Timeline for Are algebraic structures uniquely identifed by their free objects?
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16 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Jan 26, 2014 at 15:42 | comment | added | Denis | @AndrejBauer the plan was to take for $x^\omega$ all the identities that are true for the idempotent power in finite structures (axioms (A')), and maybe additionally (A) | |
| Jan 26, 2014 at 7:55 | comment | added | Andrej Bauer | @ToddTrimble: but we still have an essentially algebraic theory, don't we? | |
| Jan 26, 2014 at 7:52 | comment | added | Andrej Bauer | Are there other axioms for $x^\omega$ which you're not showing? Something that tells us $x^\omega$ is like an infinite power of $x$? | |
| Jan 26, 2014 at 2:31 | answer | added | Benjamin Steinberg | timeline score: 5 | |
| Jan 26, 2014 at 1:01 | comment | added | Todd Trimble | Oh, I am sorry DK; yes, you have Horn clauses. You can ignore my comment -- apologies. | |
| Jan 26, 2014 at 0:28 | comment | added | Denis | @ToddTrimble It is probably what I'm looking for, but I'm not very familiar with category theory. If I understand your comment, you are saying that having a free object is given almost for free according to category theory, as long as you are a monad. I tried to understand what a monad is on wikipedia, but it seems circular, because it looks like it is the structures having free objects... Also you mention varieties, but here it's not a variety, because axioms are not restricted to identities. | |
| Jan 25, 2014 at 23:57 | comment | added | Todd Trimble | It seems to me there is one pretty trivial answer to the title question, so it must not answer what you want. An algebraic theory $T$ can be construed as a monad on $Set$, and given the category of free $T$-algebras together with the forgetful functor $U: Free_T \to Set$, with left adjoint $F: Set \to Free_T$, the monad $T$ can be recovered as the composite $U F$ (with the monad data recoverable from the adjunction data). On another question: free $T$-algebras exist for any monad $T$ on $Set$, and certainly varieties defined by operations and equations of bounded rank give monads. | |
| Jan 25, 2014 at 23:15 | history | edited | Denis | CC BY-SA 3.0 |
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| Jan 25, 2014 at 22:55 | comment | added | Emil Jeřábek | As M. Shahryari mentioned, closure under products and subalgebras implies the existence of free objects over any set of generators. That’s a theorem, I don’t know what “precise” means (if anything). | |
| Jan 25, 2014 at 22:39 | comment | added | Denis | ok, is it because using "finite" as an axiom seems even worse than using implications? And is there a precise theorem stating "closed under product+ something => exists free object" ? Birkhoff seems close but I'm not sure it is the one... | |
| Jan 25, 2014 at 22:34 | comment | added | Emil Jeřábek | You do need infinite products unless the variety generated by your objects is locally finite. For example, the class of finite Heyting algebras is closed under finite products, but does not have free objects over any nonempty set of generators, as these would need to be infinite Heyting algebras. | |
| Jan 25, 2014 at 21:53 | history | edited | Denis | CC BY-SA 3.0 |
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| Jan 25, 2014 at 20:59 | answer | added | Sh.M1972 | timeline score: 0 | |
| Jan 25, 2014 at 20:56 | history | edited | Ricardo Andrade |
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| Jan 25, 2014 at 20:14 | history | asked | Denis | CC BY-SA 3.0 |