Proposition

Let $A$ be a ring, $a$ an ideal, $M$ an $A$-module. Show that $(A / a) \otimes M$ is isomorphic to $M / aM$.

Solution

$a \rightarrow A \rightarrow A / a \rightarrow 0$ is an exact sequence with the inclusion map $i: a \rightarrow A$ and the canonical quotient map $q: A \rightarrow A / a$. By Proposition 2.18 on P.28 of Atiyah, $a \otimes M \rightarrow A \otimes M \rightarrow A / a \otimes M \rightarrow 0$ is exact with $i \otimes \Id$ and $q \otimes Id$. This exact sequence gives us that

Let $\phi: A \otimes M \rightarrow M / aM$ be defined such that $x \otimes y \mapsto xy + aM$. Then $\phi$ is a composition of the isomorphism $A \otimes M$ described in Prposition 2.14 (iv) on P.26 of Atiyah and the canonical quotient map $M \mapsto M / aM$. Therefore, $\phi$ is a surjective module homomorphism.

We claim that $\ker{\phi} = a \otimes M$.

• Suppose $\phi(x \otimes y) = 0$. Then $xy + aM = 0$ and thus $xy \in aM$. This implies $xy = x’y’$ for some $x’ \in a$ and $y’ \in M$. \(\begin{align*} x \otimes y &= x(1 \otimes y) & \text{(since $A$ is a commutative ring with 1)} \\ &= 1 \otimes xy & \text{(since $M$ is an $A$-module)} \\ &= 1 \otimes x'y' \\ &= x'(1 \otimes y') & (x' \in A) \\ &= x' \otimes y' \\ &\in a \otimes M. \end{align*}\)
• On other hand, if $x \otimes y \in a \otimes M$, $\phi(x \otimes y) = xy + aM = 0$.

Therefore, $\ker{\phi} = a \otimes M$.

This implies that $(A \otimes M) / (a \otimes M) \cong M / aM$.

Hence, $(A / a) \otimes M \cong M / aM$.