Existence of a finite field element whose image under a quadratic map is a generator

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.

• What are your thoughts on the problem? Where does the question come from? What have you tried? Jul 13, 2019 at 0:12
• 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. Jul 13, 2019 at 8:08

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 $$m>2$$.
• 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. Jul 13, 2019 at 8:07
• If $q=p^n$, then a trivial upper bound for the size of the union of the proper subfields is $$\sum_{d\mid n}p^d\le\sum_{d=0}^{n/2}p^d=\frac{p^{\frac n2+1}-1}{p-1}\le p^{\frac n2+1}.$$ Jul 13, 2019 at 8:14