Problem 2.2 Atiya and MacDonald pg.31 Here is the question:
Let $A$ be a ring, $\mathfrak{a}$  an ideal, $M$ an $A$-module. Show that $(A/ \mathfrak{a}) \otimes_A M \cong M/\mathfrak{a} M.$
And here is its solution:



I have some questions about the solution:
1- Why multiplying by $- \otimes_A M$ and not by $M \otimes_A -$? Also, I believe that there is only zero on the right because the tensor functor is right exact and not left exact, am I correct?
2- Is $(A/ \mathfrak{a}) \otimes M \cong (A \otimes M)/ \operatorname{Im}j$  by the first isomorphism theorem? is this because the map on the arrow after the $j$ is surjective and its kernel equals the image of $j$?
3- I do not see how the absorption Isomorphism sends $\operatorname{Im}j$ to $\mathfrak{a} M$?
Could anyone help me in answering those questions please?
 A: *

*Tensor product is bi-linear and an equivalence class. So right tensor product or left tensor product does not matter as long as you maintain the equivalence class correctly i.e., $m \otimes a = am \otimes 1$.


*You are correct, since $a \otimes M \rightarrow A \otimes M \rightarrow A/a \otimes M \rightarrow 0$, it is first isomorphism theorem: $(A \otimes M) / (a \otimes M) \tilde{=} A/a \otimes M$.
But you need to verify that $Ker(f: A \otimes M \rightarrow A/a \otimes M) = a \otimes M$ keeping the equivalence class in mind.
$0$ in tensor product is of the form $0 \otimes m = a \otimes 0 = 0 \otimes 0$.
$f(\sum_i b_i \otimes m_i) = f(1 \otimes \sum_i b_i m_i) = f(1 \otimes m')$.
$ f(1 \otimes m') = (1+a) \otimes m'$
$(1+a) \otimes m' = 0 \otimes m'$ implies $1+(a) = 0$ which is not true.
$(1+a) \otimes m' = 1 \otimes 0$ implies $(1+b)m' = 0$ for some $b \in a$. This implies $m' = -bm'$. This implies $1 \otimes m' = 1 \otimes -bm' = -b \otimes m' \in a \otimes M$.
Hence $Ker(f) = a \otimes M$.


*This is because $b \otimes m = 1 \otimes bm \tilde{=} bm$
Note that i am assuming $1 \in A$ and $1.m = m$.
