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Let $K=\mathbb{Q}(a)$ where $a^3=d$ where $d\neq 0, \pm 1$ is a square free integer. Show that $\Delta (1, a, a^2)=-27d^2$. By calculating the traces of $\theta, a\theta, a^2\theta$ where $\theta=u+va+wa^2$ with $u,v,w\in \mathbb{Q}$ and the norm of $\theta$, show that the ring of integers $\mathcal{O}_K\subset \frac{1}{3}\mathbb{Z}[a]$.

So far I have calculated the the traces of $\theta, a\theta, a^2\theta$ are, respectively, $3u, 3wd, 3vd$ and the norm of $\theta$ is $u^3+v^3d+w^3d^2-3uvwd$. But I couldn't proceed.

I want to show that $\mathbb{Z}[a]$ has index 3 in $\mathcal{O}_K$. I think the following theorem should be useful.

Let $K$ be a number field. If $x_1, \dots, x_n\in \mathcal{O}_K$ is a basis for $K$ over $\mathbb{Q}$ and $M=x_1\mathbb{Z}+\dots+x_n\mathbb{Z}$ then $\Delta (x_1, \dots, x_n)=(\mathcal{O}_K:M)^2D_K$ where $(\mathcal{O}_K:M)$ denotes the index of a subgroup and $D_K$ is the discriminant of the number field $K$.

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  • $\begingroup$ Could we use the result that $\Delta (1,a,a²)R \subset Z[a]$, and then examining each cases? $\endgroup$ – awllower Feb 8 '13 at 4:57
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Hint/Outline: Assume that $\theta\in\mathcal{O}_K$. By calculating the trace of $\theta$ you know that $3u\in\mathbb{Z}$, $3wd\in\mathbb{Z}$, $3vd\in\mathbb{Z}$. Now, we also know that $$d\cdot3^3\cdot N_{\mathbb{Q}}^K (\theta)\in\mathbb{Z}$$ since the norm itself is in $\mathbb{Z}$. However, this tells us that $d^2 (3w)^3\in\mathbb{Z}$, and so $3w\in\mathbb{Z}$. Applying the same reasoning looking at $$3^3\cdot N_{\mathbb{Q}}^K (\theta)\in\mathbb{Z},$$ we conclude that $3v\in\mathbb{Z}$. Use this to conclude the desired result.

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