# Ideal class group of $K = \mathbb{Q}[x]/ (x^{3} + 11x + 21)$

In this question, I asked about some non-cyclic and hand-doable example of ideal class group, and Yong Hao Ng the example with minimal discriminant is $$K = \mathbb{Q}(\alpha)$$ with $$\alpha^{3} + 11\alpha + 21 = 0$$. With help of SAGE and some tedious computations, I proved that the class group is generated by two ideal classes $$[(3, \alpha-1)]$$ and $$[(3, \alpha)]$$ and both have order 2. To show that $$(3, \alpha - 1)$$ is not a principal ideal, I proved that $$\epsilon = \alpha^{2}- \alpha - 4$$ is a fundamental unit of $$K$$ by using Artin's inequality. Then if $$(3, \alpha - 1)=(\beta)$$ for some $$\beta\in \mathcal{O}_{K} = \mathbb{Z}[\alpha]$$, $$(\alpha+2) = (3, \alpha - 1)^{2} = (\beta^{2})$$ implies that $$v(\alpha + 2) = \beta^{2}$$ for some unit $$v\in U_{K} =\mathcal{O}_{K}^{\times}$$. We can assume $$v = 1$$ or $$v = \epsilon$$, and for example, if $$\alpha + 2 = \beta^{2} = (a\alpha^{2} + b\alpha +c)^{2}$$, then by expanding this, we should have $$c^{2} + 42ab = 2$$, which is impossible by viewing the equation mod 3. Similarly, we can show that there's no $$\beta\in \mathbb{Z}[\alpha]$$ s.t. $$\beta^{2} = \epsilon(\alpha + 2)= \alpha^{2}- 17\alpha - 29$$ (by mod 7). I can also prove that $$(3, \alpha)$$ is not a principal ideal since $$(3, \alpha)(13, \alpha + 3) = (\alpha + 3)$$ and $$(13, \alpha + 3)$$ is not a principal ideal by the similar method.

To prove that the class group is isomorphic to Klein 4-group, the only thing we have to do now is $$[(3, \alpha -1)] \neq [(3, \alpha)]$$, which is equivalent to that $$\mathfrak{p}_{3}\mathfrak{p}_{3}' = (3, \alpha -1)(3, \alpha)$$ is not a principal ideal. However, I can't prove this by the above method, since I can't find any contradiction.
We have $$(\mathfrak{p}_{3}\mathfrak{p}_{3}')^{2} = (-\alpha-1)$$, so we only need to show that $$(-\alpha - 1)v =\beta^{2}$$ is not possible for any $$v\in U_{K}$$ and $$\beta\in \mathcal{O}_{K}$$. As before, $$-\alpha - 1 = \beta^{2}$$ doesn't have a solution by observing the coefficient of $$\beta^{2}$$. (-1 is not a square mod 3). However, I can' prove that $$(\alpha^{2} - \alpha - 4)(-\alpha - 1) = 16\alpha + 25$$ is not a square in $$\mathcal{O}_{K}$$. This is equivalent to show that the following system of Diophantine equation

$$b^{2} + 2ac - 11a^{2} = 0 \\ 2bc + 22ab - 21a^{2} = 16\\ c^{2} + 42ab = 25$$ doesn't have any integer solution $$(a, b, c)$$. Clealy, the last equation doesn't give any contradiction modulo any prime $$p$$.

So I try to find other nontrivial relation among ideal classes using SAGE, but I failed to find any one of them that help me to prove that the above ideal is not principal. Also, I tried $$v = \epsilon^{2k+1}$$, instead of $$\epsilon$$, and nothing gives a contradiction.

Also, it is enough to show that one of the following (prime) ideals $$\mathfrak{p}_{3}'' = (3, \alpha + 1) \\ \mathfrak{p}_{17} = (17, \alpha - 2) \\ \mathfrak{p}_{23} = (23, \alpha - 8) \\ \mathfrak{p}_{29} = (29, \alpha + 4) \\ \mathfrak{p}_{29} = (29, \alpha - 10)$$ are not principal. But I failed for every above ideal. Do you have any idea? Thanks in advance.

Edit: I just find a way that I can't do without SAGE. If $$\sqrt{16\alpha+25} = \beta\in K$$, then $$\beta$$ $$\left(\frac{\beta^{2}-25}{16}\right)^{3} + 11\left(\frac{\beta^{2} - 25}{16}\right) + 21 = \frac{\beta^{6} - 75\beta^{4} + 4691\beta^{2} - 9}{4096} = 0.$$ However, according to SAGE, the polynomial $$x^{6} - 75x^{4} + 4691x^{2} - 9$$ is irreducible over $$\mathbb{Q}$$, so the degree of $$\beta$$ is 6, which contradicts to $$\beta\in \mathbb{Q}(\alpha)$$. However, I think I can't show that the degree 6 polynomial is irreducible (and even it is hard to compute it by hands.)

As you observed, $$(\mathfrak{p}_{3}\mathfrak{p}_{3}')^{2} = (\alpha+1)$$, so to prove $$\mathfrak{p}_{3}\mathfrak{p}_{3}'$$ is not principal, it suffices to show none of $$\pm(\alpha+1)$$ and $$\pm \epsilon(\alpha+1)$$ is a square in $$\mathcal{O}_K$$, where $$\epsilon = \alpha^{2}- \alpha - 4$$ is the fundamental unit you identified. There is a very efficient way to do this:

Let $$f(x)=x^3+11x+21$$, then $$101\mid f(6)$$. Since the ring of integer is monogenic, we have a homomorphism: $$\mathcal{O}_K \to \mathbb{F}_{101} \qquad \alpha \mapsto 6$$ under this, $$\pm(\alpha+1)$$ is mapped to $$\pm 7$$, but it is easy to check $$\pm 7$$ is not a quadratic residue modulo $$101$$. Hence $$\pm(\alpha+1)$$ is not a square in $$\mathcal{O}_K$$. The same prime $$101$$ does not work for $$\pm \epsilon(\alpha+1)$$, because they are mapped to $$\pm (6^2-6-4)(6+1)$$, both are quadratic residue of $$101$$.

Therefore we try another prime. Note that $$29\mid f(10)$$, use the homomorphism: $$\mathcal{O}_K \to \mathbb{F}_{29}, \alpha \mapsto 10$$, $$\pm \epsilon(\alpha+1)$$ is mapped to $$\pm (10 + 1) (10^2 - 10 - 4)$$ which are quadratic nonresidue modulo $$29$$.

• Thank you very much! You saved my time... Commented Oct 17, 2018 at 7:32
• I realized that I already used this argument to prove that some element is a fundamental unit... Commented Oct 17, 2018 at 15:24

I think you can show directly that $$\alpha\not\in (3,\alpha-1)$$ so $$(3,\alpha)$$ and $$(3,\alpha-1)$$ cannot be equal.

Suppose instead that $$\alpha \in (3,\alpha-1)$$, such that $$\alpha = 3\cdot g(\alpha) + (\alpha-1) \cdot h(\alpha)$$ for some $$g(\alpha),h(\alpha) \in \mathcal O_K = \mathbb Z[\alpha]$$. Then letting $$h(\alpha) = a + b\alpha + c\alpha^2$$ and $$a,b,c \in\mathbb Z$$ we have \begin{align} \alpha &= 3\cdot g(\alpha) + (\alpha-1)(a+b\alpha + c\alpha^2)\\ &= (-a - 21 c) + (a - b - 11 c)\alpha + (b - c) \alpha^2 +3\cdot g(\alpha)\\ \alpha &\equiv -a + (a-b-11c)\alpha + (b-c)\alpha^2 \pmod 3 \end{align} Therefore $$c \equiv b \pmod 3,\quad a\equiv 0 \pmod 3,$$ giving $$\alpha \equiv (-b-11b)\alpha \equiv 0 \pmod 3$$ which is a contradiction.

Hence we cannot have $$(3,\alpha) = (3,\alpha-1)$$

• But why it implies that their ideal class are not the same? We have to show that $(3, \alpha)(\gamma) = (3, \alpha - 1)$ is impossible for any $\gamma\in K$ and a fractional ideal $(\gamma)$. Also, there's easier way to show that they are different - if $(3, \alpha) = (3, \alpha - 1)$, then $\alpha, \alpha -1\in (3, \alpha - 1)$, so $1\in (3, \alpha - 1)$ and $(3, \alpha -1)$ should be a unit ideal - which is completely impossible. Commented Oct 17, 2018 at 5:38
• @SeewooLee Yeah, you're right. Commented Oct 17, 2018 at 6:10