# maximal vs non-maximal order in an algebraic number field

I am trying to determine whether an order in a (cubic) number field is maximal or not.

I have picked up two different fields. One has a power basis the other does not have it.

1) Let $K=\mathbb{Q}(\sqrt[3]{2})$. Since $(2)^2\not\equiv 1 \pmod 9$ we have that $\{1,\sqrt[3]{2},\sqrt[3]{2^2}\}$ is an integral basis (in fact a power basis) and $$d_{K\supset\mathbb{Q}}\{1,\sqrt[3]{2},\sqrt[3]{2^2}\}=[\mathcal{O_{K}}:\mathbb{Z}[\sqrt[3]{2}]]^2d(K)$$ $$-3^3\cdot 2^2=[\mathcal{O_{K}}:\mathbb{Z}[\sqrt[3]{2}]]^2d(K)$$

And it can be shown that $2\nmid [\mathcal{O_{K}}:\mathbb{Z}[\sqrt[3]{2}]]$ and $3\nmid [\mathcal{O_{K}}:\mathbb{Z}[\sqrt[3]{2}]]$

Hence

$$[\mathcal{O_{K}}:\mathbb{Z}[\sqrt[3]{2}]]=1$$ and $\mathcal{O_{K}}=\mathbb{Z}[\sqrt[3]{2}]$ is a maximal order in $K$

Is this a coincidence or do we always get a maximal order (index =1) when the power basis exists? Also in this case the discriminant of the field is equal to the discriminant of the minimal polynomial.

On the other hand if we take

$K=\mathbb{Q}(\sqrt[3]{10})$ Then since $10\equiv 1\pmod 9$ then the integral basis will be $\{1,\sqrt[3]{10},\frac{1+\sqrt[3]{10}+\sqrt[3]{10^2}}{3}\}$ but in this case it is not a power basis. I want again to determine the index $[\mathcal{O_{K}}:\mathbb{Z}[\sqrt[3]{10}]]$, can I use again $$d_{K\supset\mathbb{Q}}\{\alpha\}=[\mathcal{O_{K}}:\mathbb{Z}[\sqrt[3]{10}]]^2d(K)$$ or is this formula only allowed for power bases? How do I compute the index? And is $\mathbb{Z}[\sqrt[3]{10}]$ the maximal order in $K$ Here the discriminant of the field $d(K)=-3\cdot2^2\cdot5^2$ is not equal to discriminant of the minimal polynomial $d(f)=-3^52^4$

Thank you

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Doesn't "having a power basis" mean precisely that there is an algebraic integer $\alpha$ such that $\mathbb{Z}[\alpha] = \mathcal{O}_K$ is the maximal order? (In other words, yes: $\mathbb{Z}[\alpha]$ is the maximal order exactly when $1,\alpha,\ldots,\alpha^{n-1}$ forms an integral basis for $\mathcal{O}_K$.) – Pete L. Clark Feb 24 '14 at 19:49
The formula you wrote is true for any containment of orders. – Alex Youcis Feb 25 '14 at 7:59
@PeteL.Clark: How do we show for example that $\{1,\sqrt[3]{10},\frac{1+\sqrt[3]{10}+\sqrt[3]{10^2}}{3}\}$ is not a power basis for $K=\mathbb{Q}(\sqrt[3]{10})$? I have been told that is it not enough just to write that it is not of the form $\{1,\theta,\theta^{2}\}$ for some $\theta\in\mathcal{O}_{K}$ – H.E Mar 27 '14 at 12:51
J.W.S: You seem confused between "These linearly independent algebraic integers are not a power basis for $\mathcal{O}_K$" and "$\mathcal{O}_K$ does not have a power basis". Of course what you wrote down is not a power basis, just because the third number is not the square of the second one. But maybe there is some other power basis: that is probably what you have been asked. – Pete L. Clark Mar 27 '14 at 14:57
@PeteL.Clark: Actually I have found a power basis for this particular field and it is not of the form $\{1,\sqrt[3]{10},\frac{1+\sqrt[3]{10}+\sqrt[3]{10^2}}{3}\}$ I guess for him it is not enough (as an argument) to observe that the third element in a basis is not a square of the second one. Hence my question is: how to show that $\{1,\sqrt[3]{10},\frac{1+\sqrt[3]{10}+\sqrt[3]{10^2}}{3}\}$ is not a power basis? – H.E Mar 27 '14 at 16:10