generator, then it would have to be the identity. An example is F_5, where the multiplicative group is cyclic of order 4 and has a non-identity automorphism that swaps the two generators.
There remains the deeper question of whether you can have non-isomorphic additive groups for the same multiplicative group.
Regards,
Steve.
Prof. Peter Johnstone wrote:
On Tue, 4 Mar 2008, Tom Leinster wrote:
Apologies for the previous trivial question. Here is the correct version.
There's still something odd about this question. Requiring the subcategory to contain all endomorphisms of M of course requires it to contain A(M,M) as a monoid. But if you don't require it to be closed under biproducts in A, then presumably you don't require it to contain A(M,M) as a ring. It therefore raises two questions of "pure algebra":
What conditions on a monoid (with 0) are needed to ensure that it occurs as the multiplicative monoid of a ring?
Given that it does so occur, can there be several different additive group structures making it into a ring?
I suspect that a fair amount must be known about these questions, but the only result I know in this area is one which I quoted in "Stone Spaces": for a ring of the form C(X), X a compact Hausdorff space, the multiplicative monoid structure of C(X) is enough to determine the topology of X (and hence the ring structure of C(X)) uniquely.
Peter Johnstone