Timeline for Algebraic dependency over $\mathbb{F}_{2}$
Current License: CC BY-SA 3.0
23 events
| when toggle format | what | by | license | comment | |
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| Feb 26, 2020 at 18:06 | history | edited | YCor |
edited tags
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| S Dec 4, 2014 at 22:33 | history | bounty ended | eig | ||
| S Dec 4, 2014 at 22:33 | history | notice removed | eig | ||
| S Dec 1, 2014 at 15:50 | history | suggested | eig |
add tag algebraic-geometry
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| Dec 1, 2014 at 15:38 | review | Suggested edits | |||
| S Dec 1, 2014 at 15:50 | |||||
| Dec 1, 2014 at 15:17 | comment | added | Peter Arndt | @joro See my last comment: These are non-zero polynomials, which happen to represent the zero function. But the question of algebraic dependence is whether one can produce the zero polynomial. | |
| Dec 1, 2014 at 12:57 | comment | added | joro | @PeterArndt Another example $x_1^2x_2^3 + x_1^3 x_2^3$. Observe that for n>0 $x^n=x$. | |
| Dec 1, 2014 at 12:39 | comment | added | Peter Arndt | @joro The question is whether the resulting polynomial is zero itself, not whether it represents the constant function with value zero. | |
| Dec 1, 2014 at 12:00 | comment | added | joro | @Peter such polynomial is $x(x+1)$, though it is actually zero. | |
| Dec 1, 2014 at 2:48 | answer | added | David E Speyer | timeline score: 23 | |
| Nov 30, 2014 at 14:25 | comment | added | Peter Arndt | @Turbo The question is whether there is another polynomial $0 \neq g \in \mathbb{F}_2[x_1, \ldots, x_n]$ such that $g(f_1, \ldots, f_n)=0$. | |
| Nov 29, 2014 at 16:59 | comment | added | Gorav Jindal | @VítTuček, Yes. | |
| Nov 28, 2014 at 14:34 | comment | added | Vít Tuček | So an example would be $f_i(x) = x_i$, right? | |
| S Nov 28, 2014 at 13:31 | history | bounty started | eig | ||
| S Nov 28, 2014 at 13:31 | history | notice added | eig | Draw attention | |
| Nov 26, 2014 at 10:57 | comment | added | eig | @joro I think he means that for all $n$ many $f_i$'s and $2^n$ many $a$'s, $f_i(a)=a_i$. | |
| Nov 26, 2014 at 9:34 | comment | added | joro | You have $n$ $f_i$ and $2^n$ $a$. Do you take only the first $n$ $a$'s? | |
| Nov 25, 2014 at 18:30 | comment | added | Gorav Jindal | @AlexDegtyarev, I edited the question to make it clearer. | |
| Nov 25, 2014 at 18:29 | history | edited | Gorav Jindal | CC BY-SA 3.0 |
Fixed some confusion
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| Nov 25, 2014 at 18:11 | comment | added | Andrés E. Caicedo | @AlexDegtyarev $a\in\mathbb F_2^n$ is a tuple $(a_1,\dots,a_n)$. | |
| Nov 25, 2014 at 18:07 | comment | added | Alex Degtyarev | What does the restriction $f_i(a)=a_i$ mean? What are $a_i$? | |
| Nov 25, 2014 at 18:01 | review | First posts | |||
| Nov 25, 2014 at 18:19 | |||||
| Nov 25, 2014 at 18:00 | history | asked | Gorav Jindal | CC BY-SA 3.0 |