When does $\operatorname{trace} (a)=\operatorname{trace}(a^{-1})$?

Question

We work on fields with even characteristic and know that $\operatorname{trace}(a)=\operatorname{trace}(a^2)$.but when does trace of one member of field equals to its inverse?

Answer 1

I don't think there is a very useful necessary and sufficient criterion for an element and its inverse to have the same trace. For larger fields of characteristic two and a random element $a$ it is close to a 50-50 chance that this happens.

Elaborating on that. Let $\psi:\Bbb{F}_q\to\Bbb{C}^*$ be the additive character $\psi(x)=(-1)^{tr(x)}$. For $a\neq0$ we see that $tr(a)=tr(a^{-1})$ if and only if $\psi(a+1/a)=1$ and $tr(a)\neq tr(a^{-1})$ iff $\psi(a+1/a)=-1$. The frequencies are thus determined by the character sum $$ S=\sum_{x\in\Bbb{F}_q^*}\psi(x+\frac1x). $$ This is a special case, $S=S(1)$, of a so called Kloosterman sum $$ S(a)=\sum_{x\in\Bbb{F}_q^*}\psi(ax+\frac1x). $$ It is known that $|S(a)|\le2\sqrt q$ for all $a\neq0$. Because usually $2\sqrt q$ is much smaller than $q$ this means that $\psi(ax+1/x)$ takes values $+1$ and $-1$ close to equally often.

---

Using the theory of $L$-functions and Hasse-Davenport relations we can be more precise with the sum $S=S(1)$. The sum $S$ is "defined over the prime field $\Bbb{F}_2$" because the coefficient $1\in\Bbb{F}_2$. Therefore the sum $S(1)$ makes sense over any extension field $\Bbb{F}_{2^n}$. So we can consider the sequence of Kloosterman sums $$ S(1,n)=\sum_{x\in\Bbb{F}_{2^n}^*}\psi(x+\frac1x). $$ Hasse-Davenport relations imply that there exist complex numbers $\omega_1$ and $\omega_2=\overline{\omega_1}$ such that $\omega_1\omega_2=2$ and that for all $n$ we have the identity $$ S(1,n)=-\omega_1^n-\omega_2^n. $$ But, we can easily calculate that $S(1,1)=\psi(1+1)=1$. Therefore $\omega_{1,2}$ have real parts equal to $-1/2$, so we can solve that $ \omega_{1,2}=\dfrac12(-1\pm i\sqrt7). $ Hence $$ S(1,n)=-\frac{(-1+i\sqrt7)^2+(-1-i\sqrt7)^2}{2^n}.\qquad(*) $$ Examples.

• When $n=8$, i.e. the case of $\Bbb{F}_{256}$, formula $(*)$ says that $S(1,8)=31$. This implies that out of the $255$ non-zero elements $x\in\Bbb{F}_{256}$ the "equal traces" case occurs exactly $143$ times whereas the traces differ in $112$ cases ($143+112=255, 143-112=31=S(1)$).
• When $n=16$, our formula gives $S(1,16)=-449$. This means that the two traces are equal $32543$ times out of $65535$, and they differ for the remaining $32992$ choice of $x\in\Bbb{F}_{65536}$.