Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm having a hard time deciphering some old notes. The aim is to determine the sign of the Gauss sum. Paraphrasing:

Take the polynomial $$ \sum_{i=1}^{p-1}\left(\frac{i}{p}\right)x^i-\sigma\prod_{j=1}^{(p-1)/2}(x^{2j-1}-x^{p-2j+1}) $$ where $\sigma=\pm 1$. We know $x^p-1$ divides this polynomial, so write it as $(x^p-1)f(x)$. Changing variables to $1+u$, we obtain $$ \sum_{i=1}^{p-1}\left(\frac{i}{p}\right)(1+u)^i-\sigma\prod_{j=1}^{(p-1)/2}((1+u)^{2j-1}-(1+u)^{p-2j+1})=((1+u)^p-1)f(1+u). $$

Comparing coefficients, reducing modulo $p$, and applying Wilson's theorem shows $\sigma=1$.

I should mention that $\left(\frac{i}{p}\right)$stands for the Legendre symbol. The last sentence skips too many details for me to reproduce. Can anybody show in a bit more detail how to conclude $\sigma=1$? I appreciate it, thanks kindly.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

Let $A(x) = \sum_{i=1}^{p-1}\left(\frac{i}{p}\right)x^i$ and $B(x) = \prod_{j=1}^{(p-1)/2}(x^{2j-1}-x^{p-2j+1})$, so that $$A(x) - \sigma B(x) = (x^p - 1) f(x).$$

Observe that in $\mathbb{F}_p$, $x^p - 1 = (x-1)^p$, and for $j=1,2,\ldots,(p-1)/2$, $1$ is a simple root of $b_j(x) = x^{2j-1} - x^{p-2j+1}$ (i.e. $x-1\mid b_j(x)$ but $(x-1)^2\nmid b_j(x)$). Thus $$(x-1)^{(p-1)/2} \| B(x) \implies (x-1)^{(p-1)/2} \| A(x).$$

None of this is strictly necessary, but it does tell us that in order to find $\sigma$, it suffices to look at the $(p-1)/2$th derivative evaluated at $1$, or equivalently, the coefficient of $u^{(p-1)/2}$ under the substitution $x\mapsto 1+u$.

Clearly $$[u^{(p-1)/2}]((1+u)^p - 1) f(1+u) \equiv [u^{(p-1)/2}]u^p f(1+u) = 0 \pmod{p}$$ and $$[u^{(p-1)/2}] B(1+u) = \prod_{j=1}^{(p-1)/2} (2j-1-p+2j-1).$$ To compute $[u^{(p-1)/2}]A(1+u)$, recall that $\sum_{i=1}^{p-1} i^r$ is congruent to $-1\pmod{p}$ when $p-1\mid r$ and $0\pmod{p}$ otherwise, and for $i=1,2,\ldots,p-1$, $\left(\frac{i}{p}\right)\equiv i^{(p-1)/2}\pmod{p}$. (Actually, by taking derivatives, we can directly use these two facts to prove that $(x-1)^{(p-1)/2} \| A(x)$.) Hence $$[u^{(p-1)/2}]A(1+u) \equiv \sum_{i=1}^{p-1} i^{(p-1)/2}\binom{i}{(p-1)/2} \equiv \frac{-1}{((p-1)/2)!} \pmod{p}.$$ It immediately follows that $$0 \equiv -1 - \sigma((p-1)/2)!\prod_{j=1}^{(p-1)/2}(4j-2) = -1 - \sigma\prod_{j=1}^{(p-1)/2}(2j)\prod_{j=1}^{(p-1)/2}(2j-1) \equiv -1+\sigma \pmod{p},$$ where we use Wilson's theorem in the last step. Thus $\sigma \equiv 1 \pmod{p}\implies \sigma = 1$, as desired.

share|improve this answer
    
Very helpful, thanks! –  hmIII Jan 27 '13 at 3:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.