Timeline for Are algebraic structures uniquely identifed by their free objects?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jun 13, 2021 at 14:30 | comment | added | Peter Languilla | @EmilJeřábek can you help me with an example of ISP class ($\mathbf{V}$) that is not determined by $Fr_\omega(\mathbf{V})$. Also if you can help me with a hint to prove that an HSP class is determined by $Fr_\omega(\mathbf{V})$ will be very usefull. | |
| Jan 25, 2014 at 23:00 | comment | added | Denis | Thanks @EmilJeřábek. Few more questions: how can anything not be closed under isomorphic image (it's just renaming) ? What is the result stating that if it's closed under ISP then there is a free object? | |
| Jan 25, 2014 at 22:58 | comment | added | Emil Jeřábek | I = isomorphic images, S = subalgebras, P = direct products. Actually, one does not need the I as free objects are only defined up to isomorphism anyway, but usually this is the least of all worries. | |
| Jan 25, 2014 at 22:43 | comment | added | Denis | @EmilJeřábek what is an ISP-class? closed under morphism subalgebra and product? | |
| Jan 25, 2014 at 22:29 | comment | added | Emil Jeřábek | @DK: Free algebras exist in every ISP class, but this does not imply that they determine the ISP class. Any two ISP classes that generate the same variety will have the same free algebras, so free algebras only depend on the equational theory of the class. | |
| Jan 25, 2014 at 21:44 | comment | added | Denis | @M._Shahryari what is the link between free objects and infinite products? | |
| Jan 25, 2014 at 21:24 | comment | added | Denis | But still, it wouldn't be a problem that the free (A)-algebra with k generators is the same as the free (A')-algebra with k generators? | |
| Jan 25, 2014 at 21:06 | comment | added | Sh.M1972 | This is not sufficient, if you want to have free objects inside your class, you must control that is it closed under arbitrary product or not. I think that it is closed. | |
| Jan 25, 2014 at 21:04 | comment | added | Denis | I think we are only interested in closure under finite products. | |
| Jan 25, 2014 at 21:02 | comment | added | Denis | "Algebraic structure" in the title means something like "groups" or "monoids" in general, and not particular instances of them. | |
| Jan 25, 2014 at 20:59 | history | answered | Sh.M1972 | CC BY-SA 3.0 |