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Feb 10, 2016 at 19:19 comment added Mikhail Goltvanitsa I note that there are a lot of subspaces $W\neq aS$ such that the previous decomposition of $K$ is valid for one element $\theta\in A$.
Feb 10, 2016 at 16:35 comment added Ehud Meir sorry, you are right, I have a mistake there.
Feb 10, 2016 at 16:28 comment added Mikhail Goltvanitsa 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 $K\setminus S$.
Feb 10, 2016 at 16:01 history edited Ehud Meir CC BY-SA 3.0
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Feb 10, 2016 at 15:44 comment added Mikhail Goltvanitsa 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 $W$?
Feb 10, 2016 at 15:04 history answered Ehud Meir CC BY-SA 3.0