Ideals in a Quadratic Number Fields

In the literary works it is mentioned that per square illogical $\gamma=\frac{P+\sqrt{D}}{Q}$ there is an equivalent excellent $I=[|Q|/\sigma , (P+\sqrt{D})/\sigma]$, where $\sigma=1$, if $\Delta \equiv0$ mod $4$ and also $\sigma=2$, or else.

Hence, when it comes to $\frac{2+\sqrt{13}}{3}$ the linked perfect has to be $I=[3/2, (2+\sqrt{13})/2]$ that makes no feeling, as $N(I)=3/2$ is intended to be a sensible integer.

What am I doing incorrect below?

2019-05-13 04:07:08
Source Share
Answers: 1

Below is an evidence of the typical equivalences in between kinds, perfects and also numbers, excerpted from area 5.2, p. 225 of Henri Cohen is publication "A training course in computational algebraic number theory". Keep in mind that your square number is not of the kind defined in this equivalence, viz. $\rm\ \tau = (-b+\sqrt{D})/(2a)\:,\:$ and also $\rm\: 4\:a\:|\:(D-b^2)\:,\:\:$ i.e. $\rm\ a\:|\:N(a\tau)\:,\:$ a problem matching to the $\rm\mathbb Z$ - component $\rm\ a\:\mathbb Z + a\tau\ \mathbb Z\ $ being a perfect when $\rm\:D\:$ and also $\rm\:b\:$ have the very same parity, as an example see Proposition 2.8 p.18 in Franz Lemmermeyer's notes.


2019-05-17 15:25:25