$1 + \alpha+ \alpha^2+\alpha^3+ \dots + \alpha^{q-2}=$

Question

Show that for any $\alpha \in \mathbb{F}_q$

$$1 + \alpha+ \alpha^2+\alpha^3+ \dots + \alpha^{q-2}= \begin{cases} 1 &\text{if } \alpha=0 \\ -1 &\text{if } \alpha=1 \\ 0 &\text{otherwise} \end{cases} $$

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Having trouble with the otherwise statement

It looks like it works from samples like in $F_3 =\{0,1,2\}$

for $\alpha=2$

$$ p(2)=1+2=3=0$$

for $\alpha=2$ and $F_5$

$$ p(\alpha) =1+2^1+2^2+2^3=1+2+4+3$$

The argument would like to make is that the powers map to another element and the addition of that ends up being zero but need to look at theorems dont have the tools. FOr $\alpha=0$ should be obvious for $\alpha=-1$ there ends up being $q-1=-1$. It feels right to use geometric series summation too.

Answer 1

$$1 + \alpha + ... \alpha^{q-2} = \frac{1-\alpha^{q-1}}{1-\alpha} $$ Considering that $F_q^*$ is a finite (abelian) group, we have $\alpha^{q-1} = 1$.

Answer 2

For $\alpha=0$, the value is obvious.

Since $\mathbb F_q$ is an additive group of order $q$, we get that $$q\cdot 1 =\underbrace{1+\dots+1}_{q\text{ times}}=0$$ That shows your value when $\alpha = 1$:

$$1+1^1+\cdots+1^{q-2}=\underbrace{1+\cdots+1}_{q-1\text{ times}}=-1$$

Since $\mathbb F_{q}^{\times}$ is a multiplicative group of order $q-1$, when $\alpha\neq 0$, you have $\alpha^{q-1}=1$. So: $$0=\alpha^{q-1}-1=(\alpha-1)(1+\alpha+\cdots \alpha^{q-2})$$ when $\alpha\neq 0$.

When $\alpha\neq 1$ also, we can conclude the polynomial is zero, because $\alpha-1\neq 0$.