Timeline for Cotangent complex of perfect algebra over a perfect field
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 5, 2018 at 16:35 | comment | added | A Rock and a Hard Place | Oh, I see, so you are only asserting that $B^+$ and $B\otimes \mathbf Z$ are equivalent as spectra, but not necessarily as ring spectra (at least not with the "usual" ring spectrum structure on $B\otimes \mathbf Z$). Thank you for the tireless explanations! | |
| Jan 5, 2018 at 15:19 | comment | added | Jacob Lurie | The vector space $\mathbf{R}[i] \otimes_{ \mathbf{R} } \mathbf{R}[j]$ does not have a unique $\mathbf{R}$-algebra structure, even if you require the inclusion maps $\mathbf{R}[i] \hookrightarrow \mathbf{R}[i] \otimes_{ \mathbf{R} } \mathbf{R}[j] \hookleftarrow \mathbf{R}[j]$ to be maps of $\mathbf{R}$-algebras. There's a unique algebra structure in which $\mathbf{R}[i]$ and $\mathbf{R}[j]$ commute with each other, but also algebra structures where they do not (like $\mathbf{H}$). The ring spectrum $B^{+}$ is like the latter. | |
| Jan 5, 2018 at 10:38 | comment | added | A Rock and a Hard Place | Or are you saying that the forgetful functor $\operatorname{CAlg}_R\to \operatorname{Alg}_R$ isn't monoidal wrt $\otimes_R$? | |
| Jan 5, 2018 at 10:32 | comment | added | A Rock and a Hard Place | But both $\mathbf R[i]$ and $\mathbf R[j]$ are isomorphic to $\mathbf C$ as $\mathbf R$-algebras, so shouldn't it be that $\mathbf R[i]\otimes_{\mathbf R}\mathbf R[j]\simeq \mathbf C\otimes_{\mathbf R}{\mathbf C}\simeq \mathbf C^2$? | |
| Jan 4, 2018 at 9:23 | comment | added | Jacob Lurie | The ring of quaternions $\mathbf{H}$ contains commutative subfields $\mathbf{R}[i]$ and $\mathbf{R}[j]$, and the inclusion maps $\mathbf{R}[i] \hookrightarrow \mathbf{H} \hookleftarrow \mathbf{R}[j]$ induce an isomorphism $\mathbf{R}[i] \otimes_{\mathbf{R}} \mathbf{R}[j] \simeq \mathbf{H}$. But $\mathbf{H}$ is not commutative. | |
| Jan 3, 2018 at 19:38 | comment | added | A Rock and a Hard Place | Also I'm a little confused about how $B^+$ can fail to be commutative if it is equivalent to $B\otimes \mathbf Z$, and both $B$ and $\mathbf Z$ are commutative. Probably I'm missing something obvious, sorry. | |
| Jan 3, 2018 at 19:32 | comment | added | A Rock and a Hard Place | Dear Jacob, thank you very much for this very interesting answer! Is this something we can look forward to in the SAG.VII (or is it perhaps already written out somewhere?)? | |
| Jan 2, 2018 at 3:40 | history | answered | Jacob Lurie | CC BY-SA 3.0 |