Let $A$ be a ring, $\mathfrak{a}$ a right ideal of $A$, $F$ a left
$A$-module and $\mathfrak{a}F$ the sub-$\mathbf{Z}$-module of $F$
generated by the elements of the form $\lambda x$ where
$\lambda\in\mathfrak{a}$ and $x\in F$. Then there is a canonical
$\mathbf{Z}$-module isomorphism $$\pi:(A/\mathfrak{a})\otimes_A
F\rightarrow F/\mathfrak{a}F$$ such that for all $\bar{\lambda}\in
A/\mathfrak{a}$ and all $x\in F$, $\pi(\bar{\lambda}\otimes x)$ is the
class mod. $\mathfrak{a}F$ of $\lambda x$, where
$\lambda\in\bar{\lambda}$.
This result is supposed to be a corollary of the following fact:
Given two exact sequence $E'\xrightarrow{u} E\xrightarrow{v}
E''\rightarrow0$ and $F'\xrightarrow{s} F\xrightarrow{t}
F''\rightarrow 0$ of right $A$-linear and left $A$-linear mappings,
respectively, the $\mathbf{Z}$-linear mapping $$v\otimes t:E\otimes_A
F\rightarrow E''\otimes F''$$ is surjective and its kernel is equal to
$\text{Im}(u\otimes 1_F)+\text{Im}(1_E\otimes s)$.
Attempt:
We can can consider the ring $A$ as a right $A$-module. Note that
$$\mathfrak{a}F=\left\{y\ |\ (\exists\alpha)(\alpha\in\mathbf{Z}^{(\mathfrak{a}\times F)}\ \land y=\sum_{(\lambda,y)\in\mathfrak{a}\times F}\alpha_{\lambda x}\lambda x)\right\}.$$
Let $i:\mathfrak{a}\rightarrow A$ and $j:\mathfrak{a}F\rightarrow F$ be the canonical injection. On the other hand, let $p:A\rightarrow A/\mathfrak{a}$ and $q:F\rightarrow F/\mathfrak{a}F$ be the canonical surjections. We have two exact sequence $\mathfrak{a}\xrightarrow{i}A\xrightarrow{p}A/\mathfrak{a}\rightarrow0$ and $\mathfrak{a}F\xrightarrow{j}F\xrightarrow{q}F/\mathfrak{a}F\rightarrow 0$. Hence
$$p\otimes q:A\otimes_A F\rightarrow A/\mathfrak{a}\otimes_A F/\mathfrak{a}F$$
is a $\mathbf{Z}$-linear surjection with kernel $\text{Im}(i\otimes 1_F)+\text{Im}(1_A\otimes j)$.
But this is not close to what I am looking for. Should I be tensoring different linear maps? Suggestions?
Best Answer
This is essentially $(A/\mathfrak{a}) \otimes_A M \simeq M / \mathfrak{a}M$ by tensoring the canonical exact sequence.
For right $A$-modules write $$\mathfrak a \xrightarrow{\iota} A \xrightarrow{\pi} A/\mathfrak a \to 0$$ and for left $A$-modules $$0\to F \xrightarrow{1_F} F\to 0$$ Then your theorem gives a surjective map $$\pi\otimes 1_F\colon A\otimes_A F\to (A/\mathfrak a)\otimes_A F$$ which kernel is $\mathrm{im}( \iota\otimes 1_F )$. (Why?)