The narrow class group is the group of fractional ideal modulo totally positive principal ideals. In the case of quadratic fields, this is to say positive norms principal ideals.

Consider $\sqrt{7}$, which has class number one and narrow class number two. I would like to understand how to determine explicitly éléments of each class. Aren't they exactly the positive (resp. negative) fractional ideals?

I guess not: I am reading an example where bases of these ideals (see as Z-modules) are given in the form (1,w). For the principal class, they give $$w = \frac{3+\sqrt{7}}{2}$$ and for the other class (shouldn't it be the class of the different $\sqrt{28}$?) they give $$w = \frac{5+\sqrt{7}}{3}$$ I do not understand both why it is true and how to come up with such elements.


1 Answer 1


$\Bbb{Z}+w\Bbb{Z}$ is not the same as the fractional ideal $w O_K$.

$3+\sqrt{7}$ has norm $2$ so $$\Bbb{Z}+\frac{3+\sqrt{7}}2\Bbb{Z} = \frac12(3+\sqrt{7}) O_K$$ which is a totally positive principal fractional ideal.

On the other hand $$\Bbb{Z}+\frac{5+\sqrt{7}}3\Bbb{Z}=\frac13(3\Bbb{Z}+(2+\sqrt{7}))\Bbb{Z} = \frac13 (2+\sqrt7)O_K$$ is not a totally positive principal fractional ideal.

  • $\begingroup$ Thanks for the answer. Why is it natural to divide by the norm? I do not understand how you get your ideals equalities (how the first factor disappears in both?) $\endgroup$ Nov 14, 2022 at 18:17
  • $\begingroup$ @DesideriusSeverus Just check that it works. $O_K=\Bbb{Z}[\sqrt7]$ so $aO_K=a\Bbb{Z}+a \sqrt7 \Bbb{Z}$. Equivalenty check that it is a $O_K$-module containing $a$ and contained in $aO_K$. $\endgroup$
    – reuns
    Nov 14, 2022 at 18:24

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