1

Proving that $\{x\in GF(2^k)^{*}:tr^k_1(1/x)=1\}=\{u+u^{2^k}:u\in U\setminus\{1\}\}$,

where $tr^k_1(x)=x+x^{2^1}+x^{2^{2}}+ \dots+x^{2^{n-1}} $ is a trace function in finite fields and $U=\{u\in GF(2^{2k}):u^{2^k+1}=1\}$.

Given $u\in GF(2^{2k})\setminus GF(2^k)$, we can get the minimal polynomial of $u$ over $GF(2^k)$ is $f(x)=(x+u)(x+u^{2^k})=x^2+(u+u^{2^k})x+1$ if $u\in U$, I'm not sure there is a method to get $tr^k_1(1/x)$ with above equation.Or I did a wrong attempt?

garryABSv5
  • 23

1 Answers1

1

The following argument works. To save keystrokes I introduce the following notation:

  • $F=GF(2^k)=\Bbb{F}_{2^k}$.
  • $E=GF(2^{2k})=\Bbb{F}_{2^{2k}}$, the quadratic extension of $F$.
  • If $z\in E$, I will denote its only $F$-conjugate $\overline{z}=z^{2^k}$. The mapping $\sigma:z\mapsto\overline{z}$ is the only non-identity automorphism of $E$ that maps elements of $F$ to themselves. It has order two, in other words $\overline{\overline{z}}=z$ for all $z\in E$.

I need the following basic facts.

Fact 1. $U$ consists of elements of the form $u(z):=z/\overline{z}$ with $z$ ranging over $E\setminus F$.

Proof. The equation $u^{2^k+1}=1$ is equivalent to $u^{2^k}=1/u$. In other words, $u\in U$ if and only if $\overline{u}=1/u$. As $\sigma$ has order two, we immediately see that $\sigma(u(z))=\sigma(z)/\sigma(\overline{z})=\overline{z}/z=1/u(z)$. Therefore $u(z)\in U$. Clearly $u(z)=1$ if and only if $\overline{z}=z$ if and only if $z\in F$.

For the other inclusion let $g$ be a generator of the multiplicative group $E^*$, i.e. and element of order $2^{2k}-1=(2^k-1)(2^k+1)$. Then $$u(g)=g/g^{2^k}=(1/g)^{2^k-1}$$ will be of order $2^k+1$. In other words, $u(g)$ generaters the multiplicative subgroup $U\cup\{1\}$ of $E^*$. As $u(g^j)=u(g)^j$ it follows that all the elements of $U$ are of the form $u(g^j)$. QED.

Fact 2. If $a\in F$ then the roots of the polynomial $f(T)=T^2+aT+b\in F[T], a\neq0,$ are elements of $F$ if $tr(b/a^2)=0$ and elements of $E\setminus F$ if $tr(b/a^2)=1$.

Proof. This is well known. Ask, if you need help here. QED.

All is set for the proof of the main claim. Assume first that $u\in U$. By the first fact we know that $u=z/\overline{z}$ for some $z\in E\setminus F$. Here $$ x=u+u^{2^k}=u+\overline{u}=\frac{z}{\overline{z}}+\frac{\overline{z}}{z} =\frac{(z+\overline z)^2}{z\overline{z}}, $$ where in the last step I used Freshman's dream. As in your minimal polynomial calculation, we see that $z$ and $\overline{z}$ are zeros of the polynomial (with coefficients from $F$) $$ (T+z)(T+\overline{z})=T^2+(z+\overline{z})T+z\overline{z}=0. $$ Here $z,\overline{z}\in E\setminus F$, so by Fact 2. we can deduce that $$ 1=tr(\frac{z\overline{z}}{(z+\overline{z})^2})=tr(\frac1x) $$ proving one inclusion.

To get the other inclusion let's start with an element $x\in F^*$ such that $tr(1/x)=1$. By Fact 2. it follows that the zeros of the polynomial $$ T^2+T+\frac1x=0\qquad(*) $$ are elements of $E\setminus F$. Let $z$ be one of those zeros, the other is then $\overline{z}$. By Vieta relations we know that $$ z+\overline{z}=1,\qquad\text{and}\qquad z\overline{z}=\frac1x. $$ Reverting the above calculation, it follows that with $u=u(z)=z/\overline{z}\in U$ we have $$ u+u^{-1}=\frac{(z+\overline{z})^2}{z\overline{z}}=\frac{1}{1/x}=x $$ giving us the converse inclusion. QED.

There may be something simpler out there, but the above argument flows IMHO quite naturally with basic tricks of the trade in place. An alternative I came up with first was to use another parametrization for the set $U$ mentioned in this answer of mine. There $U$ was denoted by $S$, and the parametrization would have been $$U=\left\{\frac{x+z}{x+\overline{z}}\mid x\in F\right\},$$ where $z$ is a fixed element from $E\setminus F$. The details of the resulting calculation were in my opinion a bit less natural with that parametrization.

Jyrki Lahtonen
  • 140,891