I'm trying to compute the ideal class group of $\mathbb Q(\sqrt[3]{3})$, and I would like to know if my calculations are right and if I could improve my arguments.
Let $K=\mathbb Q (\sqrt[3]{3}),$ its ring of integers $\mathcal O_K$ is $ \mathbb Z[\sqrt[3]{3}]$, since Disc$(1,\sqrt[3]{3},\sqrt[3]{3^2})=-3^5 \quad $ and $f(x)=x^3-3 \ $ is Eisenstein for $3$.
The Minkowski bound is $\frac{4}{\pi}\cdot\frac{6}{27}\cdot \sqrt{3^5}<5$. Thus we need to check ideals of norm less than $5$. Set $a=\sqrt[3]{3}$ to facilitate notation.
The only ideal of norm 1 is $\mathcal{O}_K$
Every ideal contains its norm, thus ideals of norm $2$ are prime factors of $(2). \quad $ $ (2)=(a-1)(a^2+a+1) \ $, since $2=3-1=a^3-1=(a-1)(a^2+a+1). N(a-1)=2, N(a^2+a+1)=4 $.
$(3)=(a)^3$ and $(a)$ is principal.
Since $(2)=(a-1)(a^2+a+1),$ ideals of norm 4 are:
$(a-1)^2, \ (a^2+a+1).$
All of the ideals above are principal, thus the ideal class group is trivial and $\mathcal O_K$ is a PID.
Best Answer
Your computations are correct and they seem quite optimal. Here is a verification in SageMath:
Of course you meant to say $\mathcal{O}_K$ is a PID. ($K$ is also always a PID.)