Is there a classification of ideals of $\mathcal O_K$ ($K$ quadratic) over ramified and split primes depending on $d \pmod 4$?

I am unsure if the following argument is correct. I have not seen something like this in my course, so I'm a bit skeptical, since this seems like a very simple way of computing norms of ideals.

If $a \in I \trianglelefteq \mathcal O_K$ then $(a)\subseteq I$ and so $$\frac{I}{(a)}\trianglelefteq \frac{\mathcal O_K}{(a)} \implies \frac{\mathcal O_K /(a)}{I/(a)}\cong \frac{\mathcal O_K}{I} \implies \left| \frac{\mathcal O_K}{I} \right| \left| \frac{I}{(a)} \right| = \left| \frac{\mathcal O_K}{(a)} \right| \implies \left| \frac{\mathcal O_K}{I} \right| \leq \left| \frac{\mathcal O_K}{(a)} \right|$$ with equality iff $I=(a)$. Hence $N(a)$ is a proper divisor of $Norm(I)$ if $I \supsetneq (a)$. Hence, in the example $I=(7, 1+\sqrt{15})\trianglelefteq \mathcal{O}_{\mathbb Q(\sqrt {15})}$, $7 \in I \implies Norm(I)$ properly divides $49$. So it has norm $1$ or $7$. But clearly $1 \notin I$ so it has norm $7$.

I am wondering, because of this, if there is a simple classification of the prime ideals over ramified, split and inert primes. Is the following true:

Let $K=\mathbb Q(\sqrt d)$. First let $d \equiv 2,3 \pmod 4$. If $p$ is a ramified prime and $p|d$ then the unique ideal of norm $p$ in $\mathcal O_K$ is $(p,\sqrt d)$ and if $p\nmid d$ then necessarily $p=2$, $d\equiv 3 \mod 4$ and the unique ideal is $(2,1+\sqrt{d})$. If $p$ is split then the two ideals of norm $p$ are given by $(p, a \pm \sqrt d)$ where $a$ is the smallest natural number such that $p|Norm( a + \sqrt d).$

Now if $d\equiv 1 \pmod 4$ then I'm not really sure how to proceed. E.g. if $K=\mathbb Q(\sqrt{-59}), \, 3$ splits in $\mathcal O_K$ but are the two ideals of norm $3$ equal to $(3,1\pm\sqrt{-59})$ or $(3,\frac{1\pm \sqrt{-59}}{2})$? When I'm thinking of norms of ideals $I$ in terms of $|\mathcal O_K/I|$ it seems like it's probably the second pair, but when I think of the above argument, it seems like both cases could work as $3|N(1\pm\sqrt{-59})=60$ and $3|N(\frac{1\pm \sqrt{-59}}{2})=15$.

Edit: $(\frac{1\pm \sqrt{-59}}{2})\cdot 3-(1\pm \sqrt{-59})=\frac{1\pm \sqrt{-59}}{2}$, so actually $(3,1\pm\sqrt{-59})=(3,\frac{1\pm \sqrt{-59}}{2})$ from the start. Should I have been able to deduce that these ideals are the same from the above argument, without spotting this little bit of arithmetic?

Does this generalise beyond this example? What happens in the ramified case? Is it perhaps the case that the norm of an element $r$ does not equal the norm of the ideal it generates (defined as $|\mathcal O_K/(r)|$) if $d \equiv 1 \pmod 4$?

I'm sorry this question is so long, but I felt like I needed to include all of my reasoning to explain my confusion. (If someone with more experience on this site thinks I should delete the question and post this as multiple questions, then comment below and I'll do so.) Feel free to answer a small part of the question if you don't want to tackle the whole thing at once. Any help would be much appreciated.

• Consider using \pmod instead of \mod – Gregory Grant May 24 '15 at 0:11
• Thanks for the advice! I will, in future. – James May 24 '15 at 0:13
• You should look up Dedekind's criterion for splitting of primes, particularly as it relates to quadratic fields. This will tell you how to determine which primes are split, and to find generators for the primes lying over $p$. – Brandon Carter May 24 '15 at 12:00