Help with proof that $[\mathcal{O}_K/\mathfrak{B}_i^{e_i}:\mathbb{Z}/p\mathbb{Z}] = e_i[\mathcal{O}_K/\mathfrak{B}_i:\mathbb{Z}/p\mathbb{Z}]$ I'm trying to understand the following proof from Milne's Algebraic Number Theory book. The proof is on page 60 of the linked .pdf, although the text itself has page number 58.
This is part of trying to prove the formula (which doesn't appear to have a name, but I believe is well known):
$$\sum_{i = 1}^g e_i f_i = m$$
Where $\mathfrak{p}B = \mathfrak{B}_1^{e_1}\mathfrak{B}_2^{e_2}\dots\mathfrak{B}_g^{e_g}$ is how the prime ideal $\mathfrak{p}\subseteq A$ factors in $B$.
$e_i$ and $f_i$ are the ramification degree, and residue class degree respectively.
I only need to understand this in the special case of how ideal $(p)\subseteq\mathbb{Z}$ factor in the ring of integers $\mathcal{O}_K$ of a number field $K/\mathbb{Q}$.
The main part I need help with is understanding the proof of the following (I've changed the notation in the book to the special case I'm dealing with. Note that $p\mathcal{O}_K = \mathfrak{B}_1^{e_1}\mathfrak{B}_2^{e_2}\dots\mathfrak{B}_g^{e_g}$ is how $(p)\subseteq \mathbb{Z}$ factors):

... it suffices to show that $[\mathcal{O}_K/\mathfrak{B}_i^{e_i} : \mathbb{Z}/p\mathbb{Z}] = e_if_i$.
  From the definition of $f_i = [\mathcal{O}_K/\mathfrak{B}_i:\mathbb{Z}/p\mathbb{Z}]$, we know that $\mathcal{O}_K/\mathfrak{B}_i$ is a field of degree $f_i$ over $\mathbb{Z}/p\mathbb{Z}$. For each $r_i$, $\mathfrak{B}_i^{r_i}/\mathfrak{B}_i^{r_i+1}$ is a $\mathcal{O}_K/\mathfrak{B}_i$-module, and because there is no ideal between $\mathfrak{B}_i^{r_i}$ and $\mathfrak{B}_i^{r_i+1}$, it must have dimension one as a $\mathcal{O}_K/\mathfrak{B}_i$-vector space, hence dimension $f_i$ as an $\mathbb{Z}/p\mathbb{Z}$ vector space. Therefore, each quotient in the chain:
  $$B\supset \mathfrak{B}_i\supset\mathfrak{B}_{i}^2\supset\dots\supset\mathfrak{B}_i^{e_i}$$
  has dimension $f_i$  over $\mathbb{Z}/p\mathbb{Z}$, so the dimension of $\mathcal{O}_K/\mathfrak{B}_i^{e_i}$ is $e_if_i$.

My questions are:


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*How does the definition of $f_i$ imply that $\mathcal{O}_K/\mathfrak{B}$ is a field specifically?

*What do elements of $\mathfrak{B}_i^{r_i}/\mathfrak{B}_i^{r_i+1}$ intuitively look like? I understand that $I^n = \{r_1r_2\dots r_n\mid r_i\in I\}$, but find the elements in this quotient confusing.

*How do we know there's no ideal between $\mathfrak{B}_i^{r_i}$ and $\mathfrak{B}_i^{r_i+1}$. I understand how there can't be an ideal of the form $\mathfrak{B}_i^{k}$ for some $k$, but why can't there be one in general?

*How does the lack of an ideal between $\mathfrak{B}_i^{r_i}$ and $\mathfrak{B}_i^{r_i+1}$ imply that the quotient has dimension $1$ as an $\mathcal{O}_K/\mathfrak{B}_i$ module?

*Why is $\mathfrak{B}_i^{r_i}/\mathfrak{B}_i^{r_i+1}$ referred to as a module and a vector space? I understand that all vector spaces are modules, but am I missing some nuance by identifying them as vector spaces in both cases?
It could be that this source is the wrong place for me to look for a proof of this formula over $K/\mathbb{Q}$, as I have less experience with field theory/Galois Theory than I'd like. A recommendation for a more elementary source (especially for specifically $K/\mathbb{Q}$) would also be useful.
 A: Looking at another source (specifically Oggier's ANT notes here) has helped a lot.
$$
\newcommand{\O}{\mathcal{O}}\newcommand{\p}{\mathfrak{p}}
$$

*

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Prop 2.2: Consider $0\neq x\in\O/\p$. As $\p$ is prime, $\O/\p$ is an integral domain, and the map $\mu_x:\O/\p\to\O/\p$ given by $z\mapsto xz$ is injective. As the cardinality of $\O/\p$ is finite, the injection must be a bijection, so $\mu_x$ is invertible, and any $x\neq 0$ has an inverse.



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Prop 2.11: If $\p$ is a prime ideal of $\O$, and $n > 0$, then $\O/\p$ and $\p^n/\p^{n+1}$ are non-canonically isomorphic. Proof: consider the map $\phi:\O/\p\to\p^n/\p^{n+1}$ given by $\alpha\to\alpha\beta$, where $\beta\in\p^n\setminus\p^{n+1}$. Note that $\ker(\phi) = \{\alpha\mid \alpha\beta = 0\} = \{\alpha\mid \alpha\beta\in\p^{n+1}\} = \{\alpha\mid \alpha\in\p\} = \p$. Computing the image is a little more involved, but $\phi$ ends up being surjective, so $\O/\p\cong\p^n/\p^{n+1}$.



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If there were an ideal $\mathfrak{p}^{n+1}\subset B\subset\mathfrak{p}^n$, then as "to contain is to divide", we would have that $B\mid \mathfrak{p}^{n+1}$, so $B$ is of the form $\mathfrak{p}^i$ for some $i$ with $n <i < n+1$, which is spurious.



*

The isomorphism from prop 2.11 shows that it has dimension one.



*

It seems safe to call it a vector space here.

