Timeline for "Generalized theory of polynomials" for a given commutative Lawvere Theory
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 24, 2015 at 13:47 | comment | added | Georg Lehner | I'm sorry for the error, I'm new to overflow. The link I meant is of course ncatlab.org/nlab/show/tensor+product+theory | |
| Jul 24, 2015 at 13:41 | comment | added | Georg Lehner | The reason I'm asking is that, if it is; and the relationship between the resulting Lawvere theory and $T^{poly}_R$ is functorial, I can take this functor, compose it with $T^{lin}_\mathbb{Z} \otimes -$, and have a candidate for the right adjoint of the adjoint situation I was looking for. EDIT: With the tensor product I mean of course <a href="ncatlab.org/nlab/show/tensor+product+theory">the tensor product of algebraic theories</a>. | |
| Jul 24, 2015 at 13:37 | comment | added | Georg Lehner | Thank you, that is indeed helpful. The way to make this precise is the fact that for two theories $T$ and $T'$ we have $\textbf{Mod}(T \otimes T', Set) \cong \textbf{Mod}(T, \textbf{Mod}(T', Set))$. Is the slice category of Commutative $R$-Algebras over $R$ again a category of models for a Lawvere-Theory? | |
| Jul 22, 2015 at 15:51 | comment | added | Zhen Lin | Yes, that is the procedure I am thinking of. It is a fact that the category of abelian group objects in the category of algebras for a Lawvere theory is itself equivalent to the category of algebras for a Lawvere theory (and, in particular, equivalent to the category of modules for a ring). | |
| Jul 22, 2015 at 15:25 | comment | added | Omar Antolín-Camarena | Just to be explicit, I think @ZhenLin means that the category of R-modules is equivalent to the category of (Abelian) group objects in the slice over R of the category of Commutative R-algebras. | |
| Jul 22, 2015 at 13:35 | comment | added | Zhen Lin | There is a general procedure that derives the theory of $R$-modules from the theory of commutative $R$-algebras, but I do not know of a way to do the reverse. | |
| Jul 22, 2015 at 13:06 | review | First posts | |||
| Jul 22, 2015 at 13:07 | |||||
| Jul 22, 2015 at 13:03 | history | asked | Georg Lehner | CC BY-SA 3.0 |