Proof of identity on sum of powers of primitive root.

Question

Let $q = p^e$ for some prime $p$. Consider the trace function $\mathrm{Tr}_{\mathbb{F}_q/\mathbb{F}_p}:\mathbb{F}_q\to \mathbb{F}_p$ defined by $\mathrm{Tr}_{\mathbb{F}_q/\mathbb{F}_p}(x) = \sum_{i=0}^{e-1}x^{p^i}$. Let $C \subseteq \mathbb{F}_q^n$ be a linear subspace and let $\zeta = e^{2\pi i/p}$. What I would like to show is $$\sum_{c \in C}\zeta^{\mathrm{Tr}_{\mathbb{F}_q/\mathbb{F}_p}(\langle c, z\rangle)} = \begin{cases}|C|&\text{if }z\in C^\perp\\0 & \text{otherwise}\end{cases}.$$

What I know is that the trace function is surjective and each element of $\mathbb{F}_p$ is mapped to by $p^{e-1}$ elements of $\mathbb{F}_q$. Additionally, I see why the first case is true because if $z \in C^\perp$ we get $\langle c, z\rangle = 0$ for all $c$ and the result follows.

I also know that $\sum_{i=0}^{p-1}\zeta^i = \frac{\zeta^p - 1}{\zeta - 1} = 0$ so my idea was to try and show that each power of $\zeta$ is obtained the same number of times $t$. In other words, for each $i = 0, 1, \ldots, p-1$, we have $|c \in C : \mathrm{Tr}_{\mathbb{F}_q/\mathbb{F}_p}(\langle c, z\rangle) = i| = t$.

In the case of $e = 1$ and $p = 2$, we have $$\sum_{c \in C}\zeta^{\mathrm{Tr}_{\mathbb{F}_q/\mathbb{F}_p}(\langle c, z\rangle)} = \sum_{c\in C}(-1)^{c\cdot z}$$ and since $z \not \in C^\perp$, there is some $c_0 \in C$ such that $c_0 \cdot z = 1$. Then $$-\sum_{c\in C}(-1)^{c\cdot z} = (-1)^{c_0\cdot z}\sum_{c\in C}(-1)^{c\cdot z} = \sum_{c \in C}(-1)^{(c+c_0)\cdot z} = \sum_{c'\in C}(-1)^{c'\cdot z}$$ and the result follows.

Any help would be greatly appreciated!

Answer 1

We can generalize the idea you started with. Let $S$ be the following multiset:

$$ \{\langle c,z\rangle : c \in C\} $$

Since $z\notin C$, there exists $c_0\in C$ such that $\langle c_0,z\rangle=1$. Now, consider an arbitrary element $q\in \mathbb{F}_q$. $C$ is a vector space over $\mathbb{F}_q$, so $c\to c+qc_0$ is a bijection on $C$, and thus we have

$$ S=\{\langle c+qc_0,z\rangle : c\in C\}=\{\langle c,z\rangle+q : c\in C\}=\{s+q : s \in S\} $$

In other words, for any $q\in\mathbb{F}_q$, if we add $q$ to every element in $S$, we get the same multiset that we started with.

Now, we can show that any two elements in the multiset $S$ appear with equal frequencies: Consider two arbitrary $q_1,q_2\in\mathbb{F}_q$. Then, let $c$ be the frequency of $q_1$ in $S$. Then, since $x\to x+(q_2-q_1)$ is an injective function on $\mathbb{F}_q$, $q_1+(q_2-q_1)=q_2$ has frequency $c_1$ in the multiset $\{s+(q_2-q_1) : s\in S\}$. However, by the above finding, $\{s+(q_2-q_1) : s\in S\}=S$, and thus the frequency of $q_2$ in $S$ is also $c$, which shows that the frequency of $q_2$ in $S$ equals the frequency of $q_1$ in $S$.

Next, let $T$ be the following multiset:

$$ \{\mathrm{Tr}_{\mathbb{F}_q/\mathbb{F}_p}(\langle c,z\rangle) : c \in C\} $$

Since every element in $\mathbb{F}_q$ has equal frequency in $\{\langle c,z\rangle : c \in C\}$, and we know that, for every $y\in\mathbb{F}_p$, there are exactly $p^{e-1}$ elements in $\mathbb{F}_q$ such that $\mathrm{Tr}_{\mathbb{F}_q/\mathbb{F}_p}$ maps that element to $y$, it follows that every element in $\mathbb{F}_p$ has equal frequency in $T$. And, as you said in your post, this suffices to show that

$$ \sum_{c\in C} \zeta^{\mathrm{Tr}_{\mathbb{F}_q/\mathbb{F}_p}(\langle c,z\rangle)}=0 $$

because we have shown that the exponents of this sum are uniformly distributed across $\mathbb{F}_p=\{0,1,...,p-1\}$ and it is well-known that $\sum_{y\in\mathbb{F}_p}\zeta^y=0$.