I am trying to understand why, for a global extension $F \subset K$, given a place $\mu$ over $F$, we have that $$n = \sum_{\nu: \nu | \mu} n_{\nu}$$
where $n = [K:F]$, $n_{\nu} = [K_{\nu}|F_{\mu}]$ and we sum over places $\nu$ over $K$ that are extensions of $\mu$ over $F$.
Here's what I know so far. Assuming that $K = F(\alpha)$ is separable, with $\alpha$ having minimal polynomial $p(x) \in F[x]$ that factors into $p_j(x)$ over $F_{\mu}$ with roots $\alpha_j$:
- $K_{\nu} = F_{\mu}(\beta)$ for some root $\beta$ of $p(x)$, and $\{\nu \in \text{\{places of K\}} : \nu | \mu\} \cong \{ \phi_j, \text{embeddings } K \hookrightarrow \bar{F}_{\mu}, \phi_j(\alpha) = \alpha_j\}$.
- (My reasoning so far): Note that $K = F(\alpha)$, so that $n = $ degree of $p(x)$. Contrarily, if we have embedding $ \alpha \mapsto \alpha_j$ of $K \hookrightarrow \bar{F}_{\mu}$, we get corresponding place $\nu_j$ such that $\nu_j | \mu$.
However, I'm not sure what we can say about $K_{\nu_j}$. I think we want to say that $K_{\nu_j} = F_{\mu}(\alpha_j)$, which is extension of degree $=$ degree of $p_j(x)$, so that we can sum over degrees of $p_j$ above to get $n$, as desired. But why is this true (if it is)?