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Let $K$ be a number field. Then the ring of integers of $K$ is defined as $K \bigcap \mathbb{B} = \mathcal{O}_{K}$. An integral basis of $K$ is defined as a set of elements $b_{1}, b_{2},\dots, b_{n}$ such that they form a $\mathbb{Z}$-basis for $\mathcal{O}_{K}$ What exactly does a $\mathbb{Z}$-basis mean? Does it mean that these $b_{i}$'s are linearly independent and also span $\mathcal{O}_{K}$ with the coefficents being integers?

Secondly,given a ring extension $\mathbb{Z}[\alpha]$, where $\alpha$ is an integral element, and suppose that the degree of the minimal polynomial of $\alpha$ is $d$, then would the generating set of $\mathbb{Z}[\alpha]$ be $\{1,\alpha,\dots, \alpha^{d-1}\}$?

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  • $\begingroup$ Yes to both questions, but something extra needs to be said for the second. The set you mention there is a $\Bbb Z$-basis of $\Bbb Z[\alpha]$ all right, but often one says that $\{\alpha\}$ is a generating set, because it generates $\Bbb Z[\alpha]$ as a ring. $\endgroup$ – Lubin May 8 '15 at 12:34
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For the second question, the answer is yes. In particular, there is a group isomorphism $\Bbb{Z}^n \cong \mathcal{O}_K$ given by $$(a_1, \dots , a_n) \mapsto a_1 b_1 + \dots + a_nb_n$$

For the second question, the answer is yes. It is enough to show that all powers $\{ \alpha^k \}_{k \ge 0}$ are generated by $\{ 1, \alpha, \dots, a^{d-1} \}$. This can be proved easily by induction on $k$, using the fact that $$\alpha^d = - (a_{d-1}\alpha^{d-1}+ \dots +a_1 \alpha + a_0)$$

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