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The problem I am given is to find the ideal class group of $\mathbb{Q}(\sqrt{19})$, and only have remaining the issue of showing that $P = (2,\sqrt{19}+1)$ (or equivalently $Q_{\pm} = (3,\sqrt{19} \pm 1)$) is non-principal. The method I usually employ in this situation is to consider integer solutions to the equation $a^2-19b^2 = \pm 2$, and then, by taking this equation modulo some number, trying to obtain a contradiction. Although I can remove the $+2$ case by observing that $2$ is not a quadratic residue modulo 19, I cannot find a way to contradict the $-2$ case.

Also, are there any other more effective general methods for determining non-principality of prime ideals, especially those of rings of integers of real quadratic fields, I am seeming to run into similar issues quite often.

RobPratt
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Henry Dakin
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1 Answers1

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$$ 13^2 - 19 \cdot 3^2 = -2 $$

Method described by Prof. Lubin at Continued fraction of $\sqrt{67} - 4$

$$ \sqrt { 19} = 4 + \frac{ \sqrt {19} - 4 }{ 1 } $$ $$ \frac{ 1 }{ \sqrt {19} - 4 } = \frac{ \sqrt {19} + 4 }{3 } = 2 + \frac{ \sqrt {19} - 2 }{3 } $$ $$ \frac{ 3 }{ \sqrt {19} - 2 } = \frac{ \sqrt {19} + 2 }{5 } = 1 + \frac{ \sqrt {19} - 3 }{5 } $$ $$ \frac{ 5 }{ \sqrt {19} - 3 } = \frac{ \sqrt {19} + 3 }{2 } = 3 + \frac{ \sqrt {19} - 3 }{2 } $$ $$ \frac{ 2 }{ \sqrt {19} - 3 } = \frac{ \sqrt {19} + 3 }{5 } = 1 + \frac{ \sqrt {19} - 2 }{5 } $$ $$ \frac{ 5 }{ \sqrt {19} - 2 } = \frac{ \sqrt {19} + 2 }{3 } = 2 + \frac{ \sqrt {19} - 4 }{3 } $$ $$ \frac{ 3 }{ \sqrt {19} - 4 } = \frac{ \sqrt {19} + 4 }{1 } = 8 + \frac{ \sqrt {19} - 4 }{1 } $$

Simple continued fraction tableau:
$$ \begin{array}{cccccccccccccccccc} & & 4 & & 2 & & 1 & & 3 & & 1 & & 2 & & 8 & \\ \\ \frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 4 }{ 1 } & & \frac{ 9 }{ 2 } & & \frac{ 13 }{ 3 } & & \frac{ 48 }{ 11 } & & \frac{ 61 }{ 14 } & & \frac{ 170 }{ 39 } \\ \\ & 1 & & -3 & & 5 & & -2 & & 5 & & -3 & & 1 \end{array} $$

$$ \begin{array}{cccc} \frac{ 1 }{ 0 } & 1^2 - 19 \cdot 0^2 = 1 & \mbox{digit} & 4 \\ \frac{ 4 }{ 1 } & 4^2 - 19 \cdot 1^2 = -3 & \mbox{digit} & 2 \\ \frac{ 9 }{ 2 } & 9^2 - 19 \cdot 2^2 = 5 & \mbox{digit} & 1 \\ \frac{ 13 }{ 3 } & 13^2 - 19 \cdot 3^2 = -2 & \mbox{digit} & 3 \\ \frac{ 48 }{ 11 } & 48^2 - 19 \cdot 11^2 = 5 & \mbox{digit} & 1 \\ \frac{ 61 }{ 14 } & 61^2 - 19 \cdot 14^2 = -3 & \mbox{digit} & 2 \\ \frac{ 170 }{ 39 } & 170^2 - 19 \cdot 39^2 = 1 & \mbox{digit} & 8 \\ \end{array} $$

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Will Jagy
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