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Let $\mathbb{F}_q$ be a finite field of characteristic $p=2$. Let $c\in \mathbb{F}_q$. Does there always exist $a\in\mathbb{F}_q$ such that $b:=a^2+ca$ is not contained in any proper subfield of $\mathbb{F}_q$ (i.e., $\mathbb{F}_q=\mathbb{F}_p(b)$)? Thanks.

I am only interested in the case $p=2$ but feel free to generalize the statement.

Edit: Thanks for the answer. The question comes from a problem I am working on, related to stabilizers of some finite classical group action. It is not easy to explain the whole problem, but basically I want to choose $b\in\mathbb{F}_q$ not contained in any proper subfield, so that any $\mathbb{F}_q$-semilinear map in $\mathrm{\Gamma L}(V)$ fixing both a nonzero vector $v$ and $bv$ has to be $\mathbb{F}_q$-linear, i.e, in $\mathrm{GL}(V)$. And somehow I can only choose $b$ indirectly via $b=a^2+ca$ where $a\in\mathbb{F}_q$ can be any element.

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    $\begingroup$ What are your thoughts on the problem? Where does the question come from? What have you tried? $\endgroup$
    – Servaes
    Jul 13, 2019 at 0:12
  • $\begingroup$ Seconding Servaes' concern as well as the curiosity about the origin of this problem. It does not look like a homework problem to me, but who knows. $\endgroup$ Jul 13, 2019 at 8:08

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Yes, if $q>4$ such an element $a$ will always exist.

The reason is that the mapping $x\mapsto f(x)=x^2+cx$ is at most 2-to-1 (as a quadratic polynomial). In other words, the image $f(\Bbb{F}_q)$ has at least $q/2$ elements. If $q>4$ this is more than the union of all the proper subfields of $\Bbb{F}_q$, and the claim follows.

When $q=4$ we have the exception with $c=1$, in which case $f(x)=x^2+x$ is the relative trace, and $f(x)\in\Bbb{F}_2$ for all $x\in\Bbb{F}_4$. It is easy to see that $c=1$ is the only exception in $\Bbb{F}_4$.


The argument can be made more precise when $p=2$. For in that case, thanks to the Freshman's dream $(x+y)^2=x^2+y^2$, the mapping $f$ is a homomorphism of additive groups. Furthermore, the kernel consists of the elements $\{0,c\}$. So when $q=2^m$, the image of that homomorphism has size $2^{m-1}$ – well more than the union of the proper subfields when $m>2$.

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    $\begingroup$ If you increase the degree of $f$ a little, more exceptions will creep in, but the general argument should survive. The largest proper subfield has at most $\sqrt{q}$ elements, and this is the correct order for the union of the subfields. It follows that for a polynomial of a fixed degree the conclusion will hold whenever $q$ is large enough. $\endgroup$ Jul 13, 2019 at 8:07
  • $\begingroup$ If $q=p^n$, then a trivial upper bound for the size of the union of the proper subfields is $$\sum_{d\mid n}p^d\le\sum_{d=0}^{n/2}p^d=\frac{p^{\frac n2+1}-1}{p-1}\le p^{\frac n2+1}.$$ $\endgroup$ Jul 13, 2019 at 8:14

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