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https://arxiv.org/pdf/1211.6044.pdf Theorem 1.14 (2): The monomial $x^n$ is a permutation polynomial of $F_q$ if and only if $gcd(n, q − 1) = 1$.

Let's pick $n=2$ and $q=6$ so $gcd(2,5) = 1$ in this case $x^2$ should be a permutation polynomial over $F_6$. Let's put them into a table.

0 1 2 3 4 5
$x^2$ 0 1 4 3 4 1

As we see the function is not even one-to-one, it is not a permutation polynomial, what am I missing here?

[Update] Thanks to @GerryMyerson, my example definition of field is wrong here. Modulo over 6 have zero divisor. If I try it with a $F_p$ over modulo p and p is prime then it works.

$F_7$ which is modulo over 7 and $x^5$ should be permutation polynomial because of $gcd(5,6)=1$.

0 1 2 3 4 5 6
$x^5$ 0 1 4 5 2 3 6
voyager
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  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Mar 02 '23 at 20:56
  • I think things are clear. I am asking why the theorem doesn't work in this scenario and asking for help. – voyager Mar 02 '23 at 21:13
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    The notation $F_q$ stands for the field of $q$ elements. There is no $F_6$. – Gerry Myerson Mar 02 '23 at 23:20
  • That's ok, F is field such that it has {0,1,2,3,4,5} over modulo 6. – voyager Mar 03 '23 at 08:01
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    @voyager A field of finite order $q$ exists if and only if $q$ is a prime power $p^k$ – user773458 Mar 03 '23 at 08:05
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    There is no such thing as a field that has that or anything else modulo six. Modulo six, there are zero-divisors; $2\times3=0$. Fields can't have zero-divisors. Every nonzero element in a field must have a multiplicative inverse. – Gerry Myerson Mar 03 '23 at 11:46
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    @GerryMyerson you are right if I pick a $F_p$ over modulo p and p is prime it works. Thank you. – voyager Mar 03 '23 at 18:25
  • I voted to close this as a duplicate of an old question as there was a misunderstaning about the meaning of a (finite) field. As I have a dupehammer privilege on this tag, my vote took immediate effect. Now the question has been edited, and I want to check back on you. Is there still a question that needs to be answered, or are you content with the explanation? The Theorem follows immediately from the cyclicity of the multiplicative group. I'm also fairly sure that we have covered the result on our site somewhere many times. I have used it in a handful of answers I think :-) – Jyrki Lahtonen Mar 04 '23 at 06:49
  • Even the answer for the questions are same which is no $F_6$, I wouldn't close this as duplicate. Because the way question asked is still different. I would keep as related or linked instead. – voyager Mar 07 '23 at 01:50

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