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Let $p$ be a prime. Given a positive integer $n$, does there exist a $\beta$ in an extension of $F_p$ such that

1) If $F_p[\beta] = F_{p^N}$, then $N > n^n$; ( $\beta$ lies in a high extension)

2) The order of $\beta$ is at most $2^{poly(n)}$; ( $\beta$ has small order)

3) $\beta F_p + \beta^n F_p + \beta^{n^2} F_p + ... + \beta^{n^n} F_p \subseteq < \beta > $; (the subgroup generated by $\beta$ contains a linear space )

Thanks,

Qi

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1 Answer 1

Probably not, assuming $p$ is fixed and $n$ is large enough. Have a look at section 5 in my paper A49 in: http://www.integers-ejcnt.org/vol7.html (for some reason the journal doesn't allow direct links to papers although is free access).

In the notation there, let $R(x)=x^{n-1}+1$. Note that, as a consequence of your hypothesis 3), $\beta^{n-1}+1 \in <\beta>$, which implies that the order of $\beta^{n-1}+1$ is at most that of $\beta$. This will give an upper bound for $N$ in terms of $n$, using your hypothesis 2). I haven't done the calculation, so I don't know if this upper bound contradicts your hypothesis 1). Note that the bounds that I get are probably much weaker than the truth, see e.g., the conjecture of Poonen's discussed in the paper.

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