if $A$ is a module,then the family fin($A$) of all the finitely generated submodules of $A$ is a directed set and direct limit of$M_i$ is isomorphic to$A$. for prove this needed to define to injection from $M_j$ to $M_i$.is it true define inclusion map for injection?this means that $M_i$=$\sum_1^n Rm_i $ st $m_i$ are generator of $M_i$. define $\Gamma_i^j:$$M_j\to M_j$ if $M_j$$\subseteq M_i$?if we define it then the direct limit of $M_i$ is$\oplus _i$$M_i$.how to show thae $A \cong \oplus _i$$M_i$?
[Math] direct limit of finitely generated submodule
commutative-algebrahomological-algebra
Related Solutions
A direct limit of a system $(M_i, \mu_{ij})$ is an appropriate family of objects satisfying the universal property. Here Atiyah and Macdonald have constructed a $(M, \mu_i)$ which does the job. It seems like you're worried that a certain property of this entity might come from the particular construction given.
But if $N$ and $\nu_i\colon M_i \to N$ do the job just as well, then there is a (unique) isomorphism $\alpha\colon M \to N$ such that $\alpha \circ \mu_i = \nu_i$ for all $i$. If $\nu_i(x_i) = 0$ then $\alpha(\mu_i(x_i)) = 0$, and hence $\mu_i(x_i) = 0$ because $\alpha$ is an isomorphism. So you are back to exercise 15.
Here's how I would present the construction given in Wikipedia.
Suppose $(M_i,\mu_{ij})$ is a directed system of modules. We begin by taking a disjoint union of the underlying sets of the $M_i$; in order to "keep them disjoint", the usual method is to "paint" each set with its index $i$ to ensure that if $i\neq j$, then the sets are disjoint. That is, we consider the set $$\mathcal{M} = \bigcup_{i\in I}(M_i\times\{i\}).$$ The elements of $\mathcal{M}$ are pairs of the form $(x,i)$, where $i\in I$ and $x\in M_i$.
Note that $\mathcal{M}$ is not a module, at least not one with any natural structure: the operations we have on hand (the ones for the different $M_i$) are not defined on all of $\mathcal{M}$, they are only defined on proper subsets of $\mathcal{M}$.
We now define an equivalence relation on $\mathcal{M}$ as follows: $(x,i)\sim(y,j)$ if and only if there exists $k\in I$, $i,j\leq k$ such that $\mu_{ik}(x) = \mu_{jk}(y)$ in $M_k$. It is not hard to verify that this is an equivalence relation.
Let $\mathbf{M}$ be the set $\mathcal{M}/\sim$. Denote the equivalence class of $(x,i)$ by $[x,i]$.
We now define a module structure on $\mathbf{M}$: we define a sum on classes by the rule $$ [x,i] + [y,j] = [\mu_{ik}(x)+\mu_{jk}(y),k]$$ where $k$ is any element of $I$ such that $i,j\leq k$. One needs to prove that this is well defined and does not depend on the choice of the $k$. Suppose first that $k'$ is some other element with $i,j\leq k'$. Let $\ell$ be an index with $k,k'\leq \ell$; then $$\begin{align*} \mu_{i\ell}(x) + \mu_{j\ell}(y) &= \mu_{k\ell}(\mu_{ik}(x))+\mu_{k\ell}(\mu_{jk}(y))\\ &= \mu_{k\ell}(\mu_{ik}(x) + \mu_{jk}(y)). \end{align*}$$ Therefore, $[\mu_{ik}(x)+\mu_{jk}(y),k] = [\mu_{i\ell}(x)+\mu_{j\ell}(y),\ell]$. By a symmetric argument, we also have $$[\mu_{ik'}x) + \mu_{jk'}(y),k'] = [\mu_{i\ell}(x) + \mu_{j\ell}(y),\ell],$$ so the definition does not depend on the choice of $\ell$.
To show it does not depend on the representative either, suppose $[x,i]=[x',i']$ and $[y,j]=[y',j']$. There exists $m$, $i,i'\leq m$ with $\mu_{im}(x)=\mu_{i'm}(x')$, and there exists $n$, $j,j'\leq n$, with $\mu_{jn}(y)=\mu_{j'n}(y')$. Pick $k$ with $m,n\leq k$. Then $$\begin{align*} [x,i]+[y,j] &= [\mu_{ik}(x)+\mu_{jk}(y),k]\\ &= [\mu_{mk}(\mu_{im}(x)) + \mu_{nk}(\mu_{jn}(y)),k]\\ &= [\mu_{mk}(\mu_{i'm}(x')) + \mu_{nk}(\mu_{j'n}(y')),k]\\ &= [\mu_{i'k}(x') + \mu_{j'k}(y'),k]\\ &= [x',i'] + [y',j'], \end{align*}$$ so the operation is well-defined.
It is now easy to verify that $+$ is associative and commutative, $[0,i]$ is an identity (for any $i$) and that $[-x,i]$ is an inverse for $[x,i]$, so this operation turns $\mathbf{M}$ into an abelian group.
We then define a scalar multiplication as follows: given $r\in R$ and $[x,i]\in\mathbf{M}$, we let $r[x,i] = [rx,i]$. Again, one needs to show that this is well-defined (easier than the proof above), and verify that it satisfies the relevant axioms (not hard) to show that this endows $\mathbf{M}$ with the structure of a left $R$-module.
Now note that the maps $\mu_i\colon M_i\to \mathbf{M}$ given by $\mu_i(x) = [x,i]$ is a module homomorphism. The module $\mathbf{M}$ together with the maps $\mu_i$ are a direct limit of the system.
(The same construction works for Groups, Rings, etc).
There is no "linear extension" of the equivalence relation. Rather, we define an operation on $\mathcal{M}/\sim$, since $\mathcal{M}$ (being a disjoint union of the underlying set of the original modules) is not a module itself: it does not even have a total operation defined on it, just a bunch of partial operations.
Best Answer
Order the set of finitely generated sub-$A$-modules of $M$ by inclusion. This gives you a directed system $(S,\subseteq)$ of (finitely generated) sub-$A$-modules of $M$, and the direct limit of this set is isomorphic to $M$. You can see this as simply saying that the union of all finitely generated submodules of $M$ is equal to $M$, which is intuitively trivially true indeed.
Indeed, it is injective, as if you are zero in the limit/union of the $M'$'s for $M'\in S$, it means that you are zero in some $M'\in S$, and this means that you are zero. It is surjective because any element $m$ of $M$ is an $M'$ such that $M'\in S$ : just take for $M'$ the sub-$A$-module of $M$ generated by $m$.