Question about factor rings Assume $m_i$ are maximal ideals in a ring $R$.
Then I have $m_1 \cdot \dots m_{k}$ is an ideal in $m_1 \cdot \dots m_{k-1}$ hence I can quotient to get a factor ring $m_1 \cdot \dots m_{k-1} / m_1 \cdot \dots m_{k}$.
This ring is of course also an abelian group. Now I'd like to turn this into a vector space. How do I see that the field of scalars I am looking for is $R/m_k$?  
 A: If $M$ is an $R$-module and $I$ is an ideal of $R$, then $IM$ is a submodule of $M$, and there is a natural $R/I$ module structure on $M/IM$: namely, given $r+I\in R/I$ and $m+IM$ in $M/IM$, define $(r+I)(m+IM) = rm+IM$.
This is well-defined: if $r-s\in I$ then $(r-s)m\in IM$, so $rm+IM = sm+IM$. And if $n-m\in IM$, then we can express $n-m$ as a sum
$$n-m = a_1m_1+\cdots +a_km_k,\qquad a_i\in I, m_j\in M$$
so
$$r(n-m) = ra_1m_1+\cdots + ra_km_k\in IM$$
since $ra_i\in I$ for all $i$. 
Here, you have $M=m_1\cdots m_{k-1}$, $I=m_k$, so there is a natural $R/m_k$ module structure on $M/IM = m_1\cdots m_{k-1}/m_1\cdots m_k$. Since $m_k$ is maximal, $R/m_k$ is a field, so this is actually a vector space structure. So this is a natural field over which to give $M/IM$ a vector space structure, but by no means the only one.
A: The way I like to think about this is that an $R$-module is an abelian group $M$ together with a ring homomorphism $R \to \operatorname{End}(M)$. The kernel of this map is called the annihilator of $M$, and from this perspective it's easy to see that $M$ is an $R/I$-module for any ideal $I$ contained in the annihilator.
Another way of viewing your quotient is that you took the $R$-module $\mathfrak m_1 \cdots \mathfrak m_{k-1}$ and tensored with $R/\mathfrak m_k$. Remember that in general for an $R$-module $M$ and $R$-algebra $S$ the base change $M \otimes_R S$ is an $S$-module, and that for an ideal $I$ of $R$, $M \otimes_R (R/I) \simeq M/(IM)$ by right exactness of the tensor product.
