Timeline for Is every cancellative semigroup a subdirect product of subdirectly irreducible cancellative semigroups?
Current License: CC BY-SA 4.0
26 events
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| Oct 10 at 9:04 | comment | added | YCor | @SalvoTringali it was the same in group theory, say, 20 ago, but left-ordered groups have had an increasing importance since then. | |
| Oct 10 at 7:01 | comment | added | Salvo Tringali | @YCor I'm aware that group theorists tend to use the term 'bi-ordered' to mean 'totally ordered'. However, in other fields (including semigroup theory), 'ordered' is often used to mean 'partially bi-ordered'; otherwise, qualifiers like 'left' and 'right' are used, followed by terms such as 'orderable', 'totally ordered', etc. (see, for instance, Chap. 11 in the 2005 edition of Blyth's Lattices and Ordered Algebraic Structures). That's one reason why I included the definition of 'totally ordered' in my answer. | |
| Oct 10 at 6:49 | comment | added | YCor | Just on terminology: left ordered groups are important objects to so it is useful to include "bi-ordered" in the terminology. | |
| Oct 10 at 6:21 | vote | accept | Salvo Tringali | ||
| Oct 10 at 2:20 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
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| Oct 10 at 1:20 | comment | added | Salvo Tringali | @YCor That $S$ is subdirectly irreducible is a consequence of the main theorem (which, in the meanwhile, was reformulated). That being said, I agree with Benjamin Steinberg's previous comments: I thought I had specified that $S$ is the same type of semigroup as in YCor's construction, but the truth is that I had forgotten. (I deleted some of my previous comments to tidy up the comment section.) | |
| Oct 9 at 23:43 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
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| Oct 9 at 23:25 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
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| Oct 9 at 18:01 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Oct 9 at 17:26 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Oct 9 at 17:18 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Oct 9 at 17:15 | comment | added | Benjamin Steinberg | No, I think it is fine. | |
| Oct 9 at 17:12 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Oct 9 at 17:06 | comment | added | Benjamin Steinberg | I also took the liberty of simplifying the proof. | |
| Oct 9 at 17:03 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
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| Oct 9 at 16:58 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
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| Oct 9 at 16:54 | comment | added | Benjamin Steinberg | Ok. I fixed it for you but you can roll back if you don’t like the changes. | |
| Oct 9 at 16:53 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Oct 9 at 16:46 | comment | added | Benjamin Steinberg | You should begin by saying that S is the positive cone in Thompsons group F | |
| Oct 9 at 16:45 | comment | added | Benjamin Steinberg | You just say S is a positive cone in some group. The claim is that at least one factor in the subdirect product decomposition is not cancellative. This is obviously false if S is commutative. Your claim is true for positive cones which are not subdirect product of groups. @YCor has one specific example for which this is the case. He doesn’t claim it is always true. | |
| Oct 9 at 16:38 | comment | added | Benjamin Steinberg | But I do believe this works as long as S is not a subdirect product of groups. | |
| Oct 9 at 16:34 | comment | added | Benjamin Steinberg | I think you need that the cancellative semigroup is not a subdirect product of groups like in @YCor’s example | |
| Oct 9 at 16:32 | comment | added | Benjamin Steinberg | This can’t be right as written since N with + is the positive cone in Z and is a subdirect product of groups. | |
| Oct 9 at 16:29 | comment | added | YCor | Is it clear, or does it follow from your proof, that $S$ is not subdirectly irreducible? | |
| Oct 9 at 16:24 | history | edited | Salvo Tringali | CC BY-SA 4.0 |
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| Oct 9 at 16:17 | history | answered | Salvo Tringali | CC BY-SA 4.0 |