Show that the ring of integers $A$ of the cubic field $\mathbb Q[x]$ with $x^3=2$ is principal.

Question

Show that the ring of integers $A$ of the cubic field $K=\mathbb Q[x]$ with $x^3=2$ is principal.

The hint given in the book is to majorize the discriminant of $A$ by $D(1,x,x^2)$ and then use the fact that every ideal class of a number field $K$ of degree $n$ contains an ideal $b$ such that $$N(b) \leq \left(\frac {4}{\pi}\right)^{r_2}\frac{n!}{n^n}|d|^\frac{1}{2}.$$ Here $2r_2$ is the number of embeddings $\sigma:\Bbb Q[x]\to\Bbb C$ such that $\sigma(K) \not \subset \mathbb R$, and $d$ is the absolute discriminant of $K$.

I have no clue about how to proceed. Any help?

Answer 1

The hint really says it all. First note that $1, x,x^2$ are integral in your field. They satisfy $t-1=0,\, t^3-2=0,\, t^3-4=0$ which are monic, integral polynomials. They are also a $\Bbb Q$-basis for the number field, so if we have an integral basis, $\alpha_0,\alpha_1,\alpha_2$ for $K$, then you know there is an integer matrix, $M$ so that $M\alpha_i=x^i$.

The discriminant of this new basis is just

$$\operatorname{disc}(1,x,x^2)=(\det M)^2\cdot \operatorname{disc}(\alpha_0,\alpha_i,\alpha_2).$$

and since $\det M\in\Bbb Z$ we have that

$$|\operatorname{disc}(1,x,x^2)|\ge |\operatorname{disc}(\alpha_0,\alpha_i,\alpha_2)|=\Delta_K.$$

The discriminant of $\{1,x,x^2\}$ is easily computed as the determinant

$$\operatorname{disc}(1,x,x^2)=\det(\operatorname{Tr}(x^{i-1}x^{j-1}))=\begin{vmatrix} 3 & 0 & 0 \\ 0 & 0 & 6 \\ 0 & 6 & 0\end{vmatrix}=-108.$$

All right, so the hard work is over, now for the Minkowski estimate. Since $\Bbb Q(\sqrt[3]{2})$ is readily seen to have $r_1=1, r_2=1, n=3$ we compute that all ideals, $\mathfrak{b}$ possess a representative, $\xi$ of norm bound

$$N(\xi)\le \left({4\over\pi}\right)\left({6\over 27}\right)\sqrt{108}\approx 2.94.$$

This means that the only possible ideals which could not be principal are those of norm less than or equal to $2$. But we already know that $(2)=(\sqrt[3]{2})^3$ is totally ramified, so that $(\sqrt[3]{2})$ is the only ideal above $2$ in $K$ and that $(\sqrt[3]{2})$ is of norm $2$ and is principal. Hence all ideals of $K$ are principal.