$\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) $\color{red}{could\ someone\ help\ write\ more\ details\ on\ this\ map, I\ am\ confused\ on\ how\ it\ works.}$
Thank you for your help!!
Best Answer
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 $2$.