Let $p$ be an odd prime, and $\zeta$ a primitive $p$-th root of unity over a field of characteristic $0$.

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 $16k\equiv -1\mod p$.

Then, $\prod\limits_{i=1}^{\left(p-1\right) /2}\left(1-\zeta^i\right) = \zeta^k G$.

Question: Can we prove this identity purely algebraically, with no recourse to geometry and analysis?

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).

Note that my "algebraically" allows combinatorics, but I am somewhat skeptical in how far combinatorics alone can solve this. Of course, we can 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.

I am pretty sure things like this must have been done some 100 years ago.

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    $\begingroup$ Just a pedantic point. When $\zeta$ is an arbitrary primitive $p$-th root of $1$, there is no sense in talking about the sign of $G(\zeta)=\sum_{x=1}^{p-1}\left(x\over p\right)\zeta^x$, for the sign of $G(\zeta)$ changes if we replace $\zeta$ by $\zeta^u$, whenever $u$ is prime to $p$ and such that $\left(u\over p\right)=-1$. You can talk about the sign only when $\zeta$ is a specific primitive $p$-th root of $1$, such as $e^{2i\pi\over p}$, where $i$ is a 4-th root of $1$ in $\bf C$. $\endgroup$ – Chandan Singh Dalawat Sep 21 '12 at 2:14
  • $\begingroup$ Right, thank you. My $G$ is not the standard Gauss sum for $p$ but one of its two versions. $\endgroup$ – darij grinberg Sep 21 '12 at 3:30

Have you seen this blog post of mine? Summary: Use Geoff's argument to prove this up to a sign; then use $p$-adic arguments to nail down the sign.

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    $\begingroup$ Thanks - a brilliant answer! (And no, for some reason I have not been following sbseminar; I believe it is because the last times I checked it, it was too much about analysis.) This really ought to be a PDF rather than a blog post (think of the people who want to print this out; also think of the ugliness of Wordpress's latex.php generated formulas). Also, I have posted a comment with some minor errors you might want to fix. $\endgroup$ – darij grinberg Aug 19 '11 at 18:00

(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 $$p = \prod_{i=1}^{p-1}(1- \zeta^i) = (-1)^{\frac{p-1}{2}} \bar{\zeta}^{\frac{p^2-1}{8}}\prod_{i=1}^{\frac{p-1}{2}} (1- \zeta^{i})^{2}.$$

  • $\begingroup$ Now we have to rule out the "false" square roots... $\endgroup$ – darij grinberg Aug 19 '11 at 16:10
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    $\begingroup$ I would have called your post redundant if David Speyer had repeated your calculation. David did not do that, he referred to it. Now that you have removed the calculation, I ask that you restore it so that David's (and your) answer has more impact. Gerhard "Ask Me About System Design" Paseman, 2011.08.19 $\endgroup$ – Gerhard Paseman Aug 19 '11 at 21:52
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    $\begingroup$ OK, Gerhard, I will do that. The calculation is repeated in the blog linked to, but I agree there is a slight non-sequitur. $\endgroup$ – Geoff Robinson Aug 19 '11 at 22:21
  • $\begingroup$ Thank you Geoff. I like it enough that I will give it my fourth vote. Gerhard "Ask Me About System Design" Paseman, 2011.08.19 $\endgroup$ – Gerhard Paseman Aug 19 '11 at 22:35

A few historical remarks about algebraic determinations of the sign of the quadratic gaussian sum might not be out of order.

The proof in David's post was first given by Kronecker, according to Hasse's Vorlesungen. The only analytic ingredient is the determination of the sign of the sin function. This proof is reproduced in Fröhlich and Taylor, Algebraic Number Theory, pp. 228--231.

A different algebraic proof, using the same analytic ingredient, was given by Schur and can be found in Borevich and Shafarevich, Number Theory, pp. 349--353.

Hasse's Vorlesungen 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.

If you are looking for a proof using more analysis, not less, see Rohrlich's survey on Root Numbers in Arithmetic of L-functions, pp. 353--448.

Addendum. A nice (if somewhat dated) survey on The determination of Gauss sums can be found in the BAMS 5 (1981), 107-129. 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, Quadratic reciprocity and the sign of the Gauss sum via the finite Weil representation. Int. Math. Res. Not. IMRN 2010, no. 19, 3729–3745, available here.


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