Exercise 2.2

Question

Let $A$ be a ring, $\mathfrak{a}$ an ideal, $M$ an $A$-module. Show that $(A/\mathfrak{a}) \otimes_A M$ is isomorphic to $M/\mathfrak{a}M$.

Amit Awasthi · Accepted Answer

\begin{proof}
  Consider the exact sequence $0 	o \mathfrak{a} \xrightarrow{h} A \xrightarrow{\pi} A/\mathfrak{a} 	o 0$. Tensoring with $M$ over $A$ yields the right exact sequence
  \begin{equation*}
    \mathfrak{a} \otimes M \xrightarrow{h \otimes 1} A \otimes M \xrightarrow{\pi \otimes 1} (A/\mathfrak{a}) \otimes M 	o 0.
  \end{equation*}
  by Prop.~$2.18$. Prop.~$2.14$ gives the unique isomorphism $f\colon A \otimes M 	o M$, and so letting $g\coloneqq (\pi \otimes 1) \circ f^{-1} \colon M 	o A \otimes M 	o (A/\mathfrak{a}) \otimes M$, we claim the sequence
  \begin{equation*}
    0 	o \mathfrak{a}M \xrightarrow{\cdot} M \xrightarrow{g} (A/\mathfrak{a}) \otimes M 	o 0
  \end{equation*}
  is exact. $\operatorname{im}(g) = \operatorname{im}(\pi \otimes 1) = (A/\mathfrak{a}) \otimes M$, and $\operatorname{Ker}(g) = f(\operatorname{Ker}(\pi \otimes 1)) = f(\operatorname{im}(h \otimes 1)) = \mathfrak{a}M$. Thus, $(A/\mathfrak{a}) \otimes M$ is isomorphic to $M/\mathfrak{a}M$.
\end{proof}

Exercise 2.2

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Exercise 2.2

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