Lemma in definition of quadratic Gauss sums

Question

In Ireland and Rosen, we have the following lemma and proof:

(Here, $\zeta=e^{2\pi i/p}$, a primitive $p$th root of unity for some odd prime $p$.)

Lemma 1. $\sum^{p-1}_{t=0}\zeta^{at}$ is equal to $p$ if $a\equiv 0 \pmod{p}$. Otherwise it is zero.

Proof. If $a\equiv 0\pmod{p}$, then $\zeta^a=1$, and so $\sum^{p-1}_{t=0}\zeta^{at} = p$. If $a\not\equiv 0\pmod{p}$, then $\zeta^a\neq1$ and $\sum^{p-1}_{t=0}\zeta^{at} = (\zeta^{ap}-1)/(\zeta^{a}-1)=0$.

I'm a little stumped on the last statement where it says $\sum^{p-1}_{t=0}\zeta^{at} = (\zeta^{ap}-1)/(\zeta^{a}-1)=0$; I don't see the (maybe obvious) connection there.

Could someone point me in the right direction? A full explanation is not what I'm looking for, though I'd be happy to take one. Thanks.

Answer 1

It seems to be that $\;\zeta\;$ is a primitive $\;p$ -th root of unity, say $\;\zeta=e^{2\pi i/p}\;$ , then we have here simply the sum of a geometric series:

$$\sum_{t=0}^{p-1}\left(\zeta^a\right)^t=\frac{\zeta^{ap}-1}{\zeta^a-1}=0\;,\;\;\text{since}\;\;\zeta^p=1\implies (\zeta^a)^p=(\zeta^p)^a=1$$

Answer 2

Since $\zeta$ is a $p$th root of unity, we know that $\zeta^{ap} = 1$. And so $\zeta^{ap} - 1 = 0$.

And as we assume that $\zeta^a \neq 1$, we do not have a $\frac{0}{0}$ case. Thus it's $0$.