In Marcus book “Number Fields” there is an exercise which asks to calculate the Hilbert class field group of $\mathbb{Q}(\sqrt[3]{19})$ I know that the ring of integers is $R=\mathcal{O}_K=\mathbb{Z}[\alpha,\beta]$, where $\alpha=\sqrt[3]{19}$ and $\beta=\frac{\alpha^2+\alpha+1}{3}.$ After some calculations I have that $disc{K}=-3\cdot19^2$ and that the indexes of $\alpha$ and $\beta$ are, respectively $3,2.$ The Minkowski bound say that I only have to consider the following ideals that I have factored using Kummer: $$2R=(2,\alpha-1)(2,3\beta)$$ $$3R=(3,\beta)^2(3,\beta-1)$$ $$5R=(5,\alpha+1)(5,\alpha^2-\alpha+1)$$ $$7R=(7).$$ Using a preceding exercise, I know that $$N(a+b\alpha+c\alpha^2)=a^3+19b^3+19^2c^3-57abc.$$ I observe that the congruence $x^3\equiv m\mod19$ can be satisfied only if $m\equiv0,1,7,8,11,12,18.$ I can deduce that there aren’t elements of norm 2,5, and a similar arguments holds for the primes over $3.$
However, there exist elements of norm $18,45$ respectively $\alpha-1$ and $\alpha-4$.
With direct computation, I have found that $$\alpha-4\in(3\beta)(3,\beta-1)(5,\alpha+1)$$ I would like to calculate the norm of the prime ideal $(5,\alpha+1)$ I know it is a power of $5,$ more precisely $5$ or $25.$ How can I calculate this norm? For example, if I would like to calculate it computing $$|(5,\alpha+1)/(5,\alpha+1)^2|,$$ how can I proceed?
If it is $5$, in this way I should conclude that the class field group is cyclic with the class of $(3,\beta)$ as generator. Finally, how can I prove that the order of the group is divisible by $3$? I think I have to match the powers of the ideal class $(3,\beta)$ in the right way, but I’m not completely sure.
The norm of the ideal $I\subset\mathcal{O}_K$ is the order of the quotient $$\mathcal{O}_K/I=\Bbb{Z}[\alpha,\beta]/(5,\alpha+1)\cong\Bbb{F}_5[\beta],$$ which is easily verified to be $5$ because $\alpha=\sqrt[3]{19}=4\in\Bbb{F}_5$.