I am trying to understand the relation of additive and multiplicative structure of finite field of prime cardinality.
Let's say I have a set that behaves nice additively - an interval $I = \{a,a+1,\cdots ,b\} \subseteq \mathbb{F}_{p}$. I allow the size $b-a$ to be quite large - $\Omega(p)$.
What can be said about its multiplicative structure, i.e. of the discrete log of the elements (with respect to some generator)? Specifically, I am interested in the behavior of the multiset $\{(\log(a) - \log(a')) \mod p-1 | a,a' \in I \}$. Simulations show that each non-zero value appears about $|I|^2 / (p-2)$ times, as expected. Can this be proved rigorously?
I feel that this question is related to the Polya-Vinogradov inequality, which states that the sum of a character over consecutive integers behaves like a Bin(n,0.5).
EDIT (correction and a direction): It may happen for some intervals $I$ that some non-zero values modulo $p-1$ appear much less\more than expected (but the set of these values is rather small). Because of this, it seems that an average result is more fitting in this situation (instead of an absolute result).
Namely, if we let $f(x) = |I \cap Ix|$ to be the function counting the times $\log_g(x)$ appears in our multiset, then Average$(f) = \frac{1}{p-1}\sum_{x=1}^{p-1} f(x) = \frac{|I|^2}{p-1}$. From simulations, Variance$(f) = \frac{1}{p-1} \sum_{x=1}^{p-1} (f(x)-$Average$(f))^2$ is small - about $\frac{|I|^2}{p-2}$.
Writing $|Ix \cap I|$ as $\sum_{b_1, b_2 \in I} 1_{x b_1 = b_2} $, we get: $\frac{1}{p-1} \sum_{x=1}^{p-1} |Ix \cap I|^2 = \frac{1}{p-1} \sum_{x=1}^{p-1} \sum_{b_1,b_2,b_3,b_4 \in I} 1_{x = \frac{b_2}{b_1} = \frac{b_4}{b_3}} =\frac{1}{p-1} \sum_{b_1,b_2,b_3,b_4 \in I} 1_{b_2 b_3 / b_1 b_4 = 1} =$ and by using orthogonality relations of characters (similar to Jyrki's post), it simplifies to: $= \frac{1}{(p-1)^2} \sum_{\chi } \sum_{b_1, b_2, b_3, b_4 \in I} \chi(b_2 b_3 b_1^{-1} b_4^{-1})=\frac{1}{(p-1)^2} \sum_{\chi } |\sum_{b \in I} \chi(b)|^4$
The term corresponding to $\chi = 1$ equals $\frac{|I|^4}{(p-1)^2}$ and it cancels Average$(f)^2$, so the variance is:
$\frac{1}{(p-1)^2} \sum_{\chi \neq 1} |\sum_{b \in I} \chi(b)|^4$
which is an elegant expression, but why is it small? The Polya-Vinogradov gives the upper bound $O(p \log^4 p)$, but I think the $\log$ term can be made smaller.