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Let $R$ be a commutative ring and $x\in R$ be a nonzero divisor. Then i know that the direct limit of $R\mapsto R\mapsto R\mapsto\cdots $, where each map is multiplication by $x$ is $R_x$, the localization of $R$ at ${1,x, x^2,...}$.

Similarly the direct limit of $R/x^n\mapsto R/x^{n+1}\mapsto\cdots $, where maps are multiplication by $x$ is $R_x/R$.

My Question: How does one guess what the direct limit of a given direct system is, once the guess is made, then one can go about proving it using the universal property. Can someone provide an intuition for direct limits, at least in the above 2 cases? Thanks

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    $\begingroup$ You mean directed colimits. The description of $R_x$ holds without any assumption on $x$. It is purely formal, holds for arbitrary $R$-modules and just "spreads out" $x$ to make it invertible. Try to understand the case of $R=k[x]$ first. $\endgroup$ May 3, 2013 at 17:45
  • $\begingroup$ @Martin, Yes description $R_x$ holds, but i wanted to avoid $x$ being nilpotent, so i put a stronger condition. Either way, once someone describes how one could arrive at the direct limit in the above 2 examples, i would be satisfied. What do you mean by "spreads out" $x$ to make it invertible. That can be useful for me if you can say something more about it. $\endgroup$
    – messi
    May 3, 2013 at 17:54

2 Answers 2

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I think the touchstone for understanding direct limits is understanding directed unions.

A collection $C$ of sets is directed if for every $X,Y\in C$, there exists $Z\in C$ containing both $X$ and $Y$. This becomes a direct system using inclusion mappings.

Now just by using the directness of this collection, you can compare any two sets (and inductively, any finite number of sets) by finding a set that contains them all. But what if you want to compare more than finitely many? That is what the limit is going to do: the direct limit for the system above turns out to be $\cup C$, and so you get a sense that the limit is "the limit of finite approximations by compositions of the morphisms".

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  • $\begingroup$ thanks, but can you tell me how I can guess what the direct limit is in the 2 examples above? What would be the reasoning to guess $R_x$ or $R_x/R$ via directed unions? $\endgroup$
    – messi
    May 3, 2013 at 17:13
  • $\begingroup$ @messi Actually I've never heard of those two examples until you brought them up :) I will have to think about it a while. One gets the impression that at each stage in the chain, $y\in Im\phi^n$ means that $y=x^nz$ for some $z$, and so $z=y/x^n$. Somehow this should lead to the localization description. The problem is I want to start with $z$ first, not $y$... $\endgroup$
    – rschwieb
    May 3, 2013 at 18:00
  • $\begingroup$ Thanks, i think this is a more useful comment for me, and it somehow gives at least an idea that, the direct limit might be $R_x$. It would be nice if you can say something more about this and the second example. $\endgroup$
    – messi
    May 3, 2013 at 18:04
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Let $M$ be an $R$-module, $f \in R$ and let $N$ be the colimit of $M \xrightarrow{f} M \xrightarrow{f} \dotsc$. Directed colimits are easy to construct: Elements come from elements of the individual modules, and are identified if they get sent to the same element by some transition map. So in our case, if $i_n : M \to N$ denotes the $n$th colimit inclusion, every element of $N$ has the form $i_n(m)$ for some $m \in M$, and $i_n(m)=i_{n'}(m')$ iff $f^{p-n} m=f^{p-n'} m'$ for some $p \geq n,n'$. It follows that $ i_0(m) = f^n \cdot i_n(m)$ and that $f$ acts as an isomorphism on $N$. Hence, every element of $N$ has the form $i_0(m)/f^n$ for some $m \in M$ and $n \geq 0$, and we have $i_0(m)/f^n=i_0(m')/f^{n'}$ iff $f^p f^{n'} i_0(m')=f^p f^{n} i_0(m)$ for some $p \geq 0$. But this is the usual construction of the localization $M_f$, so that $N = M_f$.

Here is a more elegant and abstract explanation: By definition of a colimit, a homomorphism $\alpha : N \to T$ corresponds to a family of homomorphisms $\alpha_n : M \to T$ with $\alpha_n = \alpha_{n+1} f$ for all $n \geq 0$. Taking $\alpha_n=i_{n+1}$, we can construct a homomorphism $N \to N$ which easily seen to be inverse to $f$. If $T$ is a module on which $f$ is invertible, then the above shows that $\alpha$ is completely determined by $\alpha_0$. Hence, $N$ is the universal module over $M$ on which $f$ becomes invertible. That is, $N=M_f$. Actually the same localization works for an arbitrary endomorphism of an object in any category with directed colimits. For example, localizing the set $\mathbb{N}$ with respect to the successor function gives $\mathbb{Z}$.

Here is some intuition for this: Start with $M$, we want to make $f$ invertible on $M$. So for $m \in M$ we want to find some (unique) $m/f$ (in a module extending $M$) with $f \cdot m/f = m$. For this, just add adjoin another copy of $M$, but whose elements should behave like $m/f$. Roughly we just add some extra space in order to insert the inverses. So we have two copies $i_0(M)$ and $i_1(M)$ of $M$, but want to ensure that $f \cdot i_1(m) = i_0(m)$. So just take the corresponding quotient of $i_0(M) \oplus i_1(M)$. But with $i_1(M)$ we have to continue this way. In the end, we take the quotient of $i_0(M) \oplus i_1(M) \oplus i_2(M) \oplus \dotsc$ by $f i_n = i_{n+1}$, i.e. the directed colimit of $M \xrightarrow{f} M \xrightarrow{f} M \xrightarrow{f} \dotsc$.

Actually this is a very useful description of localizations. See Eisenbud's book on commutative algebra for some applications. For example, one immediately gets that $R_f$ is flat over $R$, since directed colimits of flat modules are flat. By another colimit argument, we get that $R_S$ is flat over $R$ for every multiplicative subset $S$.

Now for your second example: Let $M$ be an $R$-module and $f \in R$ be such that $f : M \to M$ is injective. This implies that $M \to M_f$ is injective (which can be seen, for example, from the colimit description), and therefore $M_f / M$ makes sense. Let $C$ be the colimit of $M/f^0 M \xrightarrow{f} M/f^1 M \xrightarrow{f} \dotsc$. This sequences admits an obvious epimorphism from $M \xrightarrow{f} M \xrightarrow{f} \dotsc$. This induces an epimorphism $M_f \to C$. What is the kernel? If $m/f^n$ lies in the kernel, this means that $m \bmod f^n M$ vanishes in $C$. Since all the transition maps $f : M/f^p M \to M/f^{p+1} M$ are injective, this means that $m \bmod f^n M$ vanishes in $M/f^n M$, i.e. $m \in f^n M$. Hence, $m/f^n \in M$. So the kernel is just $M$, and we see $M_f/M \cong C$.

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  • $\begingroup$ Let $Z^2\mapsto Z^2\mapsto Z^2\mapsto\cdots$, where the maps are given by the matrix with first row [1 2] and second row [1 3]. What will be the direct limit in this case? Here $Z$ is the integers. $\endgroup$
    – messi
    May 4, 2013 at 9:32

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