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Let $F$ be a finite field and $i$ an integer. Calculate the sum of all the elements of $F$, each raised to the $i$th power.

My approach so far:

Let $F=(0,1,\alpha,\alpha^2,...,\alpha^{p^n-1})$, where $p$ is a prime number and let $\sigma=1+\sum_{k=1}^{p^n-1}(α^k)^i$ be the desired sum.

Since $F$ is a ring, each element has an additive inverse. Thus, the trivial case of $i=1$ results in $\sigma=1+[\alpha+(-\alpha)+\alpha^2+(-\alpha^2)+\cdots+\alpha^{p^n-1}+(-\alpha^{p^n-1})]=1$.

We can also calculate the trivial case of $i=0$, where $\sigma=1+p^n-1=p^n$

In the general case, let $i=t (\text{mod } p^n)$.
Then: $i=t+mp^n$, where $t,m$ are integers. We take into account the fact that the order of the multiplicative group of the finite field is $p^n$, so each element raised to the $p^n$ equals $1$. Thus $σ=1+α^t+α^{2t}+\cdots+α^{t(p^n-1)}$. This is the sum of a geometric progression, plus $1$. Thus, $σ=1+α(α^{p^n-1})/α-1$ = $ 1$+($α^{p^n}-α$)/${\alpha-1}$. But according to Lagrange's theorem, for every $\alpha\in F$, the polynomial $x^{p^n}-x$ is zero. It follows that $\sigma=1$.

An important question is raised: Is it true that the characteristic of the finite field is $p$ if its order is $p^n$? If so, the case for $i=0$ leads to $σ=0≠1$ and all cases have been considered?

user26857
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MathematicianByMistake
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    The order of the multiplicative group of a finite field $F$ with $p^n$ elements is $p^n-1$. In this case, the characteristic of $F$ is indeed $p$. Also, your case $i=1$ has too many summands, but the idea is right. – Thomas Poguntke May 05 '15 at 16:07
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    $\alpha^{p^n-1}=1$, so your sum has that term twice. The largest distinct power of a primitive element you have is $\alpha^{p^n-2}$. So your sum contains $1^i$ twice, and the sum is off by one. Similarly, you should look at the congruence $i\equiv t\pmod{p^n-1}$. – Jyrki Lahtonen May 05 '15 at 16:08
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    If $\text{char } F \neq 2$, each nonzero element has an additive inverse, so for $i = 1$, $\sigma = 0$. – Travis Willse May 05 '15 at 16:10

1 Answers1

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A different approach. Assume that $i$ is such that $\alpha^i\neq1$. Let $$ \sigma_i=\sum_{x\in\Bbb{F}_{p^n}}x^i $$ be the sum. We get that $$ \alpha^i\sigma_i=\alpha^i\left(\sum_{x\in\Bbb{F}_{p^n}}x^i\right) =\sum_{x\in\Bbb{F}_{p^n}}\alpha^ix^i= \sum_{x\in\Bbb{F}_{p^n}}(\alpha x)^i. $$ Here $\alpha x$ ranges over the field when $x$ does, so the last sum is just $\sigma_i$. Thus $$ \alpha^i\sigma_i=\sigma_i. $$ Given that $\alpha^i\neq1$ I trust you to make the final deductive step. :-)

The condition $\alpha^i\neq 1$ is satisfied, iff $i\not\equiv 0\pmod{p^n-1}$. If $i$ is a multiple of $p^n-1$ your sum consists of several terms equal to $1$, which is another easy case.

Jyrki Lahtonen
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  • Certainly a much more compact and refined approach than mine! Thank you! – MathematicianByMistake May 05 '15 at 16:17
  • Excuse me but how do you define $x^i$ when $x=i=0$? It looks a bit ambiguous. – RFZ Jan 23 '23 at 23:04
  • @ZFR Some convention needed there. Sometimes it is best to leave it undefined, sometimes it makes sense to evaluate $x^0=1$ as the constant polynomial $1$. It is easy to adjust the answer according to the choice, so I'm not gonna worry about it. – Jyrki Lahtonen Jan 24 '23 at 04:50
  • Sorry but I am still confused. So we have a sum $\sum\limits_{x\in \mathbb{F}_q}x^i$, where $i\geq 0$. If $i>0$, then everything is defined. If $i=0$, then we have some issues. I am kind of new in field theory, so the natural question arises: how do we define $x^0$ for arbitrary $x\in \mathbb{F}_q$? I was not able to find it in the definition of finite fields. For some reason it remains uncovered. – RFZ Jan 25 '23 at 15:10
  • For example, if we have a group $(G,*)$ then we define $g^0=e$ for any $g\in G$, where $e$ is the identity element of group $G$. However, it seems that one needs to be careful if we want to define $x^0$ if $x\in \mathbb{F}$, where $\mathbb{F}$ be a field since field has two identity elements $0$ and $1$. – RFZ Jan 25 '23 at 16:09
  • @ZFR Depends. You can define the powers of zero as you wish. And then adjust your answer to this question as you please. Handling that single term of this sum separately is easy, and not interesting. – Jyrki Lahtonen Jan 26 '23 at 12:39