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Let $L/K$ be finite fields with elements $a,b\in L$ , with $\operatorname{tr}$ being the trace map and $f:L\to L$ given by $f(x)=x+a\operatorname{tr}(bx)$. Then,is it true that, $\forall x\in L$, if and only if $\operatorname{tr}(ab)=0$, then $f(x)=x+a\operatorname{tr}(bx)$ is a permutation?

I am having hard time thinking this problem. Specifically, since $f$ is linear, I only get that if $f$ is one-one, it is automatically a permutation. Now, for this, I get that if $f(x_1-x_2)=0=x_1-x_2-a(\operatorname{tr}(bx_1)+\operatorname{tr}(bx_2))$. Using the linearity of the trace operator, I cannot proceed any further. Any hints? Thanks beforehand.

vidyarthi
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  • An alternative proof specific to the case $tr(ab)=0$ is to observe that in this case the $p$-fold iteration of $f$ is the identity mapping. Here $p$ is the characteristic of the fields. Anyway, it is kinda cute that in that case we can further show that the permutation only has cycles of lengths $1$ and $p$. – Jyrki Lahtonen May 12 '22 at 22:06

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This was a cute exercise, and one that I don't recall having seen! I assume that $\operatorname{tr}$ stands for the relative trace sometimes denoted $\operatorname{tr}_{L/K}$. Context will make such abbreviations non-problematic.

You already made the key observations that

  • $f$ is $K$-linear, and, as $L$ is finite,
  • $f$ is a permutation if and only if it is injective.

In addition to those we need the fact from basic linear algebra that a linear transformation is injective if and only if its kernel is trivial.

You are right in that the element $z:=\operatorname{tr}(ab)$ plays a key role in deciding injectivity of $f$. But, in fact, your hunch is only half correct. If $z=0$, the function $f$ is, indeed, a permutation. But I arrived at the general observation:

The transformation $f$ is a permutation if and only if $z\neq-1$.

A way to see this is the following. Assume that $x\in L$ is in the kernel. From $f(x)=0$ we arrive at $$ x=-a\operatorname{tr}(bx).\tag{1} $$ Observe that here $y:=\operatorname{tr}(bx)\in K$. So $x=-ay$ and $$ \operatorname{tr}(bx)=\operatorname{tr}(-aby)=-y\operatorname{tr}(ab)=-yz\tag{2} $$ by $K$-linearity of the trace. The equations $(1)$ and $(2)$ thus imply that $$ 0=f(x)=f(-ay)=-ay+a\operatorname{tr}(bx)=-ay-ayz=-ay(1+z)=-x(1+z). $$ So unless $z=-1$ we see that $x=0$ is the only element of the kernel, proving the claim. Of course, if $z=-1$, the above calculation shows that every element of the form $x=-ay$, $y\in K$ is in the kernel of $f$.

A particular case is that of $K=\Bbb{F}_2$ when either $z=0$ or $z=1=-1$. In that setting your original hunch is correct and $f$ is a permutation if and only if $\operatorname{tr}(ab)=0$.

Jyrki Lahtonen
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