Consider an $\mathbb{F}\_q$-linear map $f:\mathbb{F}\_{q^n}\to \mathbb{F}\_{q^n}$ (so $f$ is a linearized polynomial). Suppose also that $f$ is not $\mathbb{F}\_{q^i}$-linear, where $i>1$.


My question is about the number of values that the function $f(x)/x$ can take, where $x$ is a nonzero element of $\mathbb{F}\_{q^n}$. The upper bound $\frac{q^n-1}{q-1}$ follows easily (since $\frac{f(\lambda x)}{\lambda x}=\frac{f(x)}{x}$ if $\lambda$ in $\mathbb{F}\_q^\*$). For the lower bound, I can use a result of S. Ball (The number of directions determined by a function over a finite field, J. Combin. Theory Ser. A, 104 (2003) 341--350), and find $q^{n-1}+1$. 


I have two questions:


(a) I believe there should be a way to avoid the heavy machinery used in the last paper to deduce the lower bound immediately. Does anyone see how?


(b) My actual question is what happens if I restrict the map $f$ to values $x$ in some $\mathbb{F}\_q$-subspace $V$ of dimension $k$ of $\mathbb{F}\_{q^n}$. I would guess that the lower and upper bounds are then $q^{k-1}+1$ and $\frac{q^k-1}{q-1}$ but I do not find a reference for this. Can someone point me in the right direction?


EDIT: As pointed out below, the lower bound can be one if there are no extra conditions. However, in practice, I assume that there is at least one direction $f(y)/y$ that is only determined $q-1$ times (namely, by $y\in V$ and its $\mathbb{F}\_q^\*$-multiples). 



The number of values of $f(x)/x$ when $f$ is a linearized polynomial
This is an answer to the second part of your question.


The upper bound seems obvious, for the same reason as in the original case. The lower bound, however, is 1.


There is some linear map that sends the chosen subspace to 0, but is not itself 0. This then gives slope 0 on that entire subspace, giving an answer of 1.


I did find something interesting when looking for an example, which the above answer does not explain. When trying for $\mathbb{F}\_8$ over $\mathbb{F}\_2$ with a subspace generated by $1, \alpha$ with $\alpha \in \mathbb{F}\_8$, the polynomial $x^4 + (\alpha^2 + \alpha + 1) x^2 + (\alpha^2 + \alpha)x$ worked no matter which $\alpha$ was chosen. This may be an artifact of a small example, but I'd be interested to see whether the coefficients of the polynomial could be chosen in terms of the subspace in a polynomial manner (or if that even makes sense).


Edited for more I figured out about the second half of my answer:


This isn't hard to prove by induction, if you take "in terms of the subspace" to mean "in terms of a basis of the sub space". The base case is the 0-dimensional point, which corresponds to the polynomial $x$. Then if we are adding $\alpha$ to a subspace with polynomial $P(X)$, the new polynomial will be $P(X)^q - P(\alpha)^{q-1}P(X)$. So for example, for the above: $P\_1 = X^2+X$, so $P\_{1, \alpha} = X^4+X^2+(\alpha^2+\alpha)(X^2+X)$, as above. This always gives a monic polynomial, and in fact is the unique monic linearized polynomial of minimal degree. More generally, define the Lang map $L$ by $L(x) = x^{q-1}$. Then the polynomial for the sub space generated by $\{\alpha\_i\}$ is $(\prod\_i (Frob - L(\alpha\_i))) X$.


