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May 8 at 15:16 history edited user267839 CC BY-SA 4.0
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May 8 at 15:14 comment added user267839 @LSpice: sure, sorry for pretty poor wording
May 8 at 15:09 comment added LSpice @AlexYoucis, re, we can even see the problem purely on the level of field points, replacing $\mathbb F_p$ and $p$ by $\mathbb Q$ and $3$. \\ @‍OP, I am having trouble parsing "Note it's a fact that if a subfunctor representable by a scheme it necessarily a fppf sheaf / satisfies fppf descent". At first I thought that the last "it" should be "is", but that doesn't read right either to me (what is the consequent of the "if"?). Could you fix it?
May 8 at 15:07 history edited LSpice CC BY-SA 4.0
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May 8 at 13:21 history edited YCor CC BY-SA 4.0
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May 8 at 13:11 history edited user267839 CC BY-SA 4.0
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May 8 at 10:23 history edited user267839 CC BY-SA 4.0
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May 8 at 10:21 comment added user267839 @AlexYoucis: oh right, I see, so seemingly this "naive" definition would already fail to declare it as well defined functor, thanks!
May 8 at 10:07 comment added Alex Youcis Is it even functorial then? For instance, consider $\mu_p$ acting on $\mathbb{A}^1_{\mathbb{F}_p}$ by $g\cdot x=gx$. Then, for your naive fixed points we would get over $\mathbb{F}_p$ the set $\{x\in \mathbb{F}_p:gx=x\text{ for all }g\in \mu_p(\mathbb{F}_p)\}$. But, $\mu_p(\mathbb{F}_p)$ is trivial, so this is just $\mathbb{F}_p$. But, if we plug in $\mathbb{F}_p[T]/(T^p-1)$ then we would get $\{x\in\mathbb{F}_p[T]/(T^p-1): gx=x\text{ for all }g\in \mu_p(\mathbb{F}_p[T]/(T^p-1)\}$. In particular, it's not true that $1\in\mathbb{F}_p$ is in here as $T\cdot 1=T\ne 1$.
May 8 at 10:01 history edited user267839 CC BY-SA 4.0
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May 8 at 9:54 history edited user267839 CC BY-SA 4.0
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May 8 at 9:47 history asked user267839 CC BY-SA 4.0