Minimal abelian subcategory (corrected)
Apologies for the previous trivial question. Here is the correct version. (The mistake was omitting to say that the subcategory must contain all endomorphisms of M.) * My colleague Walter Mazorchuk has the following question. Being abelian is a *property* of a category, not extra structure. Given an abelian category A, it therefore makes sense to define a subcategory of A to be an ABELIAN SUBCATEGORY if, considered as a category in its own right, it is abelian. Note that a priori, the inclusion need not preserve sums, kernels etc. Now let R be a ring and M an R-module. Is there a minimal abelian subcategory of Mod-R containing M and all its endomorphisms? If so, is there a canonical way to describe it? Any thoughts or pointers to the literature would be welcome. Feel free to assume hypotheses on R (it might be a finite-dimensional algebra etc), or to answer the question for full subcategories only. Thanks, Tom
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
My colleague Walter Mazorchuk has the following question.
Being abelian is a *property* of a category, not extra structure. Given an abelian category A, it therefore makes sense to define a subcategory of A to be an ABELIAN SUBCATEGORY if, considered as a category in its own right, it is abelian. Note that a priori, the inclusion need not preserve sums, kernels etc.
Now let R be a ring and M an R-module. Is there a minimal abelian subcategory of Mod-R containing M and all its endomorphisms? If so, is there a canonical way to describe it?
Any thoughts or pointers to the literature would be welcome. Feel free to assume hypotheses on R (it might be a finite-dimensional algebra etc), or to answer the question for full subcategories only.
Thanks, Tom
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
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
Oops Email is great. I cited Peter before he could correct himself. A possible remedy would be to consider not just monoids with zero, but those equipped with a homomorphism from R, so that points are retractions of that. This cuts down on the automorphisms, at least the naturally available ones. Bill On Wed Mar 5 6:22 , "Prof. Peter Johnstone" sent:
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
A message from Walter Mazorchuk: Dear Colleagues, thank you very much for your comments on my abelian envelope question. Because of my stereotype thinking I missed the point of the determination of the additive structure by the multiplicative one in the original formulation. The stereotype is based on the fact that I am a representation theorist and the origin of the question is in module categories, which are k-linear over some field k. So, the subcategory I am looking for should be a k-linear subcategory with the induced k-linear structure. Best, Walter
participants (4)
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Prof. Peter Johnstone -
Steve Vickers -
Tom Leinster -
wlawvere＠buffalo.edu