# Calculate the ideal class group of $K=\mathbb{Q}(\sqrt[3]{11})$

$$\textbf{Calculate the ideal class group of K=\mathbb{Q}(\sqrt[3]{11})}$$:
Let $$\alpha=\sqrt[3]{11}.$$ We need the fact that the ring of integer of $$K$$ is $$\mathbb{Z}[\alpha]$$.
One basis:$$\{x_1,x_2,x_3\}$$. Let $$\theta_1=\alpha, \theta_2=\alpha w, \theta_3=\alpha w^2$$. Let the $$i$$-th embedding takes $$x$$ to $$\theta_i$$. We have the following matrix: $$A=\begin{pmatrix} 1 & \alpha& \alpha^2 \\ 1 & \alpha w &\alpha^2 w^2 \\ 1 & \alpha w^2& \alpha^2 w \end{pmatrix}$$ This matrix is Vandermonde and thus $$\det(A)^2=(\theta_3-\theta_1)^2(\theta_2-\theta_1)^2(\theta_3-\theta_2)^2=-3^2\cdot 11^2\cdot (w^2-w)^2=-3^3\cdot 11^2$$
Then the Minkowski constant of $$K$$ is $$M_K=\sqrt{|\Delta_K|}\Big(\frac{4}{\pi}\Big)^{r_2}\frac{n!}{n^n}<17$$ We need to see how $$2,3,5,7,11,13$$ factor in $$\mathcal{O}_K$$:
$$2:$$ $$f(x)=x^3-11\equiv (x-1)(x^2+x+1) \pmod 2$$
$$\implies 2\mathcal{O}_K=\mathfrak{p}_2\mathfrak{p}_2', \mathfrak{p}_2=(2,\alpha-1), \mathfrak{p}_2'=(2,\alpha^2+\alpha+1)$$
$$3:$$ $$f(x)=x^3-11\equiv (x+1)^3\pmod 3$$
$$\implies 3\mathcal{O}_K=\mathfrak{p}_3^3, \mathfrak{p}_3=(3,\alpha+1)$$
$$5:$$ $$f(x)=x^3-11\equiv (x-1)(x^2+x+1) \pmod 5$$
$$\implies 5\mathcal{O}_K=\mathfrak{p}_5\mathfrak{p}_5', \mathfrak{p}_5=(5,\alpha-1), \mathfrak{p}_5'=(5,\alpha^2+\alpha+1)$$
$$7:$$ $$f(x)=x^3-11\equiv x^3-4 \pmod 7$$
$$\implies 7\mathcal{O}_K=\mathfrak{p}_7$$, $$\mathfrak{p}_7=(7)$$
$$11:$$ $$f(x)=x^3-11\equiv x^3 \pmod {11}$$
$$\implies 11\mathcal{O}_K=\mathfrak{p}_{11}^3, \mathfrak{p}_{11}=(11,\alpha)=(\alpha)$$
$$13:$$ $$f(x)=x^3-11\equiv x^3+2 \pmod {13}$$
$$\implies 13\mathcal{O}_K=\mathfrak{p}_{13},$$ $$\mathfrak{p}_{13}=(13)$$
Thus, $$Cl(\mathcal{O}_K)$$ is generated by $$[\mathfrak{p}_2],[\mathfrak{p}_3],[\mathfrak{p}_5]$$.
The minimal polynomial of $$\alpha-t(t\in \mathbb{Z})$$ is $$(x+t)^3-11$$ and thus $$N(\alpha-t)=t^3-11,(\alpha-1)=\mathfrak{p}_2\mathfrak{p}_5, (\alpha-2)=\mathfrak{p}_3, (\alpha+1)=\mathfrak{p}_2^2\mathfrak{p}_3$$ $$\color{red}{I\ understand\ that\ the\ norm\ of\ the\ ideals\ on\ each\ side\ is\ the\ same,\ so\ why\ we\ can\ deduce\ these\ ideals\ are\ the\ same,}$$ $$\color{red}{just\ because\ their\ norm\ are\ the\ same?}$$

If $$[\mathfrak{p}_2]=1,$$ then $$cl(\mathcal{O}_K)=\{0\}$$. If $$[\mathfrak{p}_2]\neq 1,$$ then $$cl(\mathcal{O}_K)\cong \mathbb{Z}/2\mathbb{Z}$$.
The question now turns out to be: whether $$\mathfrak{p}_2$$ is principal or not?
Let's compute the unit group of $$K$$ first.
Assume that $$\mathfrak{p}_2^2=(\beta)$$ for some $$\beta\in \mathcal{O}_K$$ and then $$(\beta)(\alpha-2)=(\alpha+1)$$ Take $$\beta=\frac{\alpha+1}{\alpha-2}=\alpha^2+2\alpha+5$$ and then we have $$(\beta)^2=\mathfrak{p}_2^4=(\alpha-3)$$. $$\color{red}{how\ to\ get\ this, does\ it\ follow\ from\ ((\alpha^2+2\alpha+5)^2)=(\alpha-3)?}$$
Since $$N(\alpha-3)=16$$ and $$(\alpha-3)\neq \mathfrak{p}_2^{2}\mathfrak{p}_2'$$, we have a unit in $$\mathcal{O}_K$$ $$u=-\frac{\beta^2}{\alpha-3}=18\alpha^2+40\alpha+89\approx 267$$ Let $$\epsilon$$ be the fundamental unit of $$K$$. We have $$\epsilon>\sqrt[3]{\frac{3267-24}{4}}\approx 9.3437$$ and thus $$\epsilon^2>87, \epsilon^3>815$$ which follows that $$u=\epsilon$$ or $$u=\epsilon^2$$.
If $$u=\epsilon^2,$$ we construct a homomorphism $$\phi:\mathbb{Z}[\alpha]\to \mathbb{F}_5$$ by reduction modulo $$\mathfrak{p}_5$$. Then as $$\phi(u)=2$$ and $$2$$ is not a square in $$\mathbb{F}_5$$, we get a contradiction. Thus $$\mathcal{O}_K^{*}=\{\pm u^m\}$$. (more details later) $$\color{red}{could\ someone\ help\ write\ more\ details\ on\ this\ map, I\ am\ confused\ on\ how\ it\ works.}$$

By your previous analysis, $$\alpha-1 \in \mathfrak{p}_2\cap\mathfrak{p}_5 = \mathfrak{p}_2\mathfrak{p}_5$$, so there is an ideal $$I$$ such that $$\mathfrak{p}_2\mathfrak{p}_5I=(\alpha-1)$$.

By computing norm, we see that $$I$$ has norm one. Hence $$I=\mathcal{O}_K$$ and $$(\alpha-1)=\mathfrak{p}_2\mathfrak{p}_5$$.

You can do the same for $$\alpha-2$$ and $$\alpha+1$$.

You don’t have to assume that $$\mathfrak{p}_2^2=(\beta)$$, you already know that.

The point is that by what you did previously, $$\mathfrak{p}_2^2(\alpha-2)=\mathfrak{p}_2^2\mathfrak{p}_3=(\alpha+1)$$. Hence, $$\mathfrak{p}_2^2$$ is generated by $$\beta=\frac{\alpha+1}{\alpha-2}$$.

By definition of $$\beta$$, $$\beta^2$$ generates the ideal $$(\mathfrak{p}_2^2)^2=\mathfrak{p}_2^4$$.

What is the ideal generated by $$(\alpha-3)$$? Well, $$\mathbb{Z}[\alpha]/(\alpha-3)=\mathbb{Z}[x]/(x^3-11,x-3)=\mathbb{Z}/(27-11)=\mathbb{Z}/16\mathbb{Z}$$, so $$(\alpha-3)$$ is a fourth power of a single prime ideal of norm $$2$$, hence $$(\alpha-3)=\mathfrak{p}_2^4$$.

Consider the morphism $$\phi$$ of reduction modulo $$\mathfrak{p}_5$$ (note that $$\mathbb{Z}[\alpha]=\mathbb{F}_5$$), in particular $$\phi(\alpha-1)=0$$.

Then $$\phi(u)=\phi(18\alpha^2+40\alpha+89)=3+4=2$$, so $$\pm u$$ is not a square modulo $$\mathfrak{p}_5$$, so $$\pm u$$ is not a square in $$\mathbb{Z}[\alpha]$$.

Let $$\epsilon$$ denote the unit which is positive with the smallest modulus in the real embedding.

Since $$|u| < |\epsilon|^3$$ and $$u>0$$, it follows that $$u=\epsilon$$.

But we have to get back to the question: is $$\mathfrak{p}_2$$ principal (which we only danced around)?

Because $$u$$ generates the units, the generators of $$\mathfrak{p}_2^2$$ are exactly the $$\pm \beta u^m$$ for $$m \in \mathbb{Z}$$. So the question is: is there a $$\pm \beta u^m$$ which is a square?

Since $$u,\beta>0$$, we need to decide if for some $$m \in \{0,1\}$$, $$\beta u^m$$ is a square.

Since $$\phi(\beta)=3$$, $$\beta$$ is not a square, so $$\mathfrak{p}_2$$ is principal iff $$\beta u=259\alpha^2+576\alpha+1281$$ is a square.

Let’s reduce it modulo $$\alpha-3$$ (with $$\mathbb{Z}[\alpha]/(\alpha-3)=\mathbb{Z}/16\mathbb{Z}$$). Then $$\beta u \equiv 3\cdot 3^2+1 \equiv 28\equiv 12\pmod{\alpha-3}$$. But $$12$$ is not a square modulo $$16$$, so $$\beta u$$ is not a square (you could also check that the norm of $$\beta u$$ was negative). Therefore $$\mathfrak{p}_2$$ is not principal and the class group has order $$2$$.

• Thank you for your answer. There are some points that I am still confused: How to show $\mathbb{Z}[\alpha]/(\alpha-3)\cong \mathbb{Z}/16\mathbb{Z}$ and $\mathbb{Z}[\alpha]\cong \mathbb{F}_5$? Commented Jul 16 at 9:40
• This is elementary commutative algebra. $\mathbb{Z}[\alpha]/(\alpha-3) \simeq \mathbb{Z}[x]/(x-3,x^3-11) \simeq \mathbb{Z}[x]/(x-3,3^3-11) \simeq \mathbb{Z}[x]/(x-3,16) \simeq \mathbb{Z}/(16)$. The same argument works for $\mathbb{Z}[\alpha]/(\alpha-1)$, but you get $\mathbb{Z}/(10)$ (which has a projection to $\mathbb{F}_5$). Commented Jul 16 at 9:54
• Get it, thank you for the complete answer. Commented Jul 16 at 9:59