A Question from Neukirch Regarding the Quotient Module of a Sum of Modules The Setup: Let $L/K$ be a separable extension with rings of integers $\mathcal{O}_{L}$ and $\mathcal{O}_{K}$ respectively. Let $\mathfrak{p}$ be a prime ideal of $\mathcal{O}_{K}$ and $\mathfrak{p}\mathcal{O}_{L}$ be $\mathfrak{p}$ moved upstairs in $\mathcal{O}_{L}$. Let $\omega_{1},\ldots,\omega_{n}$ be representatives of a basis for $\mathcal{O}_{L}/\mathfrak{p}\mathcal{O}_{L}$ over $\mathcal{O}_{K}/\mathfrak{p}$.
The problem: In the proof of proposition 8.2 of Neukirch's Algebraic Number Theory he starts to prove the $\omega_{i}$ are linearly independent by defining $M = \mathcal{O}_{K}\omega_{1}+\cdots+\mathcal{O}_{K}\omega_{n}$ and $N = \mathcal{O}_{L}/M$. He then says $\mathcal{O}_{L} = M+\mathfrak{p}\mathcal{O}_{L}$ implies $\mathfrak{p}\mathcal{O}_{L}N = N$. I understand the first equality,but I'm having trouble justifying the second.
My attempt: I see how in $(M+\mathfrak{p}\mathcal{O}_{L})/M$ we mod out the $M$ bit. More formally, $(M+\mathfrak{p}\mathcal{O}_{L})/M = \{p+M \mid p \in \mathfrak{p}\mathcal{O}_{L}\}$. So, we should expect to find this set equal to something like $\mathfrak{p}\mathcal{O}_{L}N$. The main issue is that I don't understand what $\mathfrak{p}\mathcal{O}_{L}N$ is formally. $\mathfrak{p}\mathcal{O}_{L}$ and $N$ are both clearly $\mathcal{O}_{K}$ modules, but we have no rigorous structure to multiply their elements since neither is a subset of the other. If we are implicitly defining the multiplication to be $p(r+M) = pr+pM$ for some $r+M \in N$ and $p \in \mathfrak{p}\mathcal{O}_{L}$ then I am still confused because $pM$ is not necessarily $M$ even under the assumption $\mathcal{O}_{L} = M+\mathfrak{p}\mathcal{O}_{L}$. Any help is appreciated!
 A: Your question can be treated in a general context: 


*

*let $A \subset B$ be an extension of commutative rings (which you think as $O_K \subset O_L$), 

*let $I \subset A$ be an ideal and $M$ be an $A$-submodule of $B$ such that $B = M + IB$.
Define the $A$-module $N = B / M$. The claim is that
$$IN=N,$$
where $IN$ is the $A$-submodule of $N$ generated by the elements $x \cdot n$ where $x \in I$ and $n \in N$.

Clearly, we have $IN \subset N$. Conversely, let $n \in N = B/M$ be the class of some $b \in B$. By assumption, we have $B = M+IB$, so that we may write
 $$b = m + \sum_{j=1}^k i_j b_j$$ for some $m \in M, i_j \in I, b_j \in B$ and $k \geq 1$.
Since $n = [b]_M = b+M \in B/M$ is the class of $b$, you get also that $n$ is the class of $\sum_{j=1}^k i_j b_j \in IB$, whence the equality
$$n = \sum_{j=1}^k i_j [b_j]_M,$$
which lies in the module $IN$, because $[b_j]_M \in B/M=N$ and $i_j \in I$ for every $j$. Therefore, we've shown $N \subseteq IN$, and thus $N = IN$ as claimed.
Elementary, isn't it? ;-)
$\hspace{9cm}\blacksquare$
