Timeline for Does the category of rings embed fully faithfully into the category of $\mathbb{F}_{1}$-algebras?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 13, 2021 at 18:28 | history | undeleted | Emily | ||
| Sep 13, 2021 at 18:27 | history | deleted | Emily | via Vote | |
| Aug 13, 2021 at 20:46 | comment | added | Emily | @BenjaminSteinberg and ＠MartinBrandenburg, Thanks! | |
| Aug 13, 2021 at 12:20 | comment | added | Benjamin Steinberg | If you Google contracted semigroup algebra you will find the ring theoretical version but it is the same for semirings | |
| Aug 13, 2021 at 12:18 | comment | added | Benjamin Steinberg | The contracted monoid algebra is the left adjoint of the forgetful functor from semirings to monoid. No fancy category theory is needed. You take the free $\mathbb N$-module (or some would say semimodule) with basis the non zero elements of the monoid M and define the product on the basis to agree with the monoid product where you interpret the zero products as the zero in the module. | |
| Aug 13, 2021 at 11:54 | comment | added | Martin Brandenburg | The forgetful functor $\mathrm{Alg}(T) \to \mathrm{Alg}(T')$ associated to a morphism of monads $T' \to T$ always has a left adjoint when $\mathrm{Alg}(T')$ has coequalizers. You write it down using presentations. A reference is Durov's thesis, Prop 3.3.19. | |
| Aug 13, 2021 at 5:00 | comment | added | Emily | @MartinBrandenburg Ah, that's a very good point! I understand that the functor $\mathsf{Semirings}\to\mathsf{Mon}$ sending a semiring to its underlying multiplicative monoid has a left adjoint, the monoid semiring functor $\mathbb{N}[-]$. Do you know if the forgetful functor $\mathsf{Semirings}\to\mathsf{Alg}_{\mathbb{F}_{1}}$ also admits a left adjoint? (I'm pretty sure it does, but I haven't worked it out; I ask because I think there's a good chance you might have already figured this out before) | |
| Aug 13, 2021 at 4:46 | comment | added | Martin Brandenburg | I don't agree with the first sentence. This would mean that $\mathbb{F}_1 \to \mathbb{Z}$ is an epimorphism. We just want to have a forgetful (left adjoint) functor. | |
| Aug 13, 2021 at 4:29 | comment | added | Emily | (This question has a similar spirit: link) | |
| Aug 13, 2021 at 4:26 | history | asked | Emily | CC BY-SA 4.0 |