Quadratic Gauss sums over finite commutative rings
Let \(R\) be finite commutative ring of odd characteristic. Let \(\lambda\) be an additive complex linear character of \(R\) and let \(G(\lambda)=\sum_{r\;\text{ in} R} \lambda(r^2)\) be the quadratic Gauss sum. The author gives an explicit determination of \(G(\lambda)\) by a series of reductions. Since \(G\) is multiplicative when \(R\) is written as a product of local rings, we can take \(R\) to be a local ring. The conductor, \(I(\lambda)\), of \(\lambda\) is the largest ideal contained in the kernel of \(\lambda\) and \(\lambda\) is primitive if \(I(\lambda)=(0)\). \(R\) has a primitive additive linear character if and only if \(R\) has a unique maximal ideal, say \({\mathfrak m}\). Assuming \(R\) is local and \(\lambda\) is a primitive linear character, \(|R|=q^{d_R}\) and we say \(R\) is of type 0 if \(R\) has an ideal which is equal to its annihilator (so \(d_R\) is even), of type 1 of \(d_R\) is odd, and of type 2 if \(d_R\) is even and \(R\) does not have an ideal \(I\) such that \(I=\operatorname {Ann}I\). If \(R\) is of type 0, then \(G(\lambda)=\sqrt{|R|}\); if \(R\) is of type 2, then \(G(\lambda)=-\sqrt{|R|}\); and finally if \(R\) is of type 1, there are again explicit formulae which split into cases according as the extension \({\mathcal F}_q/{\mathcal F}_p\) over the prime field has even or odd degree, and as \(p\equiv 1\;\text{ or} 3\bmod 4\).
scientific article; zbMATH DE number 1800285