Timeline for How to decompose an map $\phi: \mathbb{G}_m \to T$ as the product of a cocharacter $\phi'$ and a map $\phi'':\mathbb{G}_m \to T$?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| May 9, 2018 at 14:13 | history | edited | R. van Dobben de Bruyn | CC BY-SA 4.0 |
Word 'Spec' added.
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| May 9, 2018 at 14:13 | comment | added | R. van Dobben de Bruyn | The ring $k((x))$ is the fibre product: to tensor something over $k[x]$ with $k[x^{\pm 1}]$ just means inverting $x$. | |
| May 9, 2018 at 8:36 | comment | added | Jianrong Li | thank you very much. How to think Spec $k((x))$ as $k[x^{\pm 1}] \otimes_{k[x]} k[[x]]$? | |
| May 7, 2018 at 16:11 | comment | added | R. van Dobben de Bruyn | @JianrongLi: well, set-theoretically it has just one point, but writing $\{0\}$ is a bit suggestive (one might think that the point is a $k$-point, which is very false). | |
| May 7, 2018 at 14:14 | comment | added | Jianrong Li | thank you very much. I have question about $Spec k((x))$. I think that $k((x))$ is a field. Therefore $Spec k((x))$ is just $\{0\}$? | |
| May 6, 2018 at 20:21 | vote | accept | Jianrong Li | ||
| May 6, 2018 at 17:40 | history | answered | R. van Dobben de Bruyn | CC BY-SA 4.0 |