This post is a continuation of the Fractions and their Limitations post.
So this concept of localization is one that I had never encountered in my algebra course, but which I have heard come up in a lot of talks, and in particular seems to be a big deal among algebraic geometers (#shamelessplugforthatblogIlike).
The wikipedia article on localization is… okay? Like, i was mostly able to fill in the details just by using that article. But I think it had kind of a strange emphasis.
Let me back up a little bit. The point of localization is to add inverses to a ring, which may not be an integral domain. That is, the ring might have two nonzero elements $a$ and $b$ such that $ab=0$; any $a$ for which there exists such a $b$ is called a zero divisor. In this setting, the field of fractions construction fails, and in general you should not expect to be able to just haphazardly add inverses.
Localization is a verb that requires both a ring $R$ and a subset $S\subseteq R$, and we denote by $R_S$ or $S^{-1}R$ the localization of $R$ away from $S$. There are various nice ways to construct it, including one method by constructing an equivalence relation on $R\times S$ that looks fairly similar to the field of fractions equivalence.
As it says on the Wikipedia page, because we must complete $S$ with respect to the multiplication operation, we may as well take $S$ to be multiplicatively closed. More rigorously, we can show that $S^{-1}R=\overline{S}^{-1}R$, where $\overline{S}$ is the multiplicative closure of $S$ (that is, take $S$ and then throw in all products of its elements).
Some basic observations about the behavior of localization:
If $S$ does not contain a zero divisor, then actually $R_S$ is isomorphic to $R[S^{-1}]$, that is, $R$ adjoined with formal inverses of the elements in $S$.
On the other extreme, if $S$ contains a nilpotent element, then $R_S$ is the trivial ring. To anthropomorphize a bit, nilpotents really do not want to be divided by. This makes sense, because we can kind of think of nilpotents as “almost zeros”.
We might think that $S$ will also kill $R$ if it contains a zero divisor, but this isn’t actually the case. The following example, I think, is fairly instructive. Suppose that $R=\Bbb Z\times \Bbb Z$. If we try to localize away from $(1,0)$, we can think of this as adjoining $(x,y)$ so that $(1,0)(x,y)=(1,1)$, but then note that $(1,0)(x,y)=(x,0)$, and so we conclude that $(0,1)=(0,0)$. This informal treatment (see below) gives the correct answer of $R_{(0,1)}=\Bbb Z$.
So somehow when $s$ is a zero divisor, it kills off the “nilpotent part” of $R$ that $s$ can “detect”. I feel there might be a way to make what I’m saying even a little rigorous, but I’m not sure what that might be.
To be a little more upfront: if $s$ is nilpotent, adding inverses formally still gives the same result: $R[s^{-1}]=R[x]/(sx-1)$, because we can consider the polynomial $s^nx^n-1$ for sufficiently large $n$. on one hand this is $-1$, but on the other hand it has a factor of $(sx-1)$ and therefore is zero. But I still have a integrally-minded brain that doesn’t like writing down $R[a]$ to mean a ring smaller than $R$.
On the other hand, the zero divisor case is definitely just a mnemonic; since there’s no a priori reason why the localization of a direct product still needs to have a direct product structure. It happens to get the right answer, but you have to go through some other reasoning to do it formally.
Okay, and now for the elephant in the room— why the heck is this thing called localization? What is “made local”?
The answer here can be made most clear by looking at the way it seems to get used in algebraic geometry: consider a prime ideal. An equivalent condition to an ideal being prime is that its complement is multiplicatively closed, and so it is a fairly natural set to localize away from. Localizing away from the complement of a set is usually called localizing at the set, which is an intuitive picture.
When you do this, you can go through the motions to show that what you get is a ring with precisely one maximal ideal. So, when we localize at a prime ideal, we have created a ring with a particularly simple structure near the top of the ideal poset. In other words, we have “thrown out” the ideals of the ring that prevent the prime ideal from being maximal, hence “localizing” the ideal structure. When you think about this, it starts to make the behavior of localization make more sense: there are two ways to get rid of ideals, and that is either to cut out the ring elements that allow these ideals to exist, or to add more elements to make the quotient ring look more like a field.
(That reminds me; localization seems to me, in this intuition, to be something like a dual of the quotient, at least in the context of prime ideals. Modding out by a prime ideal kills off everything that isn’t above it, localizing at a prime ideal kills off everything that is above it. Not sure if there’s anything more substantial to it than that.)
