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Consider the finite filed $GF(p^q)$ such that has elements of order $r$ that means $r\mid p^r-1$. Suppose that $a>1$ is a element of $GF(p^q)$ in which it's order is $r$ that means $r$ is the smallest number that holds at the relation $a^r=1$. We define elements $a_i$, $1\leq i \leq r-1$ in the following form: $$ a_i=a^i \quad , \quad 1\leq i \leq r-1 \, . $$ It can be proved that elements $a_i$, $1\leq i \leq r-1$ are distinct. Now, my question is that how to compute $\sum_{i=1}^{r-1}\,a_i$?

My try: $$ \begin{array}{l} a_1+a_2+\cdots+a_{r-1}=a+a^2+\cdots+a^{r-1}\\ \\ =\underbrace{1+a+a^2+\cdots+a^{r-1}}+p-1=\frac{a^r+p-1}{a+p-1}+p-1\\ \\ =\frac{1+p-1}{a+p-1}+p-1==\frac{0}{a+p-1}+p-1=p-1 \end{array} $$

Is my answer correct and then Is there another method to compute $\sum_{i=1}^{r-1}\,a_i$?

Thanks for any suggestions.

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    $\begingroup$ Yes, this is ok. Mind you, I would say the sum is equal to $-1$ instead of $p-1$, but you are free to do what you want. $\endgroup$ Jul 20, 2017 at 21:08
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    $\begingroup$ Another way would be to use the trick from e.g. here. Call $$S=1+a+a^2+\cdots+a^{r-1}.$$ Then $$aS=a+a^2+\cdots+a^{r-1}+a^r=S,$$ so $(a-1)S=0$ and therefore $S=0$. $\endgroup$ Jul 20, 2017 at 21:11
  • $\begingroup$ @JyrkiLahtonen In fact, I like to find a nice answer like your answer to this qustion, I find it interesting. $\endgroup$
    – Amin235
    Jul 20, 2017 at 21:11
  • $\begingroup$ @JyrkiLahtonen Is it possible to ask you to rewrite it as an answer. Thanks $\endgroup$
    – Amin235
    Jul 20, 2017 at 21:13
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    $\begingroup$ You can ask, but I won't. I try to be as strict I can about not posting the same answer many times. If I think of something to add to it, then I will reconsider. You see, I have made it a profile point in our site policy discussions to point fingers at users who do that habitually. Therefore it would be bad :-) $\endgroup$ Jul 20, 2017 at 21:26

1 Answer 1

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Here's how I address this problem:

first, I think it's worth demonstrating that the $a^j$, $1 \le j \le r - 1$, are distinct. Suppose we had $1 \le j, k \le r -1$, $j \ne k$, and

$a^j = a^k; \tag{1}$

then we can assume without loss of generality that $j < k$. Thus

$a^{k - j} = 1 \tag{2}$

with $0 < k - j < r - 1$. But this contradicts the assumption that $r$ is the smallest such integer; so the $a^j$ are distinct for $1 \le j \le r - 1$.

Second, I think our OP Amin235's supposition that "$a > 1$" isn't really feasible in the context of finite fields, so I will suppose $a \ne 1$, which is meaningful, and a strong enough hypothesis to allow the calculation to proceed.

Having said these things, we proceed as follows:

set

$\sigma = \sum_1^{r - 1} a^j; \tag{3}$

then

$1 + \sigma = 1 + \sum_1^{r - 1} a^j = \sum_0^{r - 1} a^j, \tag{4}$

from which

$(a - 1)(1 + \sigma) = (a - 1) \sum_0^{r - 1} a^j = \sum_1^r a^j - \sum_0^{r - 1} a^j = a^r - 1 = 0, \tag{5}$

since $a^r = 1$ by hypothesis. Now since $a \ne 1$, $a - 1 \ne 0$, whence

$1 + \sigma = 0, \tag{6}$

or

$\sum_1^{r - 1} a^j = \sigma = -1, \tag{7}$

and the requisite sum is obtained.

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    $\begingroup$ It's a good proof with details. Thanks $\endgroup$
    – Amin235
    Jul 21, 2017 at 21:10

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