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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?

Watson
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odnerpmocon
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1 Answers1

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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.

Adam Hughes
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  • Shouldn't it be $\text{disc}(1,x,x^2)=(\text{det}M)^2⋅\text{disc}(α_0,α_1,α_2)$ though? – odnerpmocon Apr 13 '15 at 04:52
  • Also, I think you meant $A$ instead of $K$ in several places. But thanks, nice solution! – odnerpmocon Apr 13 '15 at 05:04
  • @odnerpmocon ah yes, the determinant should be squared, that's a typo. I always mean $K$ when I say it though: it's a common thing to refer to the field despite the objects usually being associated more directly with the integer ring, as you can see this doesn't really cause an irreparable amount of confusion, though does take a bit of getting used to. – Adam Hughes Apr 13 '15 at 06:32
  • I see. Also, I'm confused about traces in computing the discriminant - can you explain why the trace of $x$ is zero? – odnerpmocon Apr 13 '15 at 07:39
  • @odnerpmocon yes, remember that the trace of an algebraic number, $\alpha$ is the trace of the linear transformation induced by multiplication by $\alpha$, $m_\alpha$. We know the minimal polynomial for the transformation $m_x$ is equal to $p_x(t)=t^3-2$, the minimal polynomial for $x$. Since the trace is the coefficient of the second highest degree term, in this case the quadratic term, it is easily observed to be $0$. Similarly the trace of $x^2$ is $0$ since its minimal polynomial is $t^3-4$. – Adam Hughes Apr 13 '15 at 08:01
  • I am trying to unravel the last paragraph; could you see if I am reasoning this out correctly?

    The only ideals that aren't principal must have norm $\leq$ the Minkowski bound $N_0$ since otherwise, take an ideal $I$ with $N(I) > N_0$. Then the class $I \cdot F(A)$ has an ideal $I_0$ with $N(I_0) < N_0$. But $N(I) \leq N(I_0)$ since ideal norm is multiplicative.

    We know $I = (\sqrt[3]{2})$ is prime, and hence maximal, since $\mathbb{Z}[\sqrt[3]{2}]/I \cong \mathbb{Z}$, which is an integral domain. The norm is 2 since the other cosets contain $\sqrt[3]{4} + I$ and $1 + I$.

     – notes Jun 01 '15 at 17:21
  • (continued:) We also have $(2) \subset \xi$ and $(2) = I^3 \subset I$. Since $I$ is maximal, if the ideal $\xi \neq I$, then $\xi + I$ is an ideal larger than $I$ containing $I$, a contradiction. So we must have $\xi = I$. So every ideal class contains the principal ideal $I$, which means the class group is trivial. – notes Jun 01 '15 at 17:26
  • @MichaelZhao This is inaccurate, $\Bbb Z[\sqrt[3]{x}]/(\sqrt[3]{2})\cong\Bbb Z/2\Bbb Z$. Also it's excessively confusing to list $\sqrt[3]{4}\in (\sqrt[3]{2})$ as the representative for $(\sqrt[3]{2})$, just use $0$. You also mean $(\xi)+(\sqrt[3]{2})$, not $\xi+(\sqrt[3]{2})$ since the latter is a coset, not an ideal. The last paragraph's logic involves the number of possible ideals above a given ideal based on the degree, that is the logic used there. Hopefully that helps. – Adam Hughes Jun 01 '15 at 18:36