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We begin with (unfortunately, quite a bit of) notation necessary to state Stickelberger's congruence: Fix an integer $k>1$, and let $p$ be a prime number not dividing $k$. Let $r$ be the smallest positive integer such that $p^r\equiv 1 \pmod{k}$.

Let $\zeta_k=e^{2\pi i/k}$ , and let $K=\mathbb{Q}(\zeta_k)$, $M=\mathbb{Q}(\zeta_k,\zeta_p)$. Let $P\subset O_K$ be a prime above $p$ and $\mathfrak{p}\subset O_M$ be a prime above $P$.

For any integer $a$ we reduce it mod $k$, and write its $p$-adic expansion: $$\bar{a}=a_0+a_1p+a_2p^2+...+a_{r-1}p^{r-1}.$$

Now, define two functions: $$s(a)=a_0+....+a_{r-1}$$ $$t(a)=a_0!...a_{r-1}!$$

If we define the character $\chi_P$ on $(O_K/P)^\times$ by setting $\chi_P(\alpha)$, for $\alpha \not\equiv 0 \bmod P$, to be the unique $k$-th root of unity in $O_K$ satisfying $$\chi_P(\alpha)\equiv \alpha^{(p^r-1)/k} \pmod{P},$$ then define the Gauss sum $G(\chi_P^{-a})=\sum_{\alpha\in O_K/P}\chi_P^{-a}(\alpha)\zeta_p^{tr(\alpha)}$, where $tr(\alpha)$ is the trace of $\alpha$ from $O_K/P$ to $\mathbb{Z}/(p)$.

Then Stickelberger's congruence says that $$G(\chi_P^{-a})\equiv \frac{(\zeta_p-1)^{s(a)}}{t(a)} \pmod{\mathfrak{p}^{s(a)+1}}.$$

Now, I'd like to look at congruences for products of Gauss sums, and Gauss sums of products of characters, etc. Unfortunately, quantities like $t(a)t(b)$, $t(ab)$, etc are not easy to work with. So, one question would be, are there any expressions already known for more complicated expressions involving Gauss sums, or is there a more easy to compute with expression of the sort listed above? Thanks!

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Products of Gauss sums and Gauss sums of products of characters are related to each other through Jacobi sums. A congruence for Jacobi sums of several characters is given as Theorem 2 in You also might consider working with the Gross-Koblitz formula, which lifts Stickelberger's congruence to an exact p-adic formula for a Gauss sum as a product of values of the p-adic Gamma function. Perhaps working with products of values of the p-adic Gamma function first and then reducing to a congruence would be useful to you. – KConrad Jan 30 '11 at 7:17
Do you really mean $t(ab)$? I think you mean $t(a+b)$. That's at least easier to calculate - you can see there is a sort of generalized binomial coefficient appearing in $t(a+b)/(t(a)t(b))$, for instance. – Will Sawin Mar 25 at 3:37

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