Question regarding ideals and vector spaces this is my first time I am posting on this forum. 
My question is regarding a sentence I read on page 27 of "Algebraic Number Fields" by "Gerald J. Jansuz".
The set-up is as follows: Let $R \subset R'$ be Dedekind rings and $\mathfrak{p}$ be a prime ideal in $R$. Then $\mathfrak{p} R'$ has a unique factorisation $\mathfrak{p} R' = \mathfrak{B}_1^{e_1} \dots \mathfrak{B}_g^{e_g}$.
I don't really understand this statement that was written in the book: "The ring $R' / \mathfrak{B}_i^{e_i}$ is not a vector space over $R' / \mathfrak{B}_i$ unless $e_i = 1$, but the quotients $\mathfrak{B}_i^a / \mathfrak{B}_i^{a+1}$ are vector spaces over $R' / \mathfrak{B}_i$." 
It will be great if anyone can enlighten me what the book is saying. In particular I do not really understand how one multiplies the scalar with the vector in the latter case when it is a vector space. I am very sorry if this question is very elementary but I cannot seem to get my head around this.
Thank you very much in advance.
 A: It is clear that $\mathfrak{B}_i^a / \mathfrak{B}_i^{a+1}$ is an $R'$-module: $\mathfrak{B}_i^a$ is an ideal, hence a module, and it contains $\mathfrak{B}_i^{a+1}$ as a sub-module, giving $\mathfrak{B}_i^a / \mathfrak{B}_i^{a+1}$ a factor-module structure.
Now as for any $x \in \mathfrak{B}_i$ and $y \in \mathfrak{B}_i^{a}$ the element $xy$ is in $\mathfrak{B}_i^{a+1}$, the annihilator of $\mathfrak{B}_i^a / \mathfrak{B}_i^{a+1}$ contains $\mathfrak{B}_i$. This induces an $R'/\mathfrak{B}_i$-module structure on $\mathfrak{B}_i^a / \mathfrak{B}_i^{a+1}$ which is in fact a vector-space structure as $\mathfrak{B}_i$ is prime in a Dedekind ring and thus maximal. In other words: $\mathfrak{B}_i^a / \mathfrak{B}_i^{a+1}$ is an $R'/\mathfrak{B}_i$-vector space where multiplication is defined by multiplying two arbitrary representatives.
This does not work for $R'/\mathfrak{B}_i^{e_i}$ for $e_i>1$ because in Dedekind rings we have $\mathfrak{B}_i^{a+1} \subsetneq \mathfrak{B}_i^a$ and therefore not all elements of $\mathfrak{B}_i$ are in the annihilator. So multiplying by elements in $R'/\mathfrak{B}_i$ is not well-defined as it depends on the chosen representative.
A: If $A$ is a ring (e.g. $R'$ in your case) and $m$ is a maximal ideal (e.g. $\mathfrak B_i$ in your case), then $k:= A/m$ is a field.
So any $A$-module which is killed by $m$ is actually (automatically) an $A/m$-module, i.e. a $k$-module, i.e. a $k$-v.s.  (Remember --- vector space just means module for which the ring acting is a field.)
Now $A/m^2$ is not a field (typically, unless $m = m^2$, which it doesn't in your case), and so is a module over itself, but not a v.s. over anything.
But $A/m$ and $m/m^2$ are both $A$-modules killed by $m$, hence both $A/m$-modules, i.e. both $k$-vector spaces.
A: Since $\mathfrak{B}_i$ is a prime ideal in the Dedekind ring $R'$, it is a maximal ideal and the quotient $R'/\mathfrak{B}_i$ is a field. Given $r+\mathfrak{B}_i\in R'/\mathfrak{B}_i$, the scalar product of this element with $x+\mathfrak{B}_i^{a+1}$ (with $x\in \mathfrak{B}_i^a$) is the coset $rx + \mathfrak{B}_i^{a+1}$.
This is well-defined: first, since $\mathfrak{B}_i^a$ is an ideal, $rx\in \mathfrak{B}_i^a$. Next, if we choose different representatives $r'$ and $x'$ for these cosets, we can write $r = r'+b_1$ and $x = x'+b_2$ for $b_1\in \mathfrak{B}_i$ and $b_2\in \mathfrak{B}_i^{a+1}$. Then we have
$$rx - r'x' = b_1x' + b_2r'+b_1b_2\in \mathfrak{B}_i^{a+1},$$ hence
$$rx + \mathfrak{B}_i^{a+1} = r'x' + \mathfrak{B}_i^{a+1}.$$
