Members of finite field with equal Norm and Trace

Question

The setting of the problem I'm trying to solve is as follows:

Let $p$ be a prime number and $k \geq 1$. How many pairs $x, y \in \mathcal F_p^k$ satisfy $\text{Tr}(y) = N(x)$?

My first intuition tells me there are $\mathcal F_p^k$ such pairs, since for every $y \in \mathcal F_p^k$, with $\text{Tr}(y) = A$, there is an $x \in \mathcal F_p^k$ with $N(x) = B$ so that $\hat{x} = \frac{A}{B} \cdot x \in \mathcal F_p^k$, and $N(\hat{x}) = \frac{A}{B} \cdot N(x) = A = \text{Tr}(y)$

But I'm sure I'm missing something here.

Is there a better approach to this problem?

Answer 1

Here $\mathrm{Tr}:\Bbb{F}_{p^k}\to\Bbb{F}_p$ is known to be $\Bbb{F}_p$-linear and surjective. By rank-nullity its kernel has dimension $k-1$ as a vector space over the prime field, and for all $z\in\Bbb{F}_p$ we know that there are exactly $p^{k-1}$ elements $x\in\Bbb{F}_{p^k}$ such that $\mathrm{Tr}(x)=z$.

Because the norm map is also a function $N:\Bbb{F}_{p^k}\to \Bbb{F}_p$ it follows that for every $y\in\Bbb{F}_{p^k}$ there are exactly $p^{k-1}$ choices of $x\in\Bbb{F}_{p^k}$ such that the equation $$N(y)=\mathrm{Tr}(x)$$ is satisfied.

Consequently the number of pairs $(x,y)\in\Bbb{F}_{p^k}^2$ such that $N(y)=\mathrm{Tr}(x)$ is equal to $p^{k-1}\cdot p^k=p^{2k-1}$.

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Two concluding remarks:

• The same result holds for any function $g:\Bbb{F}_{p^k}\to\Bbb{F}_p$ in place of the norm map.
• In many other problems we also need to use the distribution of values of the norm map. It is well known that $y=0$ is the only element of $\Bbb{F}_{p^k}$ with $N(y)=0$. Also, when restricted to the multiplicative groups $N$ is a surjective homomorphism. Implying that it takes every possible non-zero value $z\in\Bbb{F}_p^*$ exactly $d=(p^k-1)/(p-1)=1+p+p^2+\cdots+p^{k-1}$ times. The simplest way of seeing this is to use the fact that $\Bbb{F}_{p^k}^*$ is a cyclic group, and $N(y)=y^d$ for all $y$.