Timeline for Is there a bijection between elements in algebraic closure of F2 and all bi-infinite periodic sequences made of 0 and 1, filling the properties below?
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Jun 12, 2022 at 12:45 | comment | added | David E Speyer | I agree with Emil. | |
| Jun 12, 2022 at 10:12 | comment | added | Emil Jeřábek | I don’t see a simple way of doing that. | |
| Jun 12, 2022 at 9:39 | comment | added | Ginger | @DavidESpeyer If the bijection $f$ is defined as these, how can I define the "multiplication of two periodic sequences", or the expression between $f_k(xy)$ and the entries of $f(x)$ and $f(y)$? | |
| Jun 12, 2022 at 9:39 | comment | added | Ginger | @EmilJeřábek If the bijection $f$ is defined as these, how can I define the "multiplication of two periodic sequences", or the expression between $f_k(xy)$ and the entries of $f(x)$ and $f(y)$? | |
| Jun 10, 2022 at 13:44 | comment | added | Emil Jeřábek | @DavidESpeyer Feel free to write an answer. I’m actually rather busy at the moment, and won’t be able to do it. | |
| Jun 10, 2022 at 13:15 | comment | added | David E Speyer | @EmilJeřábek One of us should write this up as an answer. I think you figured out more of it than I did, would you like to? | |
| Jun 10, 2022 at 11:54 | comment | added | Will Sawin | @DavidESpeyer Rather than use the trace pairing, it's probably slightly simpler to use the dual basis to the normal basis. | |
| Jun 10, 2022 at 11:44 | comment | added | Emil Jeřábek | @DavidESpeyer Perfect! I have meanwhile figured out I should use $\mathrm{Tr}(xy)$, but couldn’t quite put a finger on why a suitable $y$ exists. | |
| Jun 10, 2022 at 11:26 | comment | added | David E Speyer | @EmilJeřábek To see that such an $H$ exists, use the normal basis theorem to find a $y$ such that $y$, $\sigma(y)$, ... $\sigma^{n-1}(y)$ is a basis of $\mathbb{F}_{2^n}$ (where $\sigma$ is the Frobenius map). Then take the kernel of $x \mapsto \text{Tr}(xy)$. If there were some $x$ all of whose conjugates obeyed $\text{Tr}(\sigma^k(x) y) =0$, then we would have $\text{Tr}(x \sigma^{-k}(y)) =0$. But trace is a perfect pairing and the $\sigma^{-k}(y)$ span $\mathbb{F}_{2^n}$, so this implies that $x$ is $0$. | |
| Jun 10, 2022 at 11:09 | comment | added | Emil Jeřábek | Also, a simple compactness argument shows that such an $H$ exists for $\overline{\mathbb F}_2$ iff for all $n$, there exists such an $H$ for $\mathbb F_{2^n}$. | |
| Jun 10, 2022 at 11:02 | comment | added | Emil Jeřábek | Unless I missed something, the existence of $f$ is equivalent to the existence of an index-$2$ subgroup $H$ of $(\overline{\mathbb F}_2,+)$ such that every nonzero element of $\overline{\mathbb F}_2$ has a conjugate outside $H$. On the one hand, given $f$, we can define $H=\{x:f_0(x)=0\}$. On the other hand, given $H$, we can define $f$ by $f_k(x)=0\iff x^{2^{-k}}\in H$. It’s a bit tedious, but routine, to check that it satisfies all the required properties. | |
| Jun 10, 2022 at 10:24 | comment | added | მამუკა ჯიბლაძე | Is your bijection uniquely determined or you mean any bijection with these properties? | |
| S Jun 10, 2022 at 10:04 | review | First questions | |||
| Jun 10, 2022 at 10:10 | |||||
| S Jun 10, 2022 at 10:04 | history | asked | Ginger | CC BY-SA 4.0 |