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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?

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By "extend linearly", the Wikipedia page means that you add the equations $x_i = x_j$ as relations to the infinite direct sum $\bigoplus M_i$. – Jim Belk Apr 17 '12 at 4:38
@JimBelk What do you mean by equations "$x_i = x_j$"? Do you mean that if say $x_1 \sim x_2$ and $x_3 \sim x_4$ then $x_1 + x_3 \sim x_2 + x_4$? Conversely if $x_1 + x_3 \sim x_2 + x_4$ this means that $x_1 \sim x_2$ and $x_3 \sim x_4$? If you post an answer I will definitely accept it. – user38268 Apr 17 '12 at 4:40
Stated differently, you take the congruence relation generated by the individual pairs $x_i \sim x_j$. (See – Jim Belk Apr 17 '12 at 4:42
@JimBelk I know how to go from individual elements being equivalent to sums being equivalent. But what about the other way round? So am I correct in saying that in the direct sum, say $x_1 + x_2 + x_3 \sim x_4 + x_5 + x_6$ is $x_1 \sim x_4$, $x_2 \sim x_5$ and $x_3 \sim x_6$? – user38268 Apr 17 '12 at 4:50
I have posted an answer below that might help to clarify things. – Jim Belk Apr 17 '12 at 5:04
up vote 8 down vote accepted

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.

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Say we do $\bigoplus/C$ where $C$ is the submodule generated by the union of kernels of $\mu_{ik} - \mu_{jk}$ for any triple $i,j,k$ such that $i \leq k$ and $j \leq k$. Then I think what we get is not different from what you constructed above. However the problem now is how do we define kernel of $\mu_{ij} - \mu_{ik}$? The domains are not even the same....... – user38268 Apr 17 '12 at 5:19
@BenjaminLim: We do not define $C$ to be the submodule generated by "kernels of $\mu_{ik}-\mu_{jk}$". We define $C$ to be the submodule generated by all elements of the form $\mu_i(a)-\mu_j(\mu_{ij}(a))$, where $\mu_k$ is the embedding of $M_k$ in the $k$th component, and we consider all pairs $i\leq j$ and all $a\in M_i$. See this previous question. The reason you are confused is because you are mixing the two constructions. The $\mu_{ij}$ are not maps into the direct sum; they are maps between direct summands. (cont) – Arturo Magidin Apr 17 '12 at 5:26
@BenjaminLim (cont) What you need to do is relate the embeddings of the direct summands into the direct sum with the maps $\mu_{ij}$, and that is what $C$ needs to do: it provides the bridge between the maps $\mu_{ij}$ and the canonical embeddings. – Arturo Magidin Apr 17 '12 at 5:28
@BenjaminLim: I find it very frustrating that your responses seem to be basically, "Okay, but let's ignore everything you wrote and answer me this question instead." Especially when "this question" is the repetition of something that I have already pointed out is nonsensical. $\mu_{ik}-\mu_{jk}$ is not an element, it's not even a sensical expression: $\mu_{ik}$ and $\mu_{jk}$ are functions, with different domains. They are not elements that can be mapped into the direct sum, and they are not functions whose difference can be computed. The formula is nonsense. GIGO. – Arturo Magidin Apr 17 '12 at 5:38
@BenjaminLim: Then you need to keep the fact that "$a$" doesn't mean $a$, it means $(\ldots,a,\ldots)$ in mind all the time. I'm sure Atiyah and MacDonald can keep that fact straight without getting confused. I am fairly sure you cannot, given the evidence at hand. So make it explicit by using the canonical embeddings and writing them out. – Arturo Magidin Apr 17 '12 at 6:14

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.

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Thanks for your answer. Now suppose we forget the wikipedia definition for now, say I have an equivalence relation $\sim$ that says that $x_i \sim x_j$ iff there is $k\geq i,j$ such that $\mu_{ik}(x_i) = \mu_{jk}(x_j)$. I want to extend this to an equivalence relation on $\bigoplus M_i$. Now I looked up congruence relations, but they only tell you if say $a \sim b$, $c \sim d$ then $a + c \sim b + d$. What I am looking for is the reverse: How do we tell when say $a + c \sim b + d$? – user38268 Apr 17 '12 at 5:13
@BenjaminLim: Maybe this related answer can help you see how it is done. $a_i\sim b_j$ and $c_k\sim d_{\ell}$, there is $n\geq i,j$ with $\mu_{in}(a_i)=\mu_{jn}(b_j)$; there is $m\geq k,\ell$ with $\mu_{km}(c_k)=\mu_{\ellm}(d_{\ell})$. Let $t\geq n,m$. Then $\mu_{it}(a_i)+\mu_{kt}(c_k) = \mu_{nt}(\mu_{in}(a_i))+\mu_{mt}(\mu_{km}(c_k)) = \mu_{nt}(\mu_{jn}(b_j))+\mu_{mt}(\mu_{\ell m}(d_{\ell})) = \mu_{jt}(b_j)+\mu_{\ell t}(d_k)$. – Arturo Magidin Apr 17 '12 at 5:23
@BenjaminLim In general, there is no algorithm to tell whether a given pair is in the congruence relation generated by a given set of pairs. Indeed, a presentation of a group is essentially just a congruence relation generated by certain pairs on a free group, and the word problem for groups is unsolvable. – Jim Belk Apr 17 '12 at 6:06
@BenjaminLim However, the present case isn't quite as difficult. For a direct limit, you just need to keep mapping the sum forward under the $\mu_{ij}$'s. Two sums will be equal if and only if their images are eventually equal in the sequence of $M_i's$. – Jim Belk Apr 17 '12 at 6:08
@JimBelk Thanks for your all your input. – user38268 Apr 17 '12 at 6:45

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.

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I know, but how do we define whether say $x_1 + x_3 + x_4 \sim x_5 + x_6 + x_7$? The "extending linearly" bit is what is confusing me. – user38268 Apr 17 '12 at 3:32
My guess would be that since you are dealing with directed sets you can just put both sides in some bigger one (find a "max" of $A_1,A_3,A_4$ on your left side and put it in that set) – Alex Youcis Apr 17 '12 at 3:35
What do you mean by this? Can you explain a bit more? – user38268 Apr 17 '12 at 3:36
@BenjaminLim If $x\in M_i$ and $y\in M_j$ then first find $k\geq i,j$ and define $x+y= f_{ik}(x)+f_{jk}(y)$. – azarel Apr 17 '12 at 3:38
@azarel right I get that, but then how does this relate to showing whether $x + y \sim z +w$ say? – user38268 Apr 17 '12 at 3:39

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