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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6No. Not even a strategic choice of $\alpha$ helps. See Keith Conrad, Rings of integers without a power basis. – darij grinberg Dec 02 '19 at 22:22
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1As in the answers, "no", ... but, sometimes it's sufficient to know that, by free $\mathbb Z$-module ranks, you'd be off only by finite index, ... and only finitely-many primes divide the index, ... and, locally, almost everywhere you would have an integral basis for the (local) integers. :) – paul garrett Sep 30 '21 at 17:39
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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.
lhf
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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.
Lubin
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