Timeline for Quotient of a smooth curve by a finite group and differentials
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jan 11, 2017 at 0:07 | comment | added | Jason Starr | @MinseonShin. I do not actually remember the argument I had in mind, but your argument sounds great. | |
| Jan 10, 2017 at 22:11 | comment | added | Minseon Shin | OK here's one argument but there's probably an easier way. Set $y := \frac{x^{p}}{1-x^{p-1}}$. Since $k[x,\frac{1}{1-x^{p-1}}]$ is integral over $k[y]$ (namely $x^{p}+yx^{p-1}=y$ and $\frac{1}{1-x^{p-1}} = x^{p-1}+yx^{p-2}+1$) and $k[y]$ is normal, the inclusion $k[y] \subseteq k[x,\frac{1}{1-x^{p-1}}] \cap k(y)$ is an equality. Moreover $(k(x))^{\mathbb{Z}/(p)} = k(y)$. | |
| Jan 10, 2017 at 21:22 | comment | added | Minseon Shin | @jason-starr How do you conclude that the ring of invariants is $k[\frac{x^{p}}{1-x^{p-1}}]$? I see that $\frac{x^{p}}{1-x^{p-1}}$ is the product of the orbit of $x$ under the $\mathbb{Z}/p$-action, namely $\{x,\frac{x}{1+x},\dotsc,\frac{x}{1+(p-1)x}\}$. | |
| Dec 24, 2015 at 23:28 | history | edited | Jason Starr | CC BY-SA 3.0 |
added 3 characters in body
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| Dec 24, 2015 at 20:20 | comment | added | Jason Starr | @LisaS. Thank you for catching the mistake. I have now corrected the exponent. | |
| Dec 24, 2015 at 20:20 | history | edited | Jason Starr | CC BY-SA 3.0 |
Corrected exponent per suggestion of Lisa S.
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| Dec 24, 2015 at 19:09 | comment | added | Lisa S. | Thanks. In your last formula, shouldn't the numerator be $-x^{2p -2}$ instead of $-x^p$. | |
| Dec 24, 2015 at 19:01 | history | edited | Jason Starr | CC BY-SA 3.0 |
edited body
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| Dec 24, 2015 at 18:34 | history | edited | Jason Starr | CC BY-SA 3.0 |
added 120 characters in body
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| S Dec 24, 2015 at 16:17 | history | answered | Jason Starr | CC BY-SA 3.0 | |
| S Dec 24, 2015 at 16:17 | history | made wiki | Post Made Community Wiki by Jason Starr |