Generalizing Dirichlet characters Suppose I want to consider Dirichlet characters $$\chi: \mathbb{F}_p(\zeta_r)^{*} \longrightarrow \mathbb{C}$$
Can I prove something similar to the Polya Vinogradov inequality for these characters?
Basically I want look for quadratic non-residues in $\mathbb{F}_p(\zeta_r)^{*}$. What can I say about non-residues 


*

*Without assuming GRH (something similar to burgess bound ??)

*Assuming GRH (something similar to N.C Ankeny's result ??) 

 A: I think that the Polya-Vinogradov method amounts to the following. 
If $\chi$ is a multiplicative character of a finite field $K$, i.e. a homomorphism from $K^*$ to $\Bbb{C}^*$, and $M$ is some subset of $K^*$, then 
we seek to bound the character sums
$$
S(M,\chi)=\sum_{x\in M}\chi(x).
$$
We could replace the character $\chi$ with another complex valued function, say with $\chi(f(x))$ for some polynomial $f(x)\in K[x]$, but let's stick to this basic case for now.
We begin by writing the characteristic function $\chi_M:K\to\Bbb{C}$ of $M$, $\chi_M(z)=1$ if $z\in M$ and $\chi_M(z)=0$ if $z\notin M$, as a linear combination of additive characters of $K$. 
The additive characters, i.e. homomorphisms $\psi:(K,+)\to(\Bbb{C}^*,\cdot)$ have the following description. Let $tr:K\to\Bbb{Z}_p$ be the trace function
$$
tr(x)=x+x^p+x^{p^2}+\cdots+x^{p^{n-1}},
$$
where the integer $n$ is determined by $q:=|K|=p^n$. Denote $\omega=e^{2\pi i/p}$. Because the trace is a surjetive homomorphism of additive groups, to each constant $a\in K$ the function $\psi_a:K\to\Bbb{C}^*$ defined by
$$
\psi_a(x)=\omega^{tr(ax)}
$$
is also a homomorphism, i.e. an additive character. Furthermore, all the additive characters of $K$ are gotten in this way.
Assume that we can find constants $c_a\in\Bbb{C}$ such that
$$
\chi_M=\sum_{a\in K}c_a\psi_a.
$$
Then we can write the sum $S(M,\chi)$ as follows:
$$
\begin{aligned}
S(M,\chi)&=\sum_{x\in M}\chi(x)\\
&=\sum_{x\in K}\chi(x)\chi_M(x)\\
&=\sum_{x\in K}\chi(x)\sum_{a\in K}c_a\psi_a(x)\\
&=\sum_{a\in K}c_a\sum_{x\in K}\chi(x)\psi_a(x).
\end{aligned}
$$
We can make progress here provided that:


*

*We have a good idea of coefficients $c_a$. Whether this is true depends heavily on the choice of $M$.

*We have a good idea about the inner sums $\sum_{x\in K}\chi(x)\psi_a(x)$. This is true, because such sums are the well studied Gauss sums. If $a=0$ or $\chi$ is the principal character, then the sum is trivial, and otherwise it is known that the inner sum has absolute value $\sqrt q$.


So it is about the coefficients $c_a$. Any function $f:K\to\Bbb{C}$ can be written as a linear combination of the characters $\psi_a$, $a\in K$. This is just basic discrete Fourier analysis on a finite abelian group (call it using the Pontryagin dual if so inclined - my background is more on the telecommunication applications, so my past readers welcomed me calling it a DFT :-) Anyway, by orthogonality and completeness of the set of characters $\psi_a,a\in K,$ we always have
$$
f=\sum_{a\in K}\hat f_a\psi_a,
$$
where the coefficients $\hat f_a$ are defined as inner products
$$
\hat f_a=\frac1q\sum_{x\in K}f_a(x)\overline{\psi_a(x)}.
$$
Another key ingredient in Polya-Vinogradov is that when $M$ is an interval the sums $c_a=\widehat{\chi_M}_a$ are just geometric sums. Let's look at the classical case of the prime field first. When $q=p$ and 
$M=\{b,b+1,b+2,\ldots,b+N-1\}\subset [0,p-1]$ we get that whenever $a\neq0$
$$
c_a=\frac1p\sum_{x=b}^{b+N-1}\overline{\psi_a(x)}=
\frac1p\sum_{j=b}^{b+N-1}\omega^{-aj}=\frac{\overline{\omega^{ab}-\omega^{a(b+N)}}}{p(1-\overline{\omega^a})}.
$$
This worked because $c_a$ is a geometric sum with ratio $\overline{\omega^a}\neq1$, or, because
$$
\psi_a(x+1)=\omega^a\psi_a(x).\qquad(*)
$$
When $a=0$, the sum is trivial, and we get $c_0=N/p$. That will often give the main term in the end result because the other sums oscillate.
You asked specifically about sets of the type
$$
M=\{\zeta+1,\zeta+2,\zeta+3,\ldots,\zeta+k\}.
$$
The same calculations go thru. However, a possible obstruction to progress may be that this time many coefficients $c_a$ in the expansion of $\chi_M$ will share that maximum absolute value. We still have
$$
\psi_a(\zeta+m+1)=\psi_a(\zeta+m)\psi_a(1)
$$
for all natural numbers $m$. But this time the ratio of the consecutive terms,
$\psi_a(1)$ is often enough equal to $1$ even when $a\neq0$. This is because
$$
\psi_a(1)=\omega^{tr(a)}=1
$$
whenever $tr(a)=0$. By linearity of trace this happens for $q/p=p^{n-1}$ choices of $a$. I don't know, if that will ruin your day or not. 

Another situation where Polya-Vinogradov works the same way is when  $M=\{g^a,g^{a+1},\ldots,g^{a+N-1}\}$, where $g$ is a generator of $K^*$ and we are interested in evaluating a sum of the form
$$
S(M,f)=\sum_{x\in M}f(x).
$$
This time we want to write $\chi_M$ as a linear combination of the multiplicative characters (i.e. the analogues of Dirichlet characters), and perform a DFT in the group $K^*$ instead. This variant is near and dear to me for its applications in the analysis of certain pseudorandom sequences consisting of values $\psi_1(f(g^k))$ for some polynomial $f(x)\in K[x]$. Such sequences are all over the place in coding theory and such.
