It must be known that if T is a left exact theory (same as nearly equational theory) and you make a sheaf of T-algebras (meaning the stalks are T-algebras, then the set of global sections is also a T-algebra. Can anyone give me reference? Michael -- The modern conservative is engaged in one of man's oldest exercises in moral philosophy--the search for a superior moral justification for selfishness. --J.K. Galbraith [For admin and other information see: http://www.mta.ca/~cat-dist/ ]
Try D1.2.13 and D1.2.14 in the Elephant. -- Peter Johnstone On Sun, 23 Sep 2012, Michael Barr wrote:
It must be known that if T is a left exact theory (same as nearly equational theory) and you make a sheaf of T-algebras (meaning the stalks are T-algebras, then the set of global sections is also a T-algebra. Can anyone give me reference?
Michael
[For admin and other information see: http://www.mta.ca/~cat-dist/ ]
Dear Mike, Once you have that a sheaf of T-algebras is equivalent to a T-algebra in the category of sheaves, it's obvious: because the global sections functor preserves finite (in fact all) limits. But the first step is not so trivial, and in fact is not true unless you take care over "the stalks are T-algebras". For example, if the base space is Sierpinksi then a sheaf is just a function, with the domain and codomain as the stalks (over the bottom and top points respectively). You might put algebra structures on the stalks, but it won't be an algebra in the sheaf category unless the function is a homomorphism. You will get further issues if the base space is non-spatial, so there aren't enough global points to provide enough stalks. What statement of the result did you have in mind? One sufficient condition that gets round those problems is for the stalks and their T-algebra structure to be defined geometrically. Regards, Steve. Michael Barr wrote:
It must be known that if T is a left exact theory (same as nearly equational theory) and you make a sheaf of T-algebras (meaning the stalks are T-algebras, then the set of global sections is also a T-algebra. Can anyone give me reference?
Michael
[For admin and other information see: http://www.mta.ca/~cat-dist/ ]
Well, I actually know quite a bit more about it. First off, each point in the base has a local neighborhood whose sections are algebras and the restrictions are homomorphisms. I should have mentioned that all ops and partial ops are finitary, but of increasing arity. We know the answer is positive; I am just unhappy with the proof. The situation is that the stalks are integral domains. Now integral domains are not models of an LE theory. But they are models of a theory with an infinitude of LE operations (all finitary, but of increasing arity) and I want the global sections to be models of that theory (which turns out to characterize the rings in the limit closure of the domains). The space is the set of primes of some arbitrary semi-prime (no nilpotents) commutative ring with the co-Zariski topology. Mike On Mon, 24 Sep 2012, Steve Vickers wrote:
Dear Mike,
Once you have that a sheaf of T-algebras is equivalent to a T-algebra in the category of sheaves, it's obvious: because the global sections functor preserves finite (in fact all) limits.
But the first step is not so trivial, and in fact is not true unless you take care over "the stalks are T-algebras". For example, if the base space is Sierpinksi then a sheaf is just a function, with the domain and codomain as the stalks (over the bottom and top points respectively). You might put algebra structures on the stalks, but it won't be an algebra in the sheaf category unless the function is a homomorphism.
You will get further issues if the base space is non-spatial, so there aren't enough global points to provide enough stalks.
What statement of the result did you have in mind?
One sufficient condition that gets round those problems is for the stalks and their T-algebra structure to be defined geometrically.
Regards,
Steve.
Michael Barr wrote:
It must be known that if T is a left exact theory (same as nearly equational theory) and you make a sheaf of T-algebras (meaning the stalks are T-algebras, then the set of global sections is also a T-algebra. Can anyone give me reference?
Michael
-- The modern conservative is engaged in one of man's oldest exercises in moral philosophy--the search for a superior moral justification for selfishness. --J.K. Galbraith [For admin and other information see: http://www.mta.ca/~cat-dist/ ]
Dear Mike, In that case you want to look at John Kennison's 1976 paper "Integral domain type representations in sheaves and other topoi", Math. Z. 151, 35-56, and my own "Rings, fields and spectra" in J. Algebra 49 (1977), 238-260. There's a summary in section V 3.11 of "Stone Spaces". Peter Johnstone On Mon, 24 Sep 2012, Michael Barr wrote:
Well, I actually know quite a bit more about it. First off, each point in the base has a local neighborhood whose sections are algebras and the restrictions are homomorphisms. I should have mentioned that all ops and partial ops are finitary, but of increasing arity. We know the answer is positive; I am just unhappy with the proof.
The situation is that the stalks are integral domains. Now integral domains are not models of an LE theory. But they are models of a theory with an infinitude of LE operations (all finitary, but of increasing arity) and I want the global sections to be models of that theory (which turns out to characterize the rings in the limit closure of the domains). The space is the set of primes of some arbitrary semi-prime (no nilpotents) commutative ring with the co-Zariski topology.
Mike
On Mon, 24 Sep 2012, Steve Vickers wrote:
Dear Mike,
Once you have that a sheaf of T-algebras is equivalent to a T-algebra in the category of sheaves, it's obvious: because the global sections functor preserves finite (in fact all) limits.
But the first step is not so trivial, and in fact is not true unless you take care over "the stalks are T-algebras". For example, if the base space is Sierpinksi then a sheaf is just a function, with the domain and codomain as the stalks (over the bottom and top points respectively). You might put algebra structures on the stalks, but it won't be an algebra in the sheaf category unless the function is a homomorphism.
You will get further issues if the base space is non-spatial, so there aren't enough global points to provide enough stalks.
What statement of the result did you have in mind?
One sufficient condition that gets round those problems is for the stalks and their T-algebra structure to be defined geometrically.
Regards,
Steve.
Michael Barr wrote:
It must be known that if T is a left exact theory (same as nearly equational theory) and you make a sheaf of T-algebras (meaning the stalks are T-algebras, then the set of global sections is also a T-algebra. Can anyone give me reference?
Michael
[For admin and other information see: http://www.mta.ca/~cat-dist/ ]
Dear Mike, Of the results Peter reports in the Elephant, the main one about sheaves is - as he points out - restricted to point-set spaces. You may think that's enough, but morally there is a result that is both simpler and more general and applies also to point-free spaces. It is somewhat like your original phrasing: if every stalk is a model (of a given theory T), then the sheaf is an internal T-model in the category of sheaves. The proof idea is to consider the stalk for the generic point: this is the sheaf itself, and so is a model. Of course, the catch is that the generic point is internal in the topos of sheaves, not in Set. Hence the construction point |-> model has to be be not only valid in toposes, but also geometric, so that the generic construction agrees with the specific ones. If "every stalk is a model" is interpreted in this geometric way, then the result holds and applies to general geometric theories, not just lex ones. Lex comes in when you want to make a model out of the set of global sections. In your example, for a commutative ring R, the points of the co-Zariski spectrum are described geometrically as the prime ideals of R. In other words, the topos of sheaves classifies the geometric theory of prime ideals of R. (We have to be constructively careful here, since for the Zariski spectrum the points are the prime coideals.) Then the generic point is an internal prime ideal of the constant sheaf R. For every prime ideal p we can construct an integral domain R/p, and this construction is geometric. It constructs not only the sheaf (the stalks are R/p) but also the stalkwise ID structure. Applying it to the generic prime ideal of R gives us an ID in the category of sheaves, and geometricity ensures that it pulls back to the construction in Set for specific prime ideals. The theory of IDs is not lex, but it is geometric. I don't know the details of your lex theory, so there is a further question regarding the construction of models of it out of IDs: is this construction geometric? If so, then you can carry on working internally in the topos of sheaves to get a model of the lex theory. All the best, Steve. Michael Barr wrote:
Well, I actually know quite a bit more about it. First off, each point in the base has a local neighborhood whose sections are algebras and the restrictions are homomorphisms. I should have mentioned that all ops and partial ops are finitary, but of increasing arity. We know the answer is positive; I am just unhappy with the proof.
The situation is that the stalks are integral domains. Now integral domains are not models of an LE theory. But they are models of a theory with an infinitude of LE operations (all finitary, but of increasing arity) and I want the global sections to be models of that theory (which turns out to characterize the rings in the limit closure of the domains). The space is the set of primes of some arbitrary semi-prime (no nilpotents) commutative ring with the co-Zariski topology.
Mike
On Mon, 24 Sep 2012, Steve Vickers wrote:
Dear Mike,
Once you have that a sheaf of T-algebras is equivalent to a T-algebra in the category of sheaves, it's obvious: because the global sections functor preserves finite (in fact all) limits.
But the first step is not so trivial, and in fact is not true unless you take care over "the stalks are T-algebras". For example, if the base space is Sierpinksi then a sheaf is just a function, with the domain and codomain as the stalks (over the bottom and top points respectively). You might put algebra structures on the stalks, but it won't be an algebra in the sheaf category unless the function is a homomorphism.
You will get further issues if the base space is non-spatial, so there aren't enough global points to provide enough stalks.
What statement of the result did you have in mind?
One sufficient condition that gets round those problems is for the stalks and their T-algebra structure to be defined geometrically.
Regards,
Steve.
Michael Barr wrote:
It must be known that if T is a left exact theory (same as nearly equational theory) and you make a sheaf of T-algebras (meaning the stalks are T-algebras, then the set of global sections is also a T-algebra. Can anyone give me reference?
Michael
[For admin and other information see: http://www.mta.ca/~cat-dist/ ]
Hi--long time no talk....Anyway, I've been traveling and just now read your post, and I'm replying before reading any responses you may have received. Depending upon properties of "left exact theory", it's possible that this results from facts in my SLN#236 paper, now over 40 years old....time flies. Best to you and Marcia, Don [For admin and other information see: http://www.mta.ca/~cat-dist/ ]
participants (4)
-
Donovan Van Osdol -
Michael Barr -
Prof. Peter Johnstone -
Steve Vickers