[[분류:가져온 문서/오메가]]
Quadratic Gauss Sum
홀소수 <math>p</math>와 서로소인 정수 <math>a</math>, Primitive ''p''th root of unity인 <math>\zeta</math>에 대한 다음 꼴의 합을 말한다.
><math>\sum_{n=0}^{p-1} \zeta^{a\cdot n^2}=\zeta^{a\cdot0^{2}}+\zeta^{a\cdot1^{2}}+...+\zeta^{a(p-1)^{2}} </math>
== 표현 ==
일반적으로 이차 가우스 합은 [math(\displaystyle G(k;p)=\sum_{n=0}^{p-1} \zeta^{k\cdot n^2})]로 표기하며, 다음과 같은 형태를 띤다.
* <math>G(k;n)=\sum_{r=1}^n e^{2\pi ikr^2/n}</math>
하지만 [math(m)]과 [math(n)]이 서로소일 때 <math>G(k;mn)=G(km;n)G(kn;m)</math>이고,
[math(p\nmid k)]인 홀소수 [math(p)]와 [math(a\geq2)]에 대해 <math>G(k;p^a)=pG(k;p^{a-1})</math>임에서
* <math>G(k;p^a)=\begin{cases}p^{a/2}&\text{if } a \text{ is even}\\ p^{(a-1)/2}G(k;p)&\text{if } a \text{ is odd}\end{cases}</math>
가 성립하므로 [math(p\nmid k)]인 홀소수 [math(p)]에 대해 <math>G(k;p)</math> 의 성질은 매우 중요하다.
== 르장드르 기호 ==
이차 가우스 합은 다음과 같이 [[르장드르 기호]]로 나타낼 수 있다.
<math> \zeta^{a\cdot0^{2}}+\zeta^{a\cdot1^{2}}+...+\zeta^{a\cdot(p-1)^{2}} </math>
* <math> =1+2\sum_{n\in QR}^{ }\zeta^{an} </math> (<math>QR</math>: [[이차 잉여]], <math>QNR</math>: 이차 비잉여)
* <math> =1+(\sum_{n\in QR}^{ }\zeta^{an}+\sum_{n\in QR}^{ }\zeta^{an})+(\sum_{n\in QNR}^{ }\zeta^{an}-\sum_{n\in QNR}^{ }\zeta^{an}) </math>
* <math> =(\sum_{n\in QR}^{ }\zeta^{an}-\sum_{n\in QNR}^{ }\zeta^{an})+(1+\sum_{n\in QR}^{ }\zeta^{an}+\sum_{n\in QNR}^{ }\zeta^{an}) </math>
* <math> =(\sum_{n\in QR}^{ }1\cdot \zeta^{an}-\sum_{n\in QNR}^{ }(-1)\cdot \zeta^{an})+\sum_{n=0}^{p-1}\zeta^{an} </math>
* <math> =\sum_{n=1}^{p-1}\left(\frac{n}{p}\right)\zeta^{an} </math>
== 절댓값 ==
<math> g(a) = \zeta^{a\cdot0^{2}}+\zeta^{a\cdot1^{2}}+...+\zeta^{a(p-1)^{2}} =\sum_{n=1}^{p-1}\left(\frac{n}{p}\right)\zeta^{an} </math> 이라고 하자.
=== [math(g(a))] ===
<math> g(a) = \left(\frac{a}{p}\right) g(1) </math>
==== 증명 ====
<math> \left(\frac{a}{p}\right) g(a) = \left(\frac{a}{p}\right)\sum_{n=1}^{p-1}\left(\frac{n}{p}\right)\zeta^{an} = \sum_{n=1}^{p-1}\left(\frac{an}{p}\right)\zeta^{an} = \sum_{l=1}^{p-1}\left(\frac{l}{p}\right)\zeta^{l} = g(1) </math>
<math> \left(\frac{a}{p}\right)\left(\frac{a}{p}\right)g(a) = \left(\frac{a}{p}\right) g(1) </math>
* <math> \therefore g(a) = \left(\frac{a}{p}\right) g(1) </math>
=== [math(|g(1)|)] ===
<math> |g(1)| = \sqrt{(-1)^{\frac{p-1}{2}}p} </math>
==== 증명 ====
<math> \sum_{a=0}^{p-1}g(a)g(-a) </math>의 값을 두 가지 방법으로 구해 보자.
<math>\textup{(i)} \sum_{a=0}^{p-1}g(a)g(-a) = \sum_{a=0}^{p-1}\left(\frac{a}{p}\right)g(1)\cdot\left(\frac{-a}{p}\right)g(1) </math>
* <math> = \sum_{a=1}^{p-1}\left(\frac{-1}{p}\right)\{g(1)\}^{2} </math>
* <math> = (p-1)\left(\frac{-1}{p}\right)\{g(1)\}^{2} </math>
<math>\textup{(ii)} \sum_{a=0}^{p-1}g(a)g(-a) </math>
* <math> = \sum_{a=0}^{p-1}(\sum_{n=1}^{p-1}\left(\frac{n}{p}\right)\zeta^{an}\cdot\sum_{n=1}^{p-1}\left(\frac{m}{p}\right)\zeta^{am})</math>
* <math> = \sum_{a=0}^{p-1}\sum_{n=1}^{p-1}\sum_{m=1}^{p-1}\left(\frac{n}{p}\right)\left(\frac{m}{p}\right)\zeta^{a(n-m)} </math>
* <math> = \sum_{n=1}^{p-1}\sum_{m=1}^{p-1}\left(\frac{n}{p}\right)\left(\frac{m}{p}\right)\sum_{a=0}^{p-1}\zeta^{a(n-m)}</math>
* <math> = \sum_{n=1}^{p-1}\sum_{1\leq m\leq p-1, m\neq n}^{ }\left(\frac{n}{p}\right)\left(\frac{m}{p}\right)\sum_{a=0}^{p-1}\zeta^{a(n-m)}+\sum_{n=1}^{p-1}\sum_{1\leq m\leq p-1, m=n}^{ }\left(\frac{n}{p}\right)\left(\frac{m}{p}\right)\sum_{a=0}^{p-1}\zeta^{a(n-m)}</math>
* <math> = 0 +\sum_{n=1}^{p-1}\sum_{m=n}^{ }\left(\frac{m}{p}\right)\left(\frac{m}{p}\right)\sum_{a=0}^{p-1}1</math>
* <math> = 0 +\sum_{n=1}^{p-1}p</math>
* <math> = p(p-1) </math>
<math>\textup{(i), (ii)}</math>에서
<math> (p-1)\left(\frac{-1}{p}\right)\{g(1)\}^{2} = p(p-1) </math>
<math> \{g(1)\}^{2} = \left(\frac{-1}{p}\right)p = (-1)^{\frac{p-1}{2}}p</math>
* <math> \therefore |g(1)| = \sqrt{(-1)^{\frac{p-1}{2}}p} </math>
[math(g(1))]의 부호를 정하는 것은 그 절댓값을 구하는 것보다 훨씬 어려운 문제이다.
== 이차 상반법칙 ==
아래와 같은 방법으로 이차 상반법칙을 증명할 수 있다.
<math>g(a) = \zeta^{a\cdot0^{2}}+\zeta^{a\cdot1^{2}}+...+\zeta^{a(p-1)^{2}} =\sum_{n=1}^{p-1}\left(\frac{n}{p}\right)\zeta^{an} </math> 이라고 하자. <math>p</math>와 서로소인 홀소수 <math>q</math>에 대해 <math>\{g(1)\}^{q}</math>를 두 가지 방법으로 정리하자.
<math>\textup{(i)}\ \{g(1)\}^{q} = \{ \sum_{n=1}^{p-1}\left(\frac{n}{p}\right)\zeta^{n} \} ^{q}</math>
* <math> \equiv \sum_{n=1}^{p-1}\{\left(\frac{n}{p}\right)\zeta^{n}\} ^{q} = \sum_{n=1}^{p-1}\left(\frac{n}{p}\right) ^{q}\zeta^{nq} = \sum_{n=1}^{p-1}\left(\frac{n}{p}\right) \zeta^{nq} </math>
* <math>= \sum_{n=1}^{p-1}\left(\frac{q}{p}\right)\left(\frac{q}{p}\right)\left(\frac{n}{p}\right) \zeta^{nq} = \left(\frac{q}{p}\right)\sum_{n=1}^{p-1}\left(\frac{nq}{p}\right)\zeta^{nq}</math>
* <math>= \left(\frac{q}{p}\right)\sum_{l=1}^{p-1}\left(\frac{l}{p}\right)\zeta^{l} = \left(\frac{q}{p}\right)g(1) \; (mod\; q)</math>
<math>\textup{(ii)}\ \{g(1)\}^{q} = g(1)\cdot[\{g(1)\}^{2}]^{ \frac{q-1}{2} }</math>
* <math> = g(1)\cdot\{(-1)^{\frac{p-1}{2}}p\}^{ \frac{q-1}{2} }</math>
* <math> = g(1)\cdot(-1)^{\frac{p-1}{2}\frac{q-1}{2}}p^{ \frac{q-1}{2} }</math>
* <math> \equiv g(1)\cdot(-1)^{\frac{p-1}{2}\frac{q-1}{2}}\left(\frac{p}{q}\right)\; (mod\; q)</math>
<math>\textup{(i), (ii)}</math>에서
<math>\left(\frac{q}{p}\right)g(1) \equiv g(1)\cdot(-1)^{\frac{p-1}{2}\frac{q-1}{2}}\left(\frac{p}{q}\right) \; (mod\; q) </math>
<math> \left(\frac{q}{p}\right) \equiv (-1)^{\frac{p-1}{2}\frac{q-1}{2}}\left(\frac{p}{q}\right) \; (mod\; q) </math>
<math> \left(\frac{p}{q}\right)\left(\frac{q}{p}\right) \equiv (-1)^{\frac{p-1}{2}\frac{q-1}{2}} \; (mod\; q) </math>
* <math> \therefore \left(\frac{p}{q}\right)\left(\frac{q}{p}\right) = (-1)^{\frac{p-1}{2}\frac{q-1}{2}}\; \; (\because \left(\frac{p}{q}\right), \left(\frac{q}{p}\right), (-1)^{\frac{p-1}{2}\frac{q-1}{2}} \in \{1, -1\})</math>
따라서 이차 상반법칙이 증명된다.
== 보기 ==
* [[가우스 합]] [math(G(\chi)=\tau(\chi)=\tau_\chi)]
[Include(틀:가져옴2,O=오메가, C=[[https://creativecommons.org/licenses/by-nc-sa/3.0/deed.ko|CC BY-NC-SA 3.0]])]
