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Feb 12, 2016 at 13:19 comment added Lev Borisov Yes, you are right.
Feb 12, 2016 at 13:03 comment added Peter Mueller @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 $S$.
Feb 12, 2016 at 7:22 comment added Mikhail Goltvanitsa maybe you can give an advice where I can read about values of such rational functions over finite fields?
Feb 11, 2016 at 6:48 comment added Mikhail Goltvanitsa Interesting idea. But I have not full proof too)
Feb 11, 2016 at 3:10 history answered Lev Borisov CC BY-SA 3.0