Gauss sum (with sign) through algebra - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T18:30:54Z http://mathoverflow.net/feeds/question/73228 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/73228/gauss-sum-with-sign-through-algebra Gauss sum (with sign) through algebra darij grinberg 2011-08-19T15:25:34Z 2012-09-20T03:34:07Z <p>Let $p$ be an odd prime, and $\zeta$ a primitive $p$-th root of unity over a field of characteristic $0$.</p> <p>Let $G = \sum\limits_{j=0}^{p-1} \zeta^{j\left(j-1\right)/2}$ be the standard Gauss sum for $p$. (An alternative definition for $G$ is $G = \sum\limits_{j=1}^{p-1}\left(\frac{j}{p}\right)\zeta^j$, where the bracketed fraction denotes the Legendre symbol.) Denote by $k$ the element of $\left\lbrace 0,1,...,p-1\right\rbrace$ satisfying $16k\equiv -1\mod p$.</p> <p>Then, $\prod\limits_{i=1}^{\left(p-1\right) /2}\left(1-\zeta^i\right) = \zeta^k G$.</p> <p><strong>Question:</strong> Can we prove this identity purely algebraically, with no recourse to geometry and analysis?</p> <p>If we can, then we obtain an easy algebraic proof for the value - including the sign - of the Gauss sum $G$, since both the modulus and the argument of $\prod\limits_{i=1}^{\left(p-1\right) /2}\left(1-\zeta^i\right)$ are easy to find (mainly the argument - it's a matter of elementary geometry).</p> <p>Note that my "algebraically" allows combinatorics, but I am somewhat skeptical in how far combinatorics alone can solve this. Of course, we <em>can</em> formulate the question so that it asks for the number of subsets of $\left\lbrace 1,2,...,\frac{p-1}{2}\right\rbrace$ whose sum has a particular residue $\mod p$, but whether this will bring us far... On the other hand, $q$-binomial identities might be of help, since $\prod\limits_{i=1}^{\left(p-1\right) /2}\left(1-\zeta^i\right)$ is a $\zeta$-factorial.</p> <p>I am pretty sure things like this must have been done some 100 years ago.</p> http://mathoverflow.net/questions/73228/gauss-sum-with-sign-through-algebra/73229#73229 Answer by Geoff Robinson for Gauss sum (with sign) through algebra Geoff Robinson 2011-08-19T16:06:28Z 2011-08-20T08:15:06Z <p>(restored slightly simplified version of earlier post). How about: $1- \zeta^{p-1} = -\zeta^{-1}(1-\zeta)$. Doing likewise for $\zeta^{2},\ldots,\zeta^{\frac{p-1}{2}}$, and setting $\alpha = \prod_{i=1}^{\frac{p-1}{2}} (1- \zeta^{i}),$ we see that $\bar{\alpha} = (-1)^{\frac{p-1}{2}} \bar{\zeta}^{\frac{p^2-1}{8}} \alpha.$ Hence we have <code>$$p = \prod_{i=1}^{p-1}(1- \zeta^i) = (-1)^{\frac{p-1}{2}} \bar{\zeta}^{\frac{p^2-1}{8}}\prod_{i=1}^{\frac{p-1}{2}} (1- \zeta^{i})^{2}.$$</code></p> http://mathoverflow.net/questions/73228/gauss-sum-with-sign-through-algebra/73231#73231 Answer by David Speyer for Gauss sum (with sign) through algebra David Speyer 2011-08-19T16:27:19Z 2011-08-19T16:27:19Z <p>Have you seen this <a href="http://sbseminar.wordpress.com/2008/10/11/the-sign-of-the-gauss-sum/" rel="nofollow">blog post</a> of mine? Summary: Use Geoff's argument to prove this up to a sign; then use $p$-adic arguments to nail down the sign. </p> http://mathoverflow.net/questions/73228/gauss-sum-with-sign-through-algebra/107317#107317 Answer by Chandan Singh Dalawat for Gauss sum (with sign) through algebra Chandan Singh Dalawat 2012-09-16T13:26:04Z 2012-09-20T03:34:07Z <p>A few historical remarks about algebraic determinations of the sign of the quadratic gaussian sum might not be out of order. </p> <p>The proof in David's post was first given by Kronecker, according to Hasse's <em>Vorlesungen</em>. The only analytic ingredient is the determination of the sign of the sin function. This proof is reproduced in Fröhlich and Taylor, <em>Algebraic Number Theory</em>, pp. 228--231.</p> <p>A different algebraic proof, using the same analytic ingredient, was given by Schur and can be found in Borevich and Shafarevich, <em>Number Theory</em>, pp. 349--353.</p> <p>Hasse's <em>Vorlesungen</em> also contain a proof by Mordell in which the analytic ingredient is replaced by the fact that if a polynomidal $f\in{\mathbf Z}[T]$ has opposite signs at $a,b\in{\mathbf R}$, then it has a root between $a$ and $b$. This can be proved using the purely algebraic theory of Artin and Schreier.</p> <p>If you are looking for a proof using <em>more</em> analysis, not less, see Rohrlich's survey on <em>Root Numbers</em> in <em>Arithmetic of L-functions</em>, pp. 353--448.</p> <p><strong>Addendum</strong>. A nice (if somewhat dated) survey on <em>The determination of Gauss sums</em> can be found in the <a href="http://www.ams.org/journals/bull/1981-05-02/S0273-0979-1981-14930-2/" rel="nofollow">BAMS 5 (1981), 107-129</a>. I learnt there that the proof attributed by Hasse to Kronecker actually goes back to Cauchy. New proofs are still being given; see for example Gurevich, Hadani, and Howe, <em>Quadratic reciprocity and the sign of the Gauss sum via the finite Weil representation.</em> Int. Math. Res. Not. IMRN 2010, no. 19, 3729–3745, available <a href="http://www.math.wisc.edu/~shamgar/QR-IMRN.pdf" rel="nofollow">here.</a></p>