실 이차 수체의 유수와 기본 단위

개요

• 복소이차수체의 경우 가우스의 class number one 문제 는 해결되었으나, 실 이차수체의 경우는 미해결

기본 단위(fundamental unit)

• 디리클레의 단위 정리에 의하면, 실이차수체의 단위 \(\mathcal{O}_K^{\times}\)들은 군을 이루며 유일한 원소 \(\epsilon>1\)가 존재하여 다음과 같은 구조를 가진다\[\mathcal{O}_K^{\times} \simeq \{\pm 1\}\times \{\epsilon^n | n\in\mathbb{Z}\}\]
  

특수한 경우

• 소수 \(p\)가 \(p=n^2+1(n > 2)\) 꼴로 주어지는 경우, \(K=\mathbb{Q}(\sqrt{p})\) 의 기본 단위는 \(\epsilon=n+\sqrt{n^2+1}\) 로 주어진다

Ankeny-Artin-Chowla 의 정리와 추측

정리 (AAC)

소수 \(p\equiv 1 \pmod 4\)에 대하여, \(K=\mathbb{Q}(\sqrt{p})\)의 유수를 \(h\), 기본 단위를 \(\epsilon=\frac{t+u\sqrt{p}}{2}\)라 두면, \(\frac{uh}{t}\equiv B_{(p-1)/2} \pmod p\) 이 성립한다. 여기서 \(B_n\)은 베르누이 수.

테이블

$$ \begin{array}{c|c|c|c} p & h & \epsilon & \frac{\frac{h u}{t}-B_{\frac{p-1}{2}}}{p} \\ \hline 5 & 1 & \frac{1}{2}+\frac{\sqrt{5}}{2} & 1 \\ 13 & 1 & \frac{3}{2}+\frac{\sqrt{13}}{2} & 1 \\ 17 & 1 & 4+\sqrt{17} & 1 \\ 29 & 1 & \frac{5}{2}+\frac{\sqrt{29}}{2} & -1 \\ 37 & 1 & 6+\sqrt{37} & -197 \\ 41 & 1 & 32+5 \sqrt{41} & 68161 \\ 53 & 1 & \frac{7}{2}+\frac{\sqrt{53}}{2} & -1129655 \\ 61 & 1 & \frac{39}{2}+\frac{5 \sqrt{61}}{2} & -612054298165 \\ \end{array} $$

추측 (AAC)

소수 \(p\equiv 1 \pmod 4\)에 대하여,, \(K=\mathbb{Q}(\sqrt{p})\)의 기본 단위 \(\epsilon=\frac{t+u\sqrt{p}}{2}\) 은 \(u \not \equiv 0 \pmod p\)를 만족시킨다

테이블

$$ \begin{array}{c|c|c|c} p & \epsilon & u & u \pmod p \\ \hline 5 & \frac{1}{2}+\frac{\sqrt{5}}{2} & 1 & 1 \\ 13 & \frac{3}{2}+\frac{\sqrt{13}}{2} & 1 & 1 \\ 17 & 4+\sqrt{17} & 2 & 2 \\ 29 & \frac{5}{2}+\frac{\sqrt{29}}{2} & 1 & 1 \\ 37 & 6+\sqrt{37} & 2 & 2 \\ 41 & 32+5 \sqrt{41} & 10 & 10 \\ 53 & \frac{7}{2}+\frac{\sqrt{53}}{2} & 1 & 1 \\ 61 & \frac{39}{2}+\frac{5 \sqrt{61}}{2} & 5 & 5 \\ 73 & 1068+125 \sqrt{73} & 250 & 31 \\ 89 & 500+53 \sqrt{89} & 106 & 17 \\ 97 & 5604+569 \sqrt{97} & 1138 & 71 \\ 101 & 10+\sqrt{101} & 2 & 2 \\ 109 & \frac{261}{2}+\frac{25 \sqrt{109}}{2} & 25 & 25 \\ 113 & 776+73 \sqrt{113} & 146 & 33 \\ \end{array} $$

• 정규소수 (regular prime)에 대해서는 AAC 추측이 참이다

목록

• 아래의 목록은 \(K=\mathbb{Q}(\sqrt{n})\)에 대하여 각각 \(\sqrt{n}\), d는 수체의 판별식, h는 유수, 기본 단위, 기본 단위의 norm 을 나타냄
• AlgebraicNumber[Sqrt[n],{a,b}] 은 \(a+b\sqrt{n}\) 을 의미함

Sqrt[2], d=8,h=1,{AlgebraicNumber[Sqrt[2],{1,1}]}, Norm={-1}
Sqrt[3], d=12,h=1,{AlgebraicNumber[Sqrt[3],{2,1}]}, Norm={1}
Sqrt[5], d=5,h=1,{AlgebraicNumber[Sqrt[5],{1/2,1/2}]}, Norm={-1}
Sqrt[6], d=24,h=1,{AlgebraicNumber[Sqrt[6],{5,2}]}, Norm={1}
Sqrt[7], d=28,h=1,{AlgebraicNumber[Sqrt[7],{8,3}]}, Norm={1}
Sqrt[10], d=40,h=2,{AlgebraicNumber[Sqrt[10],{3,1}]}, Norm={-1}
Sqrt[11], d=44,h=1,{AlgebraicNumber[Sqrt[11],{10,3}]}, Norm={1}
Sqrt[13], d=13,h=1,{AlgebraicNumber[Sqrt[13],{3/2,1/2}]}, Norm={-1}
Sqrt[14], d=56,h=1,{AlgebraicNumber[Sqrt[14],{15,4}]}, Norm={1}
Sqrt[15], d=60,h=2,{AlgebraicNumber[Sqrt[15],{4,1}]}, Norm={1}
Sqrt[17], d=17,h=1,{AlgebraicNumber[Sqrt[17],{4,1}]}, Norm={-1}
Sqrt[19], d=76,h=1,{AlgebraicNumber[Sqrt[19],{170,39}]}, Norm={1}
Sqrt[21], d=21,h=1,{AlgebraicNumber[Sqrt[21],{5/2,1/2}]}, Norm={1}
Sqrt[22], d=88,h=1,{AlgebraicNumber[Sqrt[22],{197,42}]}, Norm={1}
Sqrt[23], d=92,h=1,{AlgebraicNumber[Sqrt[23],{24,5}]}, Norm={1}
Sqrt[26], d=104,h=2,{AlgebraicNumber[Sqrt[26],{5,1}]}, Norm={-1}
Sqrt[29], d=29,h=1,{AlgebraicNumber[Sqrt[29],{5/2,1/2}]}, Norm={-1}
Sqrt[30], d=120,h=2,{AlgebraicNumber[Sqrt[30],{11,2}]}, Norm={1}
Sqrt[31], d=124,h=1,{AlgebraicNumber[Sqrt[31],{1520,273}]}, Norm={1}
Sqrt[33], d=33,h=1,{AlgebraicNumber[Sqrt[33],{23,4}]}, Norm={1}
Sqrt[34], d=136,h=2,{AlgebraicNumber[Sqrt[34],{35,6}]}, Norm={1}
Sqrt[35], d=140,h=2,{AlgebraicNumber[Sqrt[35],{6,1}]}, Norm={1}
Sqrt[37], d=37,h=1,{AlgebraicNumber[Sqrt[37],{6,1}]}, Norm={-1}
Sqrt[38], d=152,h=1,{AlgebraicNumber[Sqrt[38],{37,6}]}, Norm={1}
Sqrt[39], d=156,h=2,{AlgebraicNumber[Sqrt[39],{25,4}]}, Norm={1}
Sqrt[41], d=41,h=1,{AlgebraicNumber[Sqrt[41],{32,5}]}, Norm={-1}
Sqrt[42], d=168,h=2,{AlgebraicNumber[Sqrt[42],{13,2}]}, Norm={1}
Sqrt[43], d=172,h=1,{AlgebraicNumber[Sqrt[43],{3482,531}]}, Norm={1}
Sqrt[46], d=184,h=1,{AlgebraicNumber[Sqrt[46],{24335,3588}]}, Norm={1}
Sqrt[47], d=188,h=1,{AlgebraicNumber[Sqrt[47],{48,7}]}, Norm={1}
Sqrt[51], d=204,h=2,{AlgebraicNumber[Sqrt[51],{50,7}]}, Norm={1}
Sqrt[53], d=53,h=1,{AlgebraicNumber[Sqrt[53],{7/2,1/2}]}, Norm={-1}
Sqrt[55], d=220,h=2,{AlgebraicNumber[Sqrt[55],{89,12}]}, Norm={1}
Sqrt[57], d=57,h=1,{AlgebraicNumber[Sqrt[57],{151,20}]}, Norm={1}
Sqrt[58], d=232,h=2,{AlgebraicNumber[Sqrt[58],{99,13}]}, Norm={-1}
Sqrt[59], d=236,h=1,{AlgebraicNumber[Sqrt[59],{530,69}]}, Norm={1}
Sqrt[61], d=61,h=1,{AlgebraicNumber[Sqrt[61],{39/2,5/2}]}, Norm={-1}
Sqrt[62], d=248,h=1,{AlgebraicNumber[Sqrt[62],{63,8}]}, Norm={1}
Sqrt[65], d=65,h=2,{AlgebraicNumber[Sqrt[65],{8,1}]}, Norm={-1}
Sqrt[66], d=264,h=2,{AlgebraicNumber[Sqrt[66],{65,8}]}, Norm={1}
Sqrt[67], d=268,h=1,{AlgebraicNumber[Sqrt[67],{48842,5967}]}, Norm={1}
Sqrt[69], d=69,h=1,{AlgebraicNumber[Sqrt[69],{25/2,3/2}]}, Norm={1}
Sqrt[70], d=280,h=2,{AlgebraicNumber[Sqrt[70],{251,30}]}, Norm={1}
Sqrt[71], d=284,h=1,{AlgebraicNumber[Sqrt[71],{3480,413}]}, Norm={1}
Sqrt[73], d=73,h=1,{AlgebraicNumber[Sqrt[73],{1068,125}]}, Norm={-1}
Sqrt[74], d=296,h=2,{AlgebraicNumber[Sqrt[74],{43,5}]}, Norm={-1}
Sqrt[77], d=77,h=1,{AlgebraicNumber[Sqrt[77],{9/2,1/2}]}, Norm={1}
Sqrt[78], d=312,h=2,{AlgebraicNumber[Sqrt[78],{53,6}]}, Norm={1}
Sqrt[79], d=316,h=3,{AlgebraicNumber[Sqrt[79],{80,9}]}, Norm={1}
Sqrt[82], d=328,h=4,{AlgebraicNumber[Sqrt[82],{9,1}]}, Norm={-1}
Sqrt[83], d=332,h=1,{AlgebraicNumber[Sqrt[83],{82,9}]}, Norm={1}
Sqrt[85], d=85,h=2,{AlgebraicNumber[Sqrt[85],{9/2,1/2}]}, Norm={-1}
Sqrt[86], d=344,h=1,{AlgebraicNumber[Sqrt[86],{10405,1122}]}, Norm={1}
Sqrt[87], d=348,h=2,{AlgebraicNumber[Sqrt[87],{28,3}]}, Norm={1}
Sqrt[89], d=89,h=1,{AlgebraicNumber[Sqrt[89],{500,53}]}, Norm={-1}
Sqrt[91], d=364,h=2,{AlgebraicNumber[Sqrt[91],{1574,165}]}, Norm={1}
Sqrt[93], d=93,h=1,{AlgebraicNumber[Sqrt[93],{29/2,3/2}]}, Norm={1}
Sqrt[94], d=376,h=1,{AlgebraicNumber[Sqrt[94],{2143295,221064}]}, Norm={1}
Sqrt[95], d=380,h=2,{AlgebraicNumber[Sqrt[95],{39,4}]}, Norm={1}
Sqrt[97], d=97,h=1,{AlgebraicNumber[Sqrt[97],{5604,569}]}, Norm={-1}
Sqrt[101], d=101,h=1,{AlgebraicNumber[Sqrt[101],{10,1}]}, Norm={-1}
Sqrt[102], d=408,h=2,{AlgebraicNumber[Sqrt[102],{101,10}]}, Norm={1}
Sqrt[103], d=412,h=1,{AlgebraicNumber[Sqrt[103],{227528,22419}]}, Norm={1}
Sqrt[105], d=105,h=2,{AlgebraicNumber[Sqrt[105],{41,4}]}, Norm={1}
Sqrt[106], d=424,h=2,{AlgebraicNumber[Sqrt[106],{4005,389}]}, Norm={-1}
Sqrt[107], d=428,h=1,{AlgebraicNumber[Sqrt[107],{962,93}]}, Norm={1}
Sqrt[109], d=109,h=1,{AlgebraicNumber[Sqrt[109],{261/2,25/2}]}, Norm={-1}
Sqrt[110], d=440,h=2,{AlgebraicNumber[Sqrt[110],{21,2}]}, Norm={1}
Sqrt[111], d=444,h=2,{AlgebraicNumber[Sqrt[111],{295,28}]}, Norm={1}
Sqrt[113], d=113,h=1,{AlgebraicNumber[Sqrt[113],{776,73}]}, Norm={-1}
Sqrt[114], d=456,h=2,{AlgebraicNumber[Sqrt[114],{1025,96}]}, Norm={1}
Sqrt[115], d=460,h=2,{AlgebraicNumber[Sqrt[115],{1126,105}]}, Norm={1}
Sqrt[118], d=472,h=1,{AlgebraicNumber[Sqrt[118],{306917,28254}]}, Norm={1}
Sqrt[119], d=476,h=2,{AlgebraicNumber[Sqrt[119],{120,11}]}, Norm={1}
Sqrt[122], d=488,h=2,{AlgebraicNumber[Sqrt[122],{11,1}]}, Norm={-1}
Sqrt[123], d=492,h=2,{AlgebraicNumber[Sqrt[123],{122,11}]}, Norm={1}
Sqrt[127], d=508,h=1,{AlgebraicNumber[Sqrt[127],{4730624,419775}]}, Norm={1}
Sqrt[129], d=129,h=1,{AlgebraicNumber[Sqrt[129],{16855,1484}]}, Norm={1}
Sqrt[130], d=520,h=4,{AlgebraicNumber[Sqrt[130],{57,5}]}, Norm={-1}
Sqrt[131], d=524,h=1,{AlgebraicNumber[Sqrt[131],{10610,927}]}, Norm={1}
Sqrt[133], d=133,h=1,{AlgebraicNumber[Sqrt[133],{173/2,15/2}]}, Norm={1}
Sqrt[134], d=536,h=1,{AlgebraicNumber[Sqrt[134],{145925,12606}]}, Norm={1}
Sqrt[137], d=137,h=1,{AlgebraicNumber[Sqrt[137],{1744,149}]}, Norm={-1}
Sqrt[138], d=552,h=2,{AlgebraicNumber[Sqrt[138],{47,4}]}, Norm={1}
Sqrt[139], d=556,h=1,{AlgebraicNumber[Sqrt[139],{77563250,6578829}]}, Norm={1}
Sqrt[141], d=141,h=1,{AlgebraicNumber[Sqrt[141],{95,8}]}, Norm={1}
Sqrt[142], d=568,h=3,{AlgebraicNumber[Sqrt[142],{143,12}]}, Norm={1}
Sqrt[143], d=572,h=2,{AlgebraicNumber[Sqrt[143],{12,1}]}, Norm={1}
Sqrt[145], d=145,h=4,{AlgebraicNumber[Sqrt[145],{12,1}]}, Norm={-1}
Sqrt[146], d=584,h=2,{AlgebraicNumber[Sqrt[146],{145,12}]}, Norm={1}
Sqrt[149], d=149,h=1,{AlgebraicNumber[Sqrt[149],{61/2,5/2}]}, Norm={-1}
Sqrt[151], d=604,h=1,{AlgebraicNumber[Sqrt[151],{1728148040,140634693}]}, Norm={1}
Sqrt[154], d=616,h=2,{AlgebraicNumber[Sqrt[154],{21295,1716}]}, Norm={1}
Sqrt[155], d=620,h=2,{AlgebraicNumber[Sqrt[155],{249,20}]}, Norm={1}
Sqrt[157], d=157,h=1,{AlgebraicNumber[Sqrt[157],{213/2,17/2}]}, Norm={-1}
Sqrt[158], d=632,h=1,{AlgebraicNumber[Sqrt[158],{7743,616}]}, Norm={1}
Sqrt[159], d=636,h=2,{AlgebraicNumber[Sqrt[159],{1324,105}]}, Norm={1}
Sqrt[161], d=161,h=1,{AlgebraicNumber[Sqrt[161],{11775,928}]}, Norm={1}
Sqrt[163], d=652,h=1,{AlgebraicNumber[Sqrt[163],{64080026,5019135}]}, Norm={1}
Sqrt[165], d=165,h=2,{AlgebraicNumber[Sqrt[165],{13/2,1/2}]}, Norm={1}
Sqrt[166], d=664,h=1,{AlgebraicNumber[Sqrt[166],{1700902565,132015642}]}, Norm={1}
Sqrt[167], d=668,h=1,{AlgebraicNumber[Sqrt[167],{168,13}]}, Norm={1}
Sqrt[170], d=680,h=4,{AlgebraicNumber[Sqrt[170],{13,1}]}, Norm={-1}
Sqrt[173], d=173,h=1,{AlgebraicNumber[Sqrt[173],{13/2,1/2}]}, Norm={-1}
Sqrt[174], d=696,h=2,{AlgebraicNumber[Sqrt[174],{1451,110}]}, Norm={1}
Sqrt[177], d=177,h=1,{AlgebraicNumber[Sqrt[177],{62423,4692}]}, Norm={1}
Sqrt[178], d=712,h=2,{AlgebraicNumber[Sqrt[178],{1601,120}]}, Norm={1}
Sqrt[179], d=716,h=1,{AlgebraicNumber[Sqrt[179],{4190210,313191}]}, Norm={1}
Sqrt[181], d=181,h=1,{AlgebraicNumber[Sqrt[181],{1305/2,97/2}]}, Norm={-1}
Sqrt[182], d=728,h=2,{AlgebraicNumber[Sqrt[182],{27,2}]}, Norm={1}
Sqrt[183], d=732,h=2,{AlgebraicNumber[Sqrt[183],{487,36}]}, Norm={1}
Sqrt[185], d=185,h=2,{AlgebraicNumber[Sqrt[185],{68,5}]}, Norm={-1}
Sqrt[186], d=744,h=2,{AlgebraicNumber[Sqrt[186],{7501,550}]}, Norm={1}
Sqrt[187], d=748,h=2,{AlgebraicNumber[Sqrt[187],{1682,123}]}, Norm={1}
Sqrt[190], d=760,h=2,{AlgebraicNumber[Sqrt[190],{52021,3774}]}, Norm={1}
Sqrt[191], d=764,h=1,{AlgebraicNumber[Sqrt[191],{8994000,650783}]}, Norm={1}
Sqrt[193], d=193,h=1,{AlgebraicNumber[Sqrt[193],{1764132,126985}]}, Norm={-1}
Sqrt[194], d=776,h=2,{AlgebraicNumber[Sqrt[194],{195,14}]}, Norm={1}
Sqrt[195], d=780,h=4,{AlgebraicNumber[Sqrt[195],{14,1}]}, Norm={1}
Sqrt[197], d=197,h=1,{AlgebraicNumber[Sqrt[197],{14,1}]}, Norm={-1}
Sqrt[199], d=796,h=1,{AlgebraicNumber[Sqrt[199],{16266196520,1153080099}]}, Norm={1}

역사

• 수학사 연표

관련된 항목들

• 펠 방정식
• 원분체 (cyclotomic field)

매스매티카 파일 및 계산 리소스

• https://docs.google.com/leaf?id=0B8XXo8Tve1cxNTdmZWE1NmItODA1Yy00NGQ5LWIwMjEtMTRlNGVlOTRiZGJl&sort=name&layout=list&num=50

사전 형태의 자료

• http://en.wikipedia.org/wiki/Fundamental_unit_(number_theory)
• http://en.wikipedia.org/wiki/Ankeny–Artin–Chowla_congruence

리뷰, 에세이, 강의노트

• Slavut\cydotskiĭ, I. Sh. “A Real Quadratic Field and the Ankeny-Artin-Chowla Conjecture.” Rossi\uı Skaya Akademiya Nauk. Sankt-Peterburgskoe Otdelenie. Matematicheski\uı\ Institut Im. V. A. Steklova. Zapiski Nauchnykh Seminarov (POMI) 286, no. Anal. Teor. Chisel i Teor. Funkts. 18 (2002): 159–68, 230–31. doi:10.1023/B:JOTH.0000035243.24060.83.

관련논문

• Riele, Herman te, and Hugh Williams. “New Computations Concerning the Cohen-Lenstra Heuristics.” Experimental Mathematics 12, no. 1 (2003): 99–113. http://www.emis.de/journals/EM/expmath/volumes/12/12.1/pp99_113.pdf
• Van der Poorten, A. J., H. J. J. te Riele, and H. C. Williams. “Computer Verification of the Ankeny-Artin-Chowla Conjecture for All Primes Less than $100\,000\,000\,000$.” Mathematics of Computation 70, no. 235 (2001): 1311–28. doi:10.1090/S0025-5718-00-01234-5.
• Ankeny, N. C., E. Artin, and S. Chowla. “The Class-Number of Real Quadratic Number Fields.” Annals of Mathematics. Second Series 56 (1952): 479–93.

관련도서

• Cohen, Henri (1993), A Course in Computational Algebraic Number Theory