Timeline for Well-foundedness of divisibility vs well-foundedness of right- and left-divisibility
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Mar 4, 2021 at 12:10 | comment | added | Benjamin Steinberg | In the transformation monoid K, Green's relations are different. The monoid K doesn't have any chain conditions on left ideals. H does since you are Just adding a maximum to all your posets. | |
| Mar 4, 2021 at 8:24 | comment | added | Salvo Tringali | Many thanks (also for the patience)! I think what I had in mind might be the following: Your 𝐻 is the unitization of a certain right ideal 𝑆 of a suitable monoid 𝐾. While the elements of 𝑆 are not left-invertible in 𝐻, they are so in 𝐾. This and 𝑆 being a right ideal of 𝐾 guarantee that, for all 𝑎,𝑏∈𝑆, the equ 𝑥𝑎=𝑏 is solvable in 𝐻 (whence ACCP, ACCPL and their DCC duals hold). And now the particular choice you make on 𝐾 (the transformation monoid of an infinite set 𝑋) and 𝑆 (the non-invertible injects 𝑓: 𝑋→𝑋 s.t. 𝑋 ∖ 𝑓(𝑋) is infinite) prevents ACCPR & DCCPR. | |
| Mar 3, 2021 at 23:48 | comment | added | Benjamin Steinberg | The point is that the equation has a solution for nonidenity elements. You can't solve ab=1. If every element was left invertibile you would have a group. | |
| Mar 3, 2021 at 22:39 | comment | added | Salvo Tringali | Whoops! Somehow, I had mindlessly translated the condition "for fixed $a$ and $b$, the equation $xa = b$ has at least one solution" to "$x = ba'$ with $a'$ a left inverse of $a$". | |
| Mar 3, 2021 at 22:17 | comment | added | Benjamin Steinberg | But you don't have left invertibility of any non unit. | |
| Mar 3, 2021 at 22:03 | comment | added | Benjamin Steinberg | It's not a monoid. It's a semigroup. All elements right divide each other but it has infinite ascending and descending chains from any element for left divisibility. If you adjoin an identity you get a monoid with 2 right divisibility classes | |
| Mar 3, 2021 at 21:56 | vote | accept | Salvo Tringali | ||
| Mar 3, 2021 at 21:51 | comment | added | Salvo Tringali | If I understand the idea correctly, you consider a monoid $H$ with the property that (i) every element is left-invertible and (ii) the ACC on principal right ideals is not satisfied: In the parlance of the OP, it follows from (i) that the divisibility preorder is artinian (equivalently, $H$ satisfies the ACC on principal ideals), and so also is the divides-from-the-right preorder (equivalently, $H$ satisfies the ACC on principal left ideals); while (ii) means that the divides-from-the-left preorder is not artinian. Nice! | |
| Mar 3, 2021 at 17:26 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Mar 3, 2021 at 15:09 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Mar 3, 2021 at 15:03 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Mar 3, 2021 at 14:52 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Mar 3, 2021 at 14:46 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Mar 3, 2021 at 14:39 | history | answered | Benjamin Steinberg | CC BY-SA 4.0 |