Timeline for Write the algebra closure of $F_p$ as union of finite fields [closed]
Current License: CC BY-SA 4.0
15 events
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| Dec 9, 2018 at 11:13 | comment | added | YCor | By the way, the "I believe that" addendum was added by the OP after it was suggested as a comment (by @reuns) on the MathSE site. This is not very fair conduct. | |
| Dec 8, 2018 at 15:14 | history | edited | Wembley Inter |
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| Dec 8, 2018 at 15:08 | comment | added | Wembley Inter | @MartinSleziak Thanks for your advice. As a new contributor, I'll learn from this. | |
| Dec 8, 2018 at 14:38 | history | closed |
YCor Andreas Thom Andreas Blass Andrés E. Caicedo Felipe Voloch |
Not suitable for this site | |
| Dec 8, 2018 at 7:06 | comment | added | Martin Sleziak | The same question on Mathematics: Write the algebra closure of $F_p$ as union of finite fields. This answer has some reasonable advice about cross-posting. Another things to keep in mind is that this site has different tags. For example, the tag (abstract-algebra) is deprecated and it is recommended to use at least one top-level tag. | |
| Dec 8, 2018 at 7:01 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
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| S Dec 8, 2018 at 1:18 | history | suggested | Amir Sagiv | CC BY-SA 4.0 |
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| Dec 7, 2018 at 23:38 | comment | added | LSpice | Certainly your condition is sufficient. For necessity, note that a primitive element of $\mathrm{GF}(q^b)$ has order $q^b - 1$, hence lies in $\mathrm{GF}(q^a)$, whose multiplicative group has order $q^a - 1$, if and only if $q^b - 1 \mid q^a - 1$, which, since $\gcd(q^b - 1, q^a - 1) = q^{\gcd(a, b)} - 1$, is true if and only if $b \mid a$. | |
| Dec 7, 2018 at 22:15 | comment | added | Lubin | Right you are. Perhaps the easiest way to remedy the author’s failing is to replace “any strictly increasing infinite sequence” with “the sequence $a_n=n!$”. | |
| Dec 7, 2018 at 22:01 | review | Suggested edits | |||
| S Dec 8, 2018 at 1:18 | |||||
| Dec 7, 2018 at 21:40 | review | Close votes | |||
| Dec 8, 2018 at 14:40 | |||||
| Dec 7, 2018 at 21:21 | comment | added | YCor | It's not a field, for instance is $a_n$ is the $n$-th prime. If $a_n=2^n$ it's a field, but it's not algebraically closed... | |
| Dec 7, 2018 at 21:21 | answer | added | Kevin Buzzard | timeline score: 2 | |
| Dec 7, 2018 at 21:15 | review | First posts | |||
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| Dec 7, 2018 at 21:12 | history | asked | Wembley Inter | CC BY-SA 4.0 |