Alternative construction of Direct Limit The construction of the direct limit that I learned from Atiyah Macdonald is the following: Suppose we have a directed system $(M_i,\mu_{ij})$ of $A$ - modules and $A$ - module homomorphims  over a directed set $I$. Let $D$ be the submodule generated by elements of the form $x_i - \mu_{ij}(x_i)$. Then 
$$\varinjlim M_i \stackrel{\text{def}}{\equiv} \bigoplus_{i\in I} M_i/ D.$$
Now I am trying to see how this is equivalent to the wikipedia definition given here. In particular, what I don't understand about the wikipedia definition is that the equivalence relation $\sim$ defined there is between individual elements, e.g. say whether $x_i \sim x_j$. How does one "extend this linearly" to an equivalence relation on $\bigsqcup M_i$ which is the set of all sums $\sum_{i \in I}x_i$ where $x_i \in M_i$ and all but finitely many $x_i$ are zero?
 A: All right, there are several things going on here.  First of all the direct limit of a sequence of algebraic objects can be defined in two equivalent ways.  The first way, the direct limit is a quotient of the direct sum of the algebraic objects:
$$
\bigoplus_{i\in I} M_i \bigg/\equiv
$$
where $\equiv$ is a certain congruence relation defined on the direct sum.  In particular, $\equiv$ is the congruence relation generated by pairs of the form $x_i \equiv \mu_{ij}(x_i)$.  (Here, the congruence relation generated by a certain set of pairs means the intersection of all congruence relations that contain the pairs).  This quotient can also be described as the quotient by the submodule $D$ which you have defined, but I have described it as a congruence relation to mimic Wikipedia's definition.
Alternatively, the set of elements of the direct limit can be described as a quotient of the disjoint union
$$
\bigsqcup_{i\in I} M_i \bigg/\sim
$$
where $\sim$ is the equivalence relation generated by all pairs of the form $x_i\sim\mu_{ij}(x_i)$.  (Again, the equivalence relation generated by a certain set of pairs means the transitive, reflexive, symmetric closure of the relation defined by the pairs.)  The Wikipedia article really does mean disjoint union here -- not direct sum.  Naively, this just defines a set of equivalence classes, but there is a natural module structure on this set inherited from the module structure on the $M_i$'s (though this requires some proof).
The first definition is more algebraic, but the second definition makes it clearer that every element of the direct limit actually comes from one of the $M_i$'s.
To show these are equivalent, you just want to construct an isomorphism directly on the level of elements.  This is basically the same as starting from the Atiyah-McDonald defintion and proving the following theorem:
Theorem. Let $L = \bigoplus_{i\in I} M_i\Big/D$ be the direct limit of the $M_i$'s.  For each $i\in I$, let $L_i$ denote the canonoical image of $M_i$ in $L$.  Then $L = \bigcup_{i\in I} L_i$.
Edit: Just in case it helps, here are a few examples of direct limits of $\mathbb{Z}$-modules (abelian groups) that are fairly illustrative.  The first is the limit of the sequence
$$
\textstyle\mathbb{Z} \;\longrightarrow\; \frac{1}{2}\mathbb{Z} \;\longrightarrow\; \frac{1}{4}\mathbb{Z} \;\longrightarrow\; \frac{1}{8}\mathbb{Z} \;\longrightarrow\; \cdots
$$
where all of the maps are inclusion maps, and $q\mathbb{Z}$ denotes the integer multiples of $q$.  The direct limit of this sequence is the group $\mathbb{Z}\bigl[\frac{1}{2}\bigr]$ of dyadic fractions.  The second example is the limit of the sequence
$$
\textstyle\mathbb{Q}\big/\mathbb{Z} \;\longrightarrow\; \mathbb{Q}\big/\frac{1}{2}\mathbb{Z} \;\longrightarrow\; \mathbb{Q}\big/\frac{1}{4}\mathbb{Z} \;\longrightarrow\; \mathbb{Q}\big/\frac{1}{8}\mathbb{Z} \;\longrightarrow\; \cdots
$$
where each of the maps is the natural quotient map.  The direct limit of this sequence is the quotient $\mathbb{Q}\big/\mathbb{Z}\bigl[\frac{1}{2}\bigr]$.
These examples are illustrative, because the first shows just the inclusion aspect of direct limits, while the second shows just the quotient aspect.  A more general direct limit will have both aspects, since a typical homomorphism is neither injective nor surjective.
A: I would guess that the "disjoint union" used in the wiki article really means coproduct in the category in question (in fact, I know it does) and it just so happens that the coproduct in $R\text{-}\mathbf{Mod}$ is direct sum. The equivalence relation is then just exactly the one you discussed.
A: 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.
