"타원곡선"의 두 판 사이의 차이

2009년 11월 13일 (금) 13:49 판

\(y^2=4x^3-g_2(\tau)x-g_3\)

\(g_2(\tau) = 60G_4=60\sum_{ (m,n) \neq (0,0)} \frac{1}{(m+n\tau )^{4}}\)

\(g_3(\tau) = 140G_6=140\sum_{ (m,n) \neq (0,0)} \frac{1}{(m+n\tau )^{6}}\)

chord-tangent method
유리수해에 대한 Mordell theorem
- 유리수체 위에 정의된 타원의 유리수해는 유한생성아벨군의 구조를 가짐
- \(\mathbb{Z}^r \oplus \mathbb{T}\)

\(y^2=x^3+ax^2+bx+c\)위의 점 \(P=(x,y)\)에 대하여,
\(2P\)의 \(x\)좌표는\(\frac{x^4-2bx^2-8cx-4ac+b^2}{4y^2}\) 로 주어진다

the only possible torsion groups for elliptic curves over Q are the cyclic groups of order 1,2,3,4,5,6,7,8,9,10,12 [sic -- 11 is not possible] and
\(\frac{\mathbb Z}{2\mathbb Z}\oplus \frac{\mathbb Z}{n\mathbb Z}\) for n=1,2,3,4
예) \(E_n : y^2=x^3-n^2x\)의 torsion은 \(\{(\infty,\infty), (0,0),(n,0),(-n,0)\}\)임

\(y^2=x^3-x\)
[/pages/2061314/attachments/2299029 MSP1975197gdf732cih44i50000361d01gd578fhc4a.gif]
유리수해
\(E(\mathbb Q)=\{(\infty,\infty), (0,0),(1,0),(-1,0)\} \simeq \frac{\mathbb Z}{2\mathbb Z}\oplus \frac{\mathbb Z}{2\mathbb Z}\)
주기
\(2\omega=4\int_0^1\frac{dx}{\sqrt{1-x^4}}=B(1/2,1/4)=\frac{\Gamma(\frac{1}{2})\Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}=\frac{\Gamma(1/4)^2}{\sqrt{2\pi}}=5.24\cdots\)
\(2\int_0^1\frac{dx}{\sqrt{x-x^3}}=B(1/2,1/4)=\frac{\Gamma(\frac{1}{2})\Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}=5.24\cdots\)

@@ 56번째 줄: / 56번째 줄: @@
 * <math>y^2=x^3-x</math><br>[/pages/2061314/attachments/2299029 MSP1975197gdf732cih44i50000361d01gd578fhc4a.gif]<br>
 *  유리수해<br><math>E(\mathbb Q)=\{(\infty,\infty), (0,0),(1,0),(-1,0)\} \simeq \frac{\mathbb Z}{2\mathbb Z}\oplus \frac{\mathbb Z}{2\mathbb Z}</math><br>
-*  주기<br><math>2\omega=4\int_0^1\frac{dx}{\sqrt{1-x^4}}=B(1/2,1/4)=\frac{\Gamma(\frac{1}{2})\Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}=5.24\cdots</math><br><math>2\omega=4\int_0^1\frac{dx}{\sqrt{1-x^4}}=B(1/2,1/4)=\frac{\Gamma(\frac{1}{2})\Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}=5.24\cdots</math><br>
+*  주기<br><math>2\omega=4\int_0^1\frac{dx}{\sqrt{1-x^4}}=B(1/2,1/4)=\frac{\Gamma(\frac{1}{2})\Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}=\frac{\Gamma(1/4)^2}{\sqrt{2\pi}}=5.24\cdots</math><br><math>2\int_0^1\frac{dx}{\sqrt{x-x^3}}=B(1/2,1/4)=\frac{\Gamma(\frac{1}{2})\Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}=5.24\cdots</math><br>
@@ 74번째 줄: / 76번째 줄: @@
-Birch and Swinnerton-Dyer conjecture
+<h5>Birch and Swinnerton-Dyer 추측</h5>