Find the ideal class group of $\mathbb{Q}(\sqrt{-5})$ by using the factorization theorem

Question

Let $K=\mathbb{Q}(\sqrt {-5})$. We have shown that $\mathcal{O}_K$ has the integral basis $1,\sqrt{-5}$ and $D=4d=-20$. By computing the Minkowski's constant:$$M_K=\sqrt{|D|}\Big(\frac{4}{\pi}\Big)^{r_2}\frac{n!}{n^n}=\Big(\frac{4\sqrt{20}}{2\pi}\Big)\approx 2.85$$ So, every ideal class in Cl($\mathcal{O}_K$) contains a nonzero ideal of norm at most $5$. Using the factorization theorem, we find that $$(2)=(2,1+\sqrt {-5})^2=\mathfrak{p}_2^{2}$$ $$(3)=(3,1+\sqrt{-5})(3,1-\sqrt{-5})=\mathfrak{p}_3\mathfrak{p}_3'$$ Since the norm of $(2)$ is $4$, we get $N(2,1+\sqrt {-5})=\pm 2$. If $\mathfrak{p}_2$ is principal, then write $(2,1+\sqrt {-5})=(\alpha)$ and $\alpha=a+bi$. $N(a+bi)=a^2+5b^2\neq \pm 2$. So it is not principal. Note that $N(1+\sqrt {-5})=6,$ it follows that $(1+\sqrt{-5})=\mathfrak{p}_2\mathfrak{p}_3$. We conclude from this that $$[\mathfrak{p}_3]=[\mathfrak{p}_2]^{-1}=[\mathfrak{p_2}]$$ where the last equality follows from $[\mathfrak{p}_2]^2=1$ in Cl($\mathcal{O}_K$). Also, $$[\mathfrak{p}_3']=[\mathfrak{p_3}]^{-1}=[\mathfrak{p}_2]^{-1}=[\mathfrak{p_2}]$$. Therefore, every ideal class coincides with either $1$ or $[\mathfrak{p}_2]$. We conclude that $Cl(\mathcal{O}_K)\cong \mathbb{Z}/2\mathbb{Z}$.

My question: 1、In the notes, it considers not only $(2),(3)$, but also $(5),(7)$. But the norm of $(5),(7)$ are $25$ and $49$ respectively. (In my procedure, $(3)$ is also not satisfied since it has the norm of $9$, but I have some questions regarding the following procedure, so I left it).

2、Why can we do it by only considering the prime ideals? If the bound is large, we only consider $(p)$ with $p$ prime, what about other ideals like $(4)$? Although we can't use the theorem to factorize it, but I think that's not the reason for not considering it.

3、How to deduce $(1+\sqrt{-5})=\mathfrak{p}_2\mathfrak{p}_3$ by using just the norm? I know $\mathfrak{p}_2\mathfrak{p}_3\subseteq \mathfrak{p}_2\cap\mathfrak{p}_3\subseteq (1+\sqrt{-5}).$

4、I have got $[\mathfrak{p}_2\mathfrak{p}_3]=1$, how to get $[\mathfrak{p}_3]=[\mathfrak{p}_2]^{-1}$? Dose it follow from $[\mathfrak{p}_3][\mathfrak{p}_2]=1,$ and then multiply $[\mathfrak{p}_2]^{-1}$ on each side? But I know it is only true when $\mathcal{O}_K$ is a Dedekind domain.

Any help is appreciated!

Answer 1

Let me try to answer your questions. If there are any questions left, please let me know.

First off, I'll answer 2. The definition of a "prime ideal" is different than what you might think, for example, the ideal $(p)$ it is not always prime ideal. Now, why don't we consider the ideal $(4)$? This is because $(4)=(2)\cdot (2)$, so that we won't get any new prime ideals here that generate the ideal class group. Now, any prime ideal $\mathfrak p$ with norm $p$ is contained in the factorization of $(p)$ (see this post). The Minkowski bound is often stated as a theorem, saying that the ideal class group $\mathrm{Cl}(\mathcal O_K)$ (where $\mathcal O_K$ is the ring of integers) is generated by ideals of norm at most $M_K$. Now, we can easily extend the theorem to say $\mathrm{Cl}(\mathcal O_K)$ is generated by prime ideals of norm at most $M_K$ (see Theorem 3.6 and this as an excellent source of example computations). As all these prime ideals $\mathfrak p$ of norm $p \leq M_K$ are in the factorization of $(p)$, we only need to factor the ideals $(p)$ in $\mathcal O_K$. And as before, if $p$ is not prime, then we can write $(p)=(a)(b)$ with $a,b>1$, giving that factoring $(p)$ does not give any new prime ideals.

Regarding your question 1, if, for example, $M_K = 13.5$, then the ideal class group $\mathrm{Cl}(\mathcal O_K)$ is generated by prime ideals with norm not exceeding $13.5$. This means that we need to factor the ideals $(2),(3),(5),(7),(11),$ and $(13)$ into prime ideals, and these prime ideals generate the ideal class group. My guess is that your notes computed that $M_K \approx 2.85$, giving $M_K^2 \approx 8.1$, so that $(2),(3),(5),$ and $(7)$ were factored. However, this is not necessary as the class group is generated by prime ideals of norm not exceeding $2.85$, and if we factor the ideal $(3)$ (or $(5)$ or $(7)$ for that matter), then we will get primes of norm a power of $3$ (or $5$ or $7$, respectively) as $(3)$ has norm $9=3^2$. But, all these norms are greater than $M_K\approx2.85$, so that we don't need to factor these ideals. Also, it is true that $(3)$ has norm $9$, but this is somewhat irrelevant. We are looking for prime ideals of norm $\leq M_K$ and $(3)$ (as you computed above) is not a prime ideal. So don't get too caught up the norm of $(p)$ which is $p^2$, but rather that $(p)$ can factor into prime ideals with norm as low as $p$.

Question 4, if $\mathcal O_K$ is the ring of integers of $K$ (then it is automatically a Dedekind domain, in fact, it is the smallest extension of $\mathbb Z$ that is a Dedekind domain in $K$), then the ideal class group is actually a finite abelian group under the operation $[I]\cdot[J] = [IJ]$ for any two ideals $I, J$. Therefore, $[\mathfrak p_2 \mathfrak p_3] = [\mathfrak p_2] [\mathfrak p_3] = [\mathfrak p_3] [\mathfrak p_2]=1$ and indeed $[\mathfrak p_3] = [\mathfrak p_2]^{-1} = [\mathfrak p_2^{-1}]$.

Question 3, this depends, if $\mathfrak p_3 = (3, 1+\sqrt{-5})$, then you are right. Again, see Theorem 3.6, which implies that (as the norm map is multiplicative) the ideal $(1+\sqrt{-5})$ is the product of an ideal of norm 2 and an ideal of norm 3. So either, $(1+\sqrt{-5}) = \mathfrak p_2 \mathfrak p_3$ or $(1+\sqrt{-5}) = \mathfrak p_2 \mathfrak p_3'$. If $\mathfrak p_3$ is defined as above, then you are right to note that $1+\sqrt{-5} \in \mathfrak p_3$, so that $(1+\sqrt{-5}) = \mathfrak p_2 \mathfrak p_3$.

Suppose you had chosen $K=\mathbb Q(\sqrt{5})$ and the order $\mathbb Z[\sqrt{5}]$ with basis $1, \sqrt{5}$. Then $\Delta(\mathbb Z[\sqrt{5}]) = 20$. Now, any order of $K$ can only be singular above some prime $p$ if $p^2$ divides the discriminant of that order. So, $\mathbb Z[\sqrt{5}]$ might be singular above $2$. Now, use the Kummer-Dedekind theorem (or an extension of it) to see that $\mathbb Z[\sqrt{5}]$ is indeed singular above $2$. Hence, $\mathbb Z[\sqrt{5}]$ is not a Dedekind domain, and therefore not equal to the ring of integers (this is $\mathbb Z[\frac{1+\sqrt{5}}{2}]$).