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Hellow. I'm sure that the following is truth, but I can't prove it.

Let $R<S<K, R=\mathrm{GF}(q),\ S= \mathrm{GF}(q^n), \ K= \mathrm{GF}(q^{mn})$ be a chain of finite fields and $A = \{\theta\in K: \mathrm{ord}\theta = q^{mn}-1\}$ be the set of primitive elements. I want to prove the following statement. If $_RW$ be a subspace of $_RK$, $\mathrm{dim}_RW = n$ such that for every $\theta \in A$ $$ W\oplus W\theta\oplus\ldots\oplus W\theta^{m-1} = K, $$ then $W = aS$ for some $a\in K$.

I will be grateful for the help.

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  • $\begingroup$ Firstly there was "...then $W=S$." $\endgroup$
    – Mikhail Goltvanitsa
    Commented Feb 10, 2016 at 16:00
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    $\begingroup$ Crossposted at Math.SE. I think that it is advisable to always crosslink, if you crosspost the same question on two sites. Some users at Math.SE want to forbid crossposting altogether. I'm not at all sure about that, but crosslinking does reduce the risk of duplicated efforts. $\endgroup$
    – Jyrki Lahtonen
    Commented Apr 3, 2016 at 7:09

2 Answers 2

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Clearly, the statement is invariant under multiplication by $a\in K$, so we may assume that $W\ni 1$. This implies that $W\supseteq R$, and we want to show that $W=S$.

Suppose that $t\in W$. I claim that $t$ can not be written as a value of a non constant rational function $f(x)/g(x)$ with coefficients in $R$ and degrees of numerator and denominator $<m$ at a primitive root $x=\theta$.

Indeed, if we have $tg(\theta)-f(\theta)=0$, then this translates into linear dependence on $W,W\theta,\ldots,W\theta^{m-1}$, since $1,t\in W$.

It remains to argue that the complement in $K$ to the set of non constant "degree $\leq m-1$" rational functions in primitive roots is contained in (and is equal to) $S$. This sounds plausible, but I don't have a proof.

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  • $\begingroup$ Interesting idea. But I have not full proof too) $\endgroup$
    – Mikhail Goltvanitsa
    Commented Feb 11, 2016 at 6:48
  • $\begingroup$ maybe you can give an advice where I can read about values of such rational functions over finite fields? $\endgroup$
    – Mikhail Goltvanitsa
    Commented Feb 12, 2016 at 7:22
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    $\begingroup$ @Lev Borisov: This approach cannot work: Take for instance $m=2$, $n=3$ and let $T$ be the quadratic extension of $R$. Then the elements of $T$ are not in the value set you described, but $T\setminus R$ is not a subset of $S$. $\endgroup$
    – Peter Mueller
    Commented Feb 12, 2016 at 13:03
  • $\begingroup$ Yes, you are right. $\endgroup$
    – Lev Borisov
    Commented Feb 12, 2016 at 13:19
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The statement obviously holds for $W=S$. Take now any nonzero element $a\in K$ and take $W=S\cdot a$. Then since everything is commutative, and multiplication by $a$ is invertible, the statement is still true for $W$ even though $W\neq S$ if $a\notin S$.

If the question is: are all subspaces $W$ are of this form, I have the following partial answer:

If for example $m$ is a prime and $n$ is a power of $m$ we can argue in the following way: there cannot be two elements $w_1$ and $w_2$ such that $w_1 = \theta w_2$ where $\theta$ is primitive, because that will contradict the assumption. This means that for a given $0\neq w\in W$, all the elements in $Ww^{-1}$ will not be primitive. But the fact that they are not primitive already means that they belong to $S$, by the assumption on $m$ and $n$. I am not sure how further it is possible to use this argument.

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  • $\begingroup$ Thank you very much. I know about this simple case, but forget to write about it. What do you think about other types of such subspaces $W$? $\endgroup$
    – Mikhail Goltvanitsa
    Commented Feb 10, 2016 at 15:44
  • $\begingroup$ If I get you right, in your assumptions on $m,n$ elements from $K\setminus S$ are primitive. But I think that it is not true. Let us see the following case. $q=2, m = 2, n = 2$. Then $R = \mathrm{GF}(2)<S = \mathrm{GF}(4)<K = \mathrm{GF}(16)$. There are $3$ non-primitive elements from $K\setminus S$. $\endgroup$
    – Mikhail Goltvanitsa
    Commented Feb 10, 2016 at 16:28
  • $\begingroup$ sorry, you are right, I have a mistake there. $\endgroup$
    – Ehud Meir
    Commented Feb 10, 2016 at 16:35
  • $\begingroup$ I note that there are a lot of subspaces $W\neq aS$ such that the previous decomposition of $K$ is valid for one element $\theta\in A$. $\endgroup$
    – Mikhail Goltvanitsa
    Commented Feb 10, 2016 at 19:19

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