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Suppose we have a finite field extension $K = \mathbb{Q(\alpha)}$ with basis $1,\alpha,\dots,\alpha^{n-1}$ where all $\alpha^i$ are integral elements. Do they form an integral basis of the ring of integers $\mathcal{O}_K$ of $K$?

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No. Take for instance $\alpha = \sqrt 5$.

Even worse, there may not exist a suitable $\alpha$. This is the case for the cubic field generated by a root of the polynomial $X^{3}-X^{2}-2X-8$, according to Wikipedia. For a discussion and a proof, see Rings of integers without a power basis by Keith Conrad.

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  • $\begingroup$ @darijgrinberg, thanks for the link. $\endgroup$ – lhf Dec 3 at 0:31
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No, not at all. Take your proposed basis $\{1,\alpha,\cdots, \alpha^{n-1}\}$, a field basis consisting of algebraic integers. Now look at $\{1,2\alpha,4\alpha^2,\cdots,2^{n-1}\alpha^{n-1}\}$, equally a field basis consisting of powers of an algebraic integer.

But since the corresponding $\Bbb Z$-modules have the property that the smaller is of index $2^{n(n-1)/2}$ in the larger, the two discriminants differ by a factor equal to the square of this, namely $2^{n(n-1)}$. Yet two integral bases of the same field must have the same discriminant. So at the very least, the second basis is not an integral basis, contrary to your conjecture.

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