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I want all the coefficients of $p(x)f(x) = \sum_{i=0}^{n-1} b_i x^i$ to be the $\mathcal{F}_q$-linear combinations of coefficients of $f(x) = \sum_{i=0}^{m-1} a_i x^i$, $m < n$. In another words, each $b_i = \sum_{j=0}^{m-1} c_{i,j} a_j$, where each $c_{i,j} \in \mathcal{F}_q$.

Mostly, I am interesting how to find the non-trivial $p(x)$ having the coefficients from $\mathcal{F}_{q^n}$ rather from $\mathcal{F}_q$.

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  • $\begingroup$ I don't understand the question. For any (fixed) $p(x)\in\Bbb{F}_{q^n}[x]$ the mapping $f(x)\mapsto p(x) f(x)$ is linear over $\Bbb{F}_{q^n}$, and consequently also over the subfield $\Bbb{F}_q$. $\endgroup$ Dec 8, 2015 at 11:19
  • $\begingroup$ @JyrkiLahtonen: I mean that all the coefficients of $p(x)f(x) = \sum_i b_i \cdot x^i$ are the $\mathcal{F}_q$-linear combinations of coefficients of $f(x) = \sum_i a_i \cdot x^i$. In another words, each $b_i = \sum_j c_j \cdot a_j$, where $c_j \in \mathcal{F}_q$. $\endgroup$ Dec 8, 2015 at 11:32
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    $\begingroup$ Ok. But if $p(x)=\sum_k p_k x^k$, then $b_i=\sum_j p_{i-j}\cdot a_j$, so I don't see that happening unless $p_i\in\Bbb{F}_q$ for all $i$. $\endgroup$ Dec 8, 2015 at 11:34
  • $\begingroup$ IOW $c_{i,j}=p_{i-j}$ (or $0$ if $i-j$ is out of range). And those coefficients are uniquely determined by the linear transformation. $\endgroup$ Dec 8, 2015 at 11:37
  • $\begingroup$ @JyrkiLahtonen So all the $p(x)$ that I want are only $\mathcal{F}_q[x]$? How it can be proven that there are no $p(x) \in \mathcal{F}_{q^r}[x] \setminus \mathcal{F}_q[x]$ satisfying my requirements? $\endgroup$ Dec 9, 2015 at 8:54

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