Ideals in a Quadratic Number Fields

In the literature it is stated that to each quadratic irrational $\gamma=\frac{P+\sqrt{D}}{Q}$ there is a corresponding ideal $I=[|Q|/\sigma , (P+\sqrt{D})/\sigma]$, where $\sigma=1$, if $\Delta \equiv0$ mod $4$ and $\sigma=2$, otherwise.

Thus, in the case of $\frac{2+\sqrt{13}}{3}$ the associated ideal must be $I=[3/2, (2+\sqrt{13})/2]$ which makes no sense, as $N(I)=3/2$ is supposed to be a rational integer.

What am I doing wrong here?

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3/2 is not integral –  stankewicz Sep 30 '10 at 0:10
The literature also usually states that Q is even and positive. Also, if you take $\Delta$ to be a fundamental discriminant, then the associated ideal is $(Q, (P+\sqrt{\Delta})/2)$. –  Dror Speiser Sep 30 '10 at 1:32
This question has also been asked on math.SE and has received a perfectly good answer there. I have voted to close. –  Pete L. Clark Sep 30 '10 at 7:13