I would appreciate help, please, with Exercise 6.11 in "Ireland and Rosen" (self-study).
By evaluating $(1)$ $\sum_{t} (1+(t/p))\zeta^t$ in two ways, prove $(2)$ $g=\sum_{t} \zeta^{t^2}$
Here $g$ is defined as the quadratic Gauss sum $(3)$ $g_1=g:=\sum_{t}(\frac{t}{p})\zeta^t$.
I tried a specific example with $p=7$.
$1+(t/p)$ is either $0$ when $t$ is a non-residue or $2$ when $t$ is a residue.
For $p=7$ the residues are $1,2,4$. And the sum $(1)$ as above is $2(\zeta+\zeta^2+\zeta^4)$ or twice the sum of $\zeta$s raised to powers of the quadratic residues.
Evaluating the RHS of sum $(2)$, gives $\zeta+\zeta^4+\zeta^9+\zeta^{16}+\zeta^{25}+\zeta^{36}$ or $2(\zeta+\zeta^2+\zeta^4)$. Where the exponents in the last expression are the squares of each of the exponents between $1$ and $p-1$ taken modulo $p$, thus the quadratic residues.
Lastly using the RHS of the definition of $(3)$ $g$, the sum is plus $\zeta$s to the powers of the QRs and minus the powers of $\zeta$s to the NRs. But each of the latter are equivalent to the plus $\zeta$ to the power of a QR, so again the sum is as desired.
So I am comfortable that these three expressions are equivalent. But I would appreciate help in solving the problem as was highlighted above.
I would guess the "two ways" are to show:
$\sum_{t} (1+(t/p))\zeta^t=\sum_{t}(\frac{t}{p})\zeta^t$
The LHS $=\sum_{t}\zeta^{t}+ \sum_{t}(\frac{t}{p})\zeta^t$. But $\sum_{t}\zeta^{t}=-1$, not $0$ as would be convenient. So I think I am making some mistake here.
and
$\sum_{t} (1+(t/p))\zeta^t = \sum_{t} \zeta^{t^2}$
Thanks
Your idea is right - the only issue, I think, was that the sum $ \sum_t \zeta^{t^2}$ includes $\zeta^0$ .
Assume $p\ne 2$, of course.
On the one hand, $$ 1+ \sum_{t=1}^{p-1}( 1 + (t/p))\zeta^t = 1+ (-1 + g) =g.$$
On the other, $$\begin{align} 1+\sum_{t=1}^{p-1}( 1 + (t/p))\zeta^t &=1+ 2 \sum_{t \in (\mathbb F_p^*)^2 } \zeta^t\\ &= \zeta^0 + \sum_{t=1}^{p-1}\zeta^{t^2}, \\ &= \sum_{t=0}^{p-1} \zeta^{t^2}. \end{align}$$ The point of course is that for $a\ne0$, $x^2=a$ has two distinct solutions, if it is has any at all, and $x^2=0$ has only one solution, for $x$ and $a \in \mathbb F_p$.