Timeline for Confusion about two statements about cohomology of curves with automorphisms
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Sep 10, 2013 at 1:49 | vote | accept | Darius Math | ||
| Sep 7, 2013 at 0:54 | comment | added | Darius Math | @S.Carnahan. I did not wrie this. I just copied what Will had wirtten. But it's now edited in his answer. | |
| Sep 6, 2013 at 19:31 | history | edited | Will Sawin | CC BY-SA 3.0 |
added 4 characters in body
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| Sep 6, 2013 at 18:41 | comment | added | S. Carnahan♦ | When you write $\mathbb{Q}[x]/x^m$, do you really want $x$ to be nilpotent? | |
| Sep 6, 2013 at 17:20 | comment | added | Will Sawin | There is an action of that on the cohomology when $m$ is prime, or in certain other special cases. I think the references you see are either in those cases, or using notation where $\mathbb Q[\xi]\neq \mathbb Q(\xi)$. But the second vector space structure does not exist in general. | |
| Sep 6, 2013 at 16:27 | comment | added | Darius Math | Of course I did not claim $\mathbb{Q}[x]/(x^{m}-1)$ is a field, but it is not equal to $\mathbb{Q}[\xi]$, my meaning was that $\mathbb{Q}[\xi]=\mathbb{Q}[x]/ \Phi_{m-1}(x)=\mathbb{Q}(\xi)$ is a field. And it is normally said that there is an action of this on the cohomology. But now this should give (at least) two $\mathbb{Q}$-vector space structures. | |
| Sep 6, 2013 at 16:21 | comment | added | Will Sawin | $Q(\xi)$ is the field extenction generated by a primitive $m$th root of unity. $\mathbb Q[x]/(x^m-1)$ is not a field, since it has zero divisors: $(x-1)(\sum_{k=1}^{m-1} x^k)=0$. The first is a $\phi(m)$-dimensional vector space over $\mathbb Q$, and the second is $m$-dimensional. | |
| Sep 6, 2013 at 16:14 | comment | added | Darius Math | Thanks but I did not get your meaning by the last sentence! As far as I know: $\mathbb{Q}(\xi)=\mathbb{Q}[\xi]$. And also, $\mathbb{Q}[x]/{x^{m}}$ is just an $m$-times product of $\mathbb{Q}$. | |
| Sep 6, 2013 at 16:07 | history | answered | Will Sawin | CC BY-SA 3.0 |