In the proof of Proposition 5.2 in Neukirch's Algebraic number theory, it is stated that $\alpha_1, \ldots, \alpha_n$ is a $\mathbb{Z}$-basis of an ideal of $o_K$ with $K$ an algebraic number field of degree $n$.
Why does the number of basis elements in a $\mathbb{Z}$-basis of an ideal in $o_K$ match the degree of the field extension?
Any $\alpha\in k$ is algebraic over $\mathbb{Q}$; that is, $P(\alpha) = 0$ for some nonzero $P = X^n + \cdots + a_1 X + a_0 \in \mathbb{Q}[X]$. Clearing out denominators, we then have $Q(m \alpha) = 0$ for some nonzero $m\in \mathbb{Z}$ such that $Q(X) = m^n P(X/m) = X^n + \cdots + m^{n-1} a_1 + m^n a_0 $ lies in $\mathbb{Z}[X]$. Hence $m\alpha\in \mathcal{O}_k$. We've thus shown that $\mathcal{O}_k \otimes \mathbb{Q} = k$, from which your result follows.
First of all, it is known that the additive group of $\mathcal{O}_K$ is a free $\mathbb{Z}$-module of rank $n=[K:\mathbb{Q}]$. It follows by general theorems on free $\mathbb{Z}$-modules that the additive group of any ideal $\mathfrak{a}\subseteq\mathcal{O}_K$ is a free $\mathbb{Z}$-module of rank $\leq n$. Now assume $\mathfrak{a}\neq 0$ and choose an element $a\in\mathfrak{a}-\{0\}$. Let $\beta_1,...,\beta_n$ and integral basis of $K$. It is obvious that $$G:=\{k_1a\beta_1+...+k_na\beta_n \ : \ k_i\in\mathbb{Z}\}$$ is a free $\mathbb{Z}$-module of rank $n$ (a $\mathbb{Z}$-basis is clearly $a\beta_1,...,a\beta_n$). But $G$ is also a subgroup of $\mathfrak{a}$ because each $a\beta_i$ is an element of $\mathfrak{a}$. Again by general theorems on free $\mathbb{Z}$-modules, we have $$n=\text{rank } G \leq \text{rank } \mathfrak{a} \leq n\text{.}$$ Hence $\text{rank }\mathfrak{a}=n$, so any $\mathbb{Z}$-basis of $\mathfrak{a}$ consists of $n$ elements.
