Exercise 2.17

Question

Let $(M_i)_{i \in I}$ be a family of submodules of an $A$-module, such that for each pair of indices $i,j$ in $I$ there exists $k \in I$ such that $M_i + M_j \subseteq M_k$. Define $i \le j$ to mean $M_i \subseteq M_j$ and let $\mu_{ij} \colon M_i 	o M_j$ be the embedding of $M_i$ in $M_j$. Show that
  \begin{equation*}
    \lim_{\longrightarrow} M_i = \sum M_i = \bigcup M_i.
  \end{equation*}
  In particular, any $A$-module is the direct limit of its finitely generated submodules.

Amit Awasthi · Accepted Answer

\begin{proof}
  For the second equality, we first see that trivially $\bigcup M_i \subset
  \sum M_i$, and so it suffices to show the other inclusion.
  But this follows since
  $m_{i_1} + \cdots + m_{i_n} \in M_{i_1} + \cdots + M_{i_n} \subset
  M_j \subset \bigcup M_i$, where $j$ is chosen such that $i_1,\ldots,i_n \le
  j$.

\par To show $\displaystyle\lim_{\longrightarrow} M_i = \bigcup M_i$, it
  suffices to show that $\bigcup M_i$ satisfies the universal property for the
  direct limit in Exercise 2.16.
  Let $N$ be an $A$-module and $\alpha_i\colon M_i 	o N$ an $A$-module
  homomorphism such that $\alpha_i = \alpha_j \circ \mu_{ij}$ when $i \le j$.
  Then, defining $\alpha\colon \bigcup M_i 	o N$ such that for $m \in \bigcup
  M$, we choose $i$ such that $m \in M_i$, and then let $\alpha(m) =
  \alpha_i(m)$.
  If $m \in M_i \cap M_j$, then we can choose $k \ge i,j$
  by hypothesis, and so $\alpha_k(m) = \alpha_i(m) = \alpha_j(m)$ since the
  $\mu_{ij}$'s are canonical embeddings.
  $\alpha$ is therefore well-defined, and is an $A$-module homomorphism such
  that $\alpha_i = \alpha \circ \mu_i$ for all $i \in I$.
  Moreover, since this is uniquely determined by the $\alpha_i$, we see that
  $\bigcup M_i$ satisfies the universal property, and so
  $\displaystyle\lim_{\longrightarrow} M_i \cong \bigcup M_i$.
  
  \par Now suppose $M$ is an $A$-module.
  If $\mathscr{F}$ is the family of finitely-generated submodules of $A$,
  we can define the direct limit of them since $M_1,M_2$ finitely generated
  implies $M_1 + M_2$ is finitely generated, and so $M_1 + M_2 \in \mathscr{F}$.
  Clearly $\displaystyle\lim_{\longrightarrow} \mathscr{F} \subset M$; we
  now show the reverse inclusion.
  So suppose $x \in M$.
  Then, $Ax \in \mathscr{F}$ since it is generated by $x$, and so
  $x \in \bigcup \mathscr{F}$.
  By the paragraph above, this shows
  $x \in \displaystyle\lim_{\longrightarrow} \mathscr{F}$.
\end{proof}

Exercise 2.17

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Exercise 2.17

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