I am trying to understand the proof of a proposition stating that if a ring $A$ contains some maximal ideal $I$, then the rank of a free $A$-module, $M$, is defined. We define the submodule $M_I = \big\{\sum_{i=1}^n \alpha_im_i: \alpha_i \in I, m_i \in M, n \in \mathbb{N} \big\}$. We assume everything is commutative.
A key part of the proof is realising the quotient module $M/M_I$ is actually an $A/I$ module. I can't figure out why this is true. In my notes it says: "$M/M_I$ is an $A$-module, but it is also naturally an $A/I$ module, because $I$ kills it". I understand why it's an $A$-module, but not why it's an $A/I$ module. I'm not sure what "kills it" means, so I'm just trying to prove this rigorously.
Specifically, the product operation must be a map $\cdot: A/I \times M/M_I \rightarrow M/M_I$. I.e., the product of a quotient ring element and a quotient module element map back into the quotient module. I assume this map has to be defined as $(a+I)\cdot (m+M_I) = (am+M_I)$, where $a\in A, m\in M$, and the product $am$ is taken as the product in the module $M$.
I am particularly struggling with showing this map is actually well defined. I.e., to show that for any two representations of the same quotient ring element and quotient module element, the defined product is equal. I know this equates to having to prove that if $a_1 + I = a_2 + I \text{ and } m_1+M_I = m_2 + M_I$, then $a_1m_1+M_I = a_2m_2+M_I$. This is equivalent to showing that given $a_1-a_2 \in I \text{ and } m_1 – m_2 \in M_I$, then $a_1m_1 – a_2m_2 \in M_I$.
I really can't figure out how to prove this however. I tried writing out combinations of elements that are in $M_I$, such as variations on $a_1(m_1-m_2)$, and also using that since $a_1-a_2 \in I$, then by definition of $M_I$, $(a_1-a_2)\cdot m \in M_I$ for any $m$. I tried combining different expressions to arrive at $a_1m_1 – a_2m_2 \in M_I$, but the closet I got was that term plus another in $M_I$. I also tried using the definition of $M_I$ as finite linear sums of products over the ideal $I$, but I am getting tripped up by the fact that by assumption $a_{1,2}\in A$, but not necessarily in $I$. (to this point I also realised I can actually assume neither $a_{1,2}$ are in $I$, as then the result is immediate).
Can someone point me in the right direction, or to a reasonably simple result I could use to help me prove this?
Best Answer
$a_1m_1-a_2m_2=a_1m_1-a_1m_2+a_1m_2-a_2m_2=a_1(m_1-m_2)+(a_1-a_2)m_2$
The first term belongs to $M_I$ because $m_1-m_2\in M_I$, and an $A$-module is closed under scalar multiplication. The second term belongs to $M_I$ by definition, because $a_1-a_2\in I$. So their sum is also in $M_I$.
It is still good to understand the explanation given in the proof. In general, if $M$ is an $A$-module and $I$ acts trivially on $M$ (i.e $am=0$ for all $a\in I, m\in M$. This is the meaning of "$I$ kills $M$") then you can naturally define an $A/I$-module structure on $M$.