Timeline for The number of values of $f(x)/x$ when $f$ is a linearized polynomial
Current License: CC BY-SA 3.0
13 events
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May 10, 2017 at 2:26 | history | edited | user440858 | CC BY-SA 3.0 |
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May 10, 2017 at 2:25 | comment | added | user440858 | Yes, I'll modify the statement a little. | |
May 9, 2017 at 22:46 | comment | added | user44191 | This direction that has a unique output would have to be in the sub space, right? Otherwise, my answer below still does work. | |
May 7, 2017 at 22:48 | history | edited | user440858 | CC BY-SA 3.0 |
added 257 characters in body
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May 7, 2017 at 22:15 | vote | accept | user440858 | ||
May 4, 2017 at 9:20 | answer | added | user44191 | timeline score: 1 | |
Apr 27, 2017 at 8:34 | comment | added | user440858 | Yes, I know that $f$ is a linearized polynomial. I do not know how this helps to answer my question, unfortunately. | |
Apr 27, 2017 at 6:01 | comment | added | user44191 | This may help: a $\mathbb{F}_q$-linear function $f: \mathbb{F}_{q^n} \rightarrow \mathbb{F}_{q^n}$ has a unique expression of the form $\sum_{i = 0}^{n - 1} a_i x^{q^i}, a_i \in \mathbb{F}_{q^n}$. | |
Apr 27, 2017 at 4:44 | comment | added | user44191 | Ah, I misunderstood; I thought you were talking about restricting the output, not the input. My apologies. | |
Apr 27, 2017 at 4:42 | comment | added | user440858 | I do not really see why that is impossible. If $x$ doesn't run through the whole field, you have less points (of the form (x,f(x)), so this smaller set of points will determine less directions. No? | |
Apr 27, 2017 at 3:43 | comment | added | user44191 | Your second question implies that the lower bound becomes lower when adding an extra constraint; shouldn't that be impossible? | |
Apr 27, 2017 at 0:54 | review | First posts | |||
Apr 27, 2017 at 1:00 | |||||
Apr 27, 2017 at 0:54 | history | asked | user440858 | CC BY-SA 3.0 |