Intuition behind Direct limits

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

Solution 1:

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".

Solution 2:

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 $M_f/M \cong C$.