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Can we rewrite $Tr(ab^2)$ as $Tr(f(a)b)$ where $Tr:F_{2^{kn}}\rightarrow F_{2^{k}} $ is trace map, $k \neq 1$, $f$ is a function just depends to $a$.

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  • $\begingroup$ Assume there are more terms inside.. Then also assume we get $tr(g(a)b)$ for some $g$. Now we can more easily talk about distrubition.. This answer can be a good trick for such ideas. // For base field, we can use such a trick for all 2-powers.. $\endgroup$
    – vudu vucu
    Mar 22, 2015 at 11:34
  • $\begingroup$ In this case, i dont think it will help, i just need to find $f$. I am trying to solve my problem in different sense but i really want to find this answer, because it will be really helpful. $\endgroup$
    – vudu vucu
    Mar 22, 2015 at 12:14

1 Answer 1

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No.

The reason is that the function $B_1:(a,b)\mapsto tr(f(a)b)$ from $\Bbb{F}_{2^{kn}}\times\Bbb{F}_{2^{kn}}$ to $\Bbb{F}_{2^k}$ is $\Bbb{F}_{2^k}$-linear in the variable $b$. Irrespective of what the function $f$ might be.

On the other hand the function $B_2:(a,b)\mapsto tr(ab^2)$, while additive, is not $\Bbb{F}_{2^k}$-linear in the variable $b$.

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More concretely, for fixed $a,b$ such that $tr(ab^2)\neq0$ we run into the following problem. If $e\in\Bbb{F}_{2^{k}}$ is not an element of the prime field we get $$ B_1(a,eb)= eB_1(a,b) $$ while $$ B_2(a,eb)=e^2B_2(a,b). $$ So if $B_1(a,b)=c=B_2(a,b)$ for some choice of $f$, then $B_1(a,eb)=ec\neq e^2c=B_2(a,eb).$

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  • $\begingroup$ Thanks for answering. It is really right. $\endgroup$
    – vudu vucu
    Mar 26, 2015 at 14:51

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