Notion of divisibility in a proof about number fields I don't understand a point in the proof of the theorem 9.4 in Anthony Scholl's Number Fields lecture notes.

Theorem: Let $K$ be a number field, $\mathcal{O}_K$ its ring of integers and $\alpha \in \mathcal{O}_K$. Then $\text{Tr}_{K/\mathbb{Q}}(\alpha^p) \equiv \text{Tr}_{K/\mathbb{Q}}(\alpha) \pmod p$.
Proof: By Fermat's little theorem $\text{Tr}_{K/\mathbb{Q}}(\alpha)^p \equiv \text{Tr}_{K/\mathbb{Q}}(\alpha) \pmod p$ and
$$\text{Tr}_{K/\mathbb{Q}}(\alpha)^p - \text{Tr}_{K/\mathbb{Q}}(\alpha^p) = \left( \sum_{i=1}^n \sigma_i(\alpha) \right)^p-\left( \sum_{i=1}^n \sigma_i(\alpha)^p \right) = \sum_{0 <k_1  \ldots  k_n<p \\ k_1+ \cdots + k_n=p} \frac{p!}{k_1! \cdots k_n!}\sigma_1(\alpha)^{k_1} \cdots \sigma_n(\alpha)^{k_n}$$
by the generalised binomial theorem where $\sigma_i$ are the complex embedings of $K$. Each of the terms is divisible by $p$, so the expression is $0 \pmod p$.

What I don't understand is how this can possibly be a valid argument when the conjugates $\sigma_i(\alpha)$ are complex numbers and divisibility is a notion in $\mathbb{Z}$. What does dividing mean here? Is it that they belong to the ideal $(p)$?
 A: You're right in that one has to be a bit more precise here. Hint: Write down the division of the RHS by $p$ in ${\mathbb C}$ explicitly, and convince yourself that it's (a) invariant under all complex embeddings (or rather their extensions to ${\mathbb C}$), and (b) still an algebraic integer. Deduce.
A: The $\sigma(\alpha)$ are more than random complex numbers- they are integral over $\mathbb{Q}$ (since $\alpha \in \mathcal{O}_K$) and so they are elements of the ring $\mathcal{O}$ of integers of the Galois closure of $K/\mathbb{Q}$. Then obviously $p \mid p\sigma(\alpha)$ in $\mathcal{O}$. And $\text{Tr}_{K/\mathbb{Q}}(\alpha)^p - \text{Tr}_{K/\mathbb{Q}}(\alpha^p) \in \mathbb{Z}$ so we get divisibility in $\mathbb{Z}$ 
A: Divisibility is a notion in any ring $(A,+,\cdot)$. For $a,b\in A$, we say $a\mid b$ iff there exists $d\in A$ with $b=da$. You know this for $A=\Bbb Z$, here it is used in $\mathcal O_K$. (Of course, in a field such as $K$, the concept would we almost worthless, as we can ake $d=\frac ab$ except when $b=0$).
