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$n\ge 2$ is an integer $$tr : \mathbb F_{2^n} \to \mathbb F_{2} \\ tr(x)=x+x^2+x^{2^2}+\cdots x^{2^{n-1}}$$ is the absolute trace map

For fixed constants $\alpha, \theta \in \mathbb F_{2^n} $ define the following map

$$\phi : \mathbb F_{2^n} \to\mathbb F_{2^n}\\ \phi(y)= y+\alpha tr(\theta y)$$

My aim is to show that $\phi(y)$ is a permutation i.e. set bijection if and only if $tr(\alpha \theta)=0$

1.) $tr(\alpha \theta)=0$ iff the map is a permutation of $\mathbb F_{2^n}$

2.) Also $\phi \circ \phi = Identity$ iff the map is a permutation of $\mathbb F_{2^n}$ (Right to left implication is trivial but having bad time about the other implication)

For 1.) I tried to manupule the equation putting the fixed values $\alpha, \theta \in \mathbb F_{2^n} $

and gained $$\phi(\alpha)=\alpha+\alpha tr(\alpha \theta)$$

also tried a lot of variables with additional inverses. The trick I think to think that trace function only has two results $0, 1$ but couldnot finish the proof.

The same manipulations are also tried for 2.) but here for the inverse direction when writing for an arbitrary $x$ $$\phi \phi(x)=x+\alpha tr(x \theta)+\alpha tr(x \theta)+\alpha tr\left( \alpha tr(x \theta)\right)$$

since trace function only has two results $0, 1$ I think I can conclude the last line is equal only to x which shows the statement is true.

Any help, proof, hint would be highly appreciated.

User not found
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1 Answers1

2

  1. $y\ne 0\in \ker \phi$ iff $y = \alpha$ and $tr(\theta \alpha)=1$.

  2. If $tr(\theta \alpha)=0$ then $\phi(\phi(y))= y+\alpha\ tr(y\theta)+\alpha tr((y+\alpha\ tr(y\theta))\theta) = y+\alpha tr(\alpha tr(y\theta) \theta)= y$

reuns
  • 79,880
  • thank you, 2.) is okay using 1.) however I cannot see the iff statement in 1.) can you elaborate as much as you can since I am very confused about the topic :) – User not found May 12 '22 at 21:33
  • $tr(y\theta)$ is either $0$ or $1$, if it is $0$ then $\phi(y)=y$ if it is $1$ then $\phi(y)=y+\alpha$ @Daianne'deSouza – reuns May 12 '22 at 21:38