# Class Number of $\mathbb{Q}(\sqrt[3]{19})$ and Hilbert class field

Finding the class number of $\mathbb{Q}(\sqrt[3]{19})$ is an exercise from Marcus 'Number Field'. This question was uploaded by some other user, but it was removed by now. I have worked on details and have some more questions.

Following are the steps in the exercise:

Let $K=\mathbb{Q}(\alpha)$, $R=\mathcal{O}_K$, and $3R=P^2Q$, where $\alpha=\sqrt[3]{19}$, and $P$, $Q$ are prime ideals in $R$.

(a) Prove that the ideal class group is cyclic, generated by the class containing $P$.

(b) Prove that the number of ideal classes is a multiple of $3$.

(c) Prove that there are either three or six ideal classes.

(d) Prove that there are three ideal classes. (Suggestion: Suppose there were six. Show that none of the ideals $J$ with $||J||\leq 9$ are in the same class with $P^3$. )

After solving this problem, it seems that the calculations from my solution to (a) are enough to conclude that the class number is $3$.

To do (a), we find the Minkowski's bound: $$\frac{3!}{3^3}\frac 4{\pi} \sqrt{1083} \approx 9.3$$ This shows that each ideal class in the ideal class group contains an ideal $J$ with $||J||\leq 9$. Also, each ideal class generating the class group contains a prime ideal lying above $2$, $3$, $5$, $7$.

So, we find prime factorizations of $2R$, $3R$, $5R$, $7R$: $$2R=P_1P_2,\ \ \ 3R=P^2 Q,$$ $$5R= P_3P_4, \ \ \ 7R = P_5,$$ with $P_1=(2, \alpha-1)$, $P=(3, \frac{\alpha^2+\alpha+1}3)$, $P_3=(5,\alpha+1)$.

With a help of SAGE, I found that $$P_1= \frac{\alpha-1}3 P, \ \ \ P_3= \frac{-\alpha+4}3 P, \ \ \ P_5=\frac{7\alpha+14}9 P^3.$$

Then the prime ideals appear above (as factorizations of $2$, $3$, $5$, $7$) belong to one of the three ideal classes of $P$, $P^2$, $I$ (principal).

My questions are

1. Are these calculations enough to conclude that the ideal class group has order $3$?

2. If the answer to 1 is no, then how are the rest of parts solved?

3. How do we determine the Hilbert class field of $\mathbb{Q}(\sqrt[3]{19})$?

• And how are you concluding that all the ideal class aren't in fact the principal ideal class? – John M Jul 29 '16 at 2:36
• I have to prove that the ideal class of $P$ is not principal, so the class of $P$ has order $3$. Okay, so part (b) is really necessary to do that. I see. – Sungjin Kim Jul 29 '16 at 2:42
• Except for showing that at least one ideal is nonprincipal, I think you’re OK. Since $P_5=(7)$, principal, and also equals $(z)P^3$ for a number $z$, $P^3$ is principal. $P_2$ has norm $4$, but since $P_1P_2\sim1$, you have $P_2\sim P^2$. $Q$ also hs norm $3$, but it’s equiv to $P$. $P_3$ has norm $5$, so $P_4$ has norm $25$, so it’s not a problem. Seems to me that you’ve covered all bases for the ideal class group to be of order $3$. – Lubin Jul 29 '16 at 2:46
• My comment above may be worthless, but for the Class Field, I’d first look at the cubic subfield of $\Bbb Q(\zeta_{19})$. It’s unramified outside $19$, and you’d have a good chance of showing that its compositum with $K$ is unramified above $19$, over $K$. – Lubin Jul 29 '16 at 3:21

It turns out that the answer to 1 is no as commented by @JohnM. We need to prove that $P$ is not principal. To do this, assume that $P=\beta R$. Then $||P||=||\beta||=3$. So, now the question becomes whether there exists an element $\beta\in R$ with $||\beta||=3$.

Since any element in R can be expressed as $\beta=a+b\alpha+c\alpha^2$ with $a, b, c\in \mathbb{Q}$ and $3a, 3b, 3c \in \mathbb{Z}$, $3a\equiv 3b\equiv 3c \ \mathrm{mod} \ 3$, we see that $$||\beta||=3=a^3+19b^3+19^2c^3-3\cdot 19 abc$$ and $$27\cdot 3 =81= (3a)^3 + 19(3b)^3 + 19^2 (3c)^3 - 3 \cdot 19 (3a)(3b)(3c)\equiv (3a)^3 \ \mathrm{mod} \ 19$$ However, $81 \equiv 5 \ \mathrm{mod} \ 19$ is not a cubic residue modulo $19$ (list of cubic residues mod $19$ is $0, 1, 7, 8, 11, 12, 18$). Therefore, there is no element $\beta\in R$ of norm $3$. This proves that $P$ is not principal.

Then the rest of argument will follow as @Lubin commented. Thus, we needed to solve (b) to conclude, but (c) and (d) are not necessary. This answers my question 2.

To find the Hilbert class field, we use a cubic subfield $K'$ of the cyclotomic field $\mathbb{Q}(\zeta_{19})$ as commented by @Lubin. This method was already discussed by @franzlemmermeyer in the answer to this question.

Let $L$ be composite field of $K'$ and $K=\mathbb{Q}(\alpha)$. Let $\mathcal{P}$ be a prime in $L$. If $\mathcal{P}$ lies over rational prime $p\neq 3, 19$, then $p$ is unramified in both extensions $K$ and $K'$. Thus, the prime $\mathcal{P}\cap K$ lying under $\mathcal{P}$ is unramified in $L$.

Suppose that $\mathcal{P}$ lies over rational prime $3$. Since the rational prime $3$ is inert in $K'$, the residue field degree must be $3$. By the decomposition $3R=P^2Q$, the residue field degree for $P$ and $Q$ over rational prime $3$ must be both $1$. This gives the residue field degree for $\mathcal{P}$ over both $P$ and $Q$ must be $3$, yielding that the ramification indices for $\mathcal{P}$ over both $P$ and $Q$ are $1$.

Now assume that $\mathcal{P}$ is over rational prime $19$. As @franzlemmermeyer answered in the above link, we use Abhyankar's lemma. Since $19$ is totally ramified in both $K$ and $K'$, we have by Abhyankar's lemma that the ramification index for $\mathcal{P}$ over $\mathcal{P}\cap K$ is $1$.

Hence, the extension $L$ over $K$ is unramified over $K$, yielding that $L$ is the Hilbert class field of $K$. This answers my question 3.

Acknowledgement @JohnM, @Lubin, Thank you very much for the helpful comments with great insights.