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Does $\mathbb{Z}[\sqrt{-p}]$ have class number $h(-4p)$?

I am currently reading this paper on isogenies between elliptic curves over $\mathbb{F}_p$ and got a question on the proof of Proposition 2.5:

Context: Let $p\equiv 1\text{ mod }4$ be a prime, then in $K=\mathbb{Q}(\sqrt{-p})$, the maximal order $\mathcal{O}_K$ equals $\mathbb{Z}[\sqrt{-p}]$. However, for $p\equiv 3\text{ mod }4$, the maximal order $\mathcal{O}_K$ equals $\mathbb{Z}[\frac{1+\sqrt{-p}}{2}]$ which is strictly larger than $\mathbb{Z}[\sqrt{-p}]$. (Notice that if $p\equiv 1\text{ mod 4}$, then $-p\equiv 3\text{ mod } 4$ and vice versa).

Now in the context of the paper, it is claimed that the endomorphism ring of an elliptic curve over $\mathbb{F}_p$ either equals $O_K$ in the first case, and equals either $O_K$ or $\mathbb{Z}[\sqrt{-p}]$ in the second case. Proposition 2.5 now wants to count the number of ideal classes in these cases:

For $p\equiv 1\text{ mod }4$, the discriminant equals $-4p$ and thus by definition there are $h(-4p)$ ideal classes. However, for $p\equiv 3\text{ mod }4$, it is claimed that there are $h(-4p)+h(-p)$ many ideal classes. Why is that?

I assume that one summand counts the ideal classes in $\mathbb{Z}[\sqrt{-p}]$, and the other one counts the classes in $\mathcal{O}_K$. However, we know that the corresponding discriminant is just $-p$, and we know that the ring $\mathcal{O}_K$ has $h(-p)$ classes. Consequently, this leads to the conclusion that $\mathbb{Z}[\sqrt{-p}]$ must correspond to $h(-4p)$, thus have $h(-4p)$ ideal classes. Is this true? If so, why?

Till now I was not even aware of the fact that this is defined. Consider for example $p=7$, then the class group has $h(-7)=1$ element. However, is $h(-28)$ even defined? For example, it is not listed in this list of class numbers by wolfram alpha. Clearly, $-28=-7\cdot2^2$ is a discriminant with corresponding fundamental discriminant $-7$, but what is its meaning in this context?

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We certainly can define a ring with characteristic discriminant $-28$, and that would be $\mathbb{Z}[\sqrt{-7}]$ whose minimal associated polynomial equation is $x^2+7=0$.

Like all $\mathbb{Z}[\sqrt{-d}]$ rings with $d>2$, this one has easily demonstrated nonunique factorization, and thus class number greater than $1$. A simple calculation with ideals reveals that the correct class number is $2$, so we can regain UF by augmenting our ring with an additional lattice having the same density in the complex plane. Going to $\mathbb{Z}[(1+\sqrt{-7})/2]$ does exactly that.

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  • $\begingroup$ Thank you. So is it generally true that for $p\equiv 3 \text{ mod }4$, $\mathbb{Z}[\sqrt{-p}]$ has $h(-4p)$ ideal classes? $\endgroup$ Commented Dec 5, 2024 at 17:17
  • $\begingroup$ I will not go that far. $\endgroup$ Commented Dec 5, 2024 at 17:18
  • $\begingroup$ Thinking about your answer, doesn't $\mathbb{Z}[\sqrt{-7}]$ essentially define $h(-28)$? Because for $p\equiv 1\text{ mod }4$, the ring always have discriminant $-4p$? I mean, what alternative interpretation does $h(-28)$ have? $\endgroup$ Commented Dec 5, 2024 at 18:22
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    $\begingroup$ Yup, it does that. $\endgroup$ Commented Dec 5, 2024 at 18:48

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