Timeline for Multiplicative group of a ring as a morphism of theories
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 14, 2023 at 10:24 | vote | accept | Arshak Aivazian | ||
| Jun 14, 2023 at 4:31 | comment | added | Mike Shulman | @FernandoMuro Generally "the theory of X" means "the theory whose algebras are X". | |
| Jun 14, 2023 at 4:03 | answer | added | Zhen Lin | timeline score: 8 | |
| Jun 14, 2023 at 2:34 | comment | added | Arshak Aivazian | @ZhenLin Really! After dualization, the group ring functor $\mathbb{Z}[-]^{\mathrm{op}} \colon \mathrm{Group}_{\mathrm{fin}}^{\mathrm{op}} \to \mathrm{Ring}_{\mathrm{fin}}^{\mathrm{op}}$ preserves finite limits (initially it is left adjoint, hence after dualization it preserves all limits). And the functor induced by it on the categories of algebras sends the ring $R$ to the group whose presheaf $G \mapsto \mathrm{Hom}(\mathbb{Z}[G], R) = \mathrm{Hom}(G, R^{\times})$. So this is the group of invertible elements $R$. Thank you! If you post this answer, I will accept it. | |
| Jun 14, 2023 at 2:21 | comment | added | Arshak Aivazian | @FernandoMuro But what theories do you mean? If we talk about algebraic theories (= Lawvere theories), then there is no morphism that induces the desired functor, do you agree? | |
| Jun 13, 2023 at 21:13 | comment | added | Fernando Muro | @ArshakAivazian you said the theory of commutative rings not the theory whose algebras are commutative rings. | |
| Jun 13, 2023 at 14:54 | comment | added | Arshak Aivazian | @FernandoMuro I don't quite understand what you mean: we need a functor between theories (preserving the product, yes), not between categories of algebras. | |
| Jun 13, 2023 at 11:16 | comment | added | Zhen Lin | No need to go to geometric. This phenomenon can already be seen at the level of cartesian (= finite limit) theories. | |
| Jun 13, 2023 at 10:04 | comment | added | Fernando Muro | A morphism of theories is a product preserving functor, and $(R\times S)^\times =R^\times\times S^\times$. | |
| Jun 13, 2023 at 1:28 | history | asked | Arshak Aivazian | CC BY-SA 4.0 |