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!
Best Answer
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 $\mathbb Z[\frac{1+\sqrt{5}}{2}]$).