# Is field-basis of integral elements an integral basis?

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 $$K$$?

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.

• @darijgrinberg, thanks for the link. – lhf Dec 3 at 0:31

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.