Some properties of elements of the finite field GF(q)

Question

Consider the finite field $GF(q)$ and an element $w$ of order $n$ which is an $n$th root of unity in this field.

Could someone give me explanation or reference to the following questions regarding properties of the elements of $GF(q)$:

1. Why does $w$ exist only if $n$ divides $q-1$?
2. Why is $w^r$ a zero of the left side for all $r$? Actually what does it mean $w$ to be a zero of the left side for all $r$? This is the context: $$\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}w^{(k-i)j},$$ where $i$ is fixed. It seems that $r$ is $k-i$. I can't understand the conclusion - "Hence for all $r \neq 0 \pmod n$, $w^r$ is a zero of the last term."
3. In any field, $x^n - 1 = (x - 1)(x^{n-1} + x^{n-2} + ... + x + 1)$. Why is this statement true?

Thank you!

Answer 1

1) $(GF \setminus \lbrace 0 \rbrace)^{\times} $ is a cyclic group of order $q-1$, so this is an application of Lagrange theorem

2) This point depends on $i$ , infact if $ i \leq n-1 $ $$\sum_{k=0}^{n-1}\sum_{j=0}^{n-1}w^{(k-i)j} = \sum_{k = 0, k \neq i}^{n-1} \frac{(\omega^{k-i})^n -1}{\omega^{k-i} -1} + \sum_{j = 0}^{n-1} 1 = \sum_{k = 0, k \neq i}^{n-1} \frac{(\omega^{n})^{k-i} -1}{\omega^{k-i} -1} + n = \sum_{k = 0, k \neq i}^{n-1} 0+ n = n $$

If $ i > n-1 $ the sum is $0$.

This remains true if you substitute $\omega$ with $\omega^r$, because $\omega^r$ is still a $n$-th root of unit.

3) Here you simply have to calculate, in every field you can do sums and products

Answer 2

$GF(q)$ contains $0$ and a bunch of units. Any time you address a question about multiplicative properties, you are working in the multiplicative subgroup, $GF(q)^\times$. You'll notice that many arguments split as: "for an element that is a unit, ...argument..., giving result. Note also that for zero, the same result holds", reflecting this partition of elements.

Restricting attention to $GF(q)^\times$, $\langle w \rangle = \{1,w,w^2, w^3, \dots, w^{n-1}\}$ is the subgroup generated by $w$, so is subject to Lagrange's Theorem, which addresses question 1.

I am unable to understand your question 2.

Your third question has answer "yes". Fields have addition and multiplication and therefore the expression on the right evaluates to the same expression as on the left.

Answer 3

I hazard an educated guess that the somewhat unclear part 2 actually means the following. Let $w$ be of order $n$ in a finite field. Let $r\not\equiv0\pmod{n}$ be some integer. Then in $GF(q)$ we have $$ S(r):=\sum_{j=0}^{n-1}w^{rj}=0. $$

Proof. That sum is geometric with ratio of successive terms equal to $w^r$. The assumption about $r$ means that $w^r\neq1$. Therefore the formula for a geometric sum works, and $$ S(r)=\frac{1-w^{rn}}{1-w^r}=\frac{1-1}{1-w^r}=0. $$