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.


1 Answer 1


Your computations are correct and they seem quite optimal. Here is a verification in SageMath:

sage: x = polygen(QQ)
sage: f = x^3 - 3
sage: f.is_irreducible()
sage: K.<a> = NumberField(f); K
Number Field in a with defining polynomial x^3 - 3
sage: OK = K.ring_of_integers(); OK
Maximal Order in Number Field in a with defining polynomial x^3 - 3
sage: OK.basis()
[1, a, a^2]
sage: OK.discriminant().factor()
-1 * 3^5
sage: M = K.minkowski_bound(); M
sage: bool(M == 4/pi * 6/27 * sqrt(3^5))
sage: bool(M < 5)
sage: OK.ideal(2).factor()
(Fractional ideal (a^2 + a + 1)) * (Fractional ideal (a - 1))
sage: OK.ideal(3).factor()
(Fractional ideal (a))^3
sage: K.class_number()
sage: OK.class_group()
Class group of order 1 of Number Field in a with defining polynomial x^3 - 3

Of course you meant to say $\mathcal{O}_K$ is a PID. ($K$ is also always a PID.)

  • $\begingroup$ Thanks a lot for your help! I should definetely learn how to use sage, and thanks for spotting the typo! $\endgroup$
    – Alessandro
    Feb 14, 2021 at 20:28

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