What is the relationship between local ring and localization?

Question

As I am learning localization, it is easy to see that the localization of a ring $ R $ at $ p $, where $ p $ is a prime ideal of $ R $, is a local ring. And I also know that the localization of a local ring is actually isomorphic to itself hence also a local ring. I am looking for more statements involving both local ring and localization, can someone give me such examples?

Answer 1

It is not true that the localization of local ring is local: only if you localize with respect to a prime ideal.

In fact this is a general result: the localization of a commutative ring $R$ with respect a multiplicative subset $S$ is a local ring (so a non-zero ring with a unique maximal ideal) if and only if you are localizing with respect to a prime ideal $P$, so $S^{-1}R=R_P$.

Comment: A previous version I said that $S=R\setminus P$, which is not true (I really wanted to say what it is written now, but I wrote wrongly). What it is true is that the saturation of $S$ is the complement of a prime ideal, so you get the same localization. I thank Msrc Paul and barto for this correction.

Answer 2

The question deserves an answer which is probably too broad to fit in a single answer, as the topic is so vast and has so many ramifications that becomes very difficult to talk about all of them.

I think a good starting point, is to reflect on the nature of localization by reviewing the relevant "local properties". The term is borrowed from algebraic geometry, as you certainly know: if $R$ is the ring of regular functions of an affine variety, then its localization $R_p$ at a prime ideal $p$ is an open neighbourhood of the point $x_p$ corresponding to the ideal $p$. So, passing to $R_p$ is like looking around the surroundings of that particular point.

Now, the structure of local ring is that such rings have an unique maximal ideal; this goes accordingly with the geometric picture, if you get that maximal ideals are points in a variety. So, in the local ring $R_p$, the only maximal ideal represents the centre of the neighbourhood, while the extra material is the local data around this point.

It is crucial, in Algebraic geometry, to be able to discern those properties that hold "locally" (so around a point) but not "globally" (in the whole variety); this, at least for affine varieties, is completely equivalent to understand which properties holding for $R_p$ fail when passing to $R$.

More refined properties of algebraic varieties that have a strong local flavour are properties of singularities. Indeed, when an algebraic variety is regular at a point $p$, its local ring at that point is a discrete valuation ring. The presence of a valuation is essential to determine local equations for that variety (centered at the point).

Anyway, I feel like I'm just touching many deep topics. If you are interested in learning more about the "local" behaviour of local rings (pun intended!) you should definitely look at Eisenbud's Commutative algebra with a view towards algebraic geometry.

Answer 3

It turns out that modern algebraic geometry is considerably more general than the "geometric" case referred to in the discussion. It provides a dictionary between algebraic notions in any ring $R$, and geometric notions in a certain object called its "spectrum", denoted $\text{Spec}\,R$. This object is a topological space whose points are all prime ideals of $R$---not just maximal ones!---and whose topology is a generalization of the Zariski topology that we might have seen on the maximal ideals. It also comes equipped with a notion of its "sheaf of functions", which ties together the elements of many of the localizations of $R$. The operation of inverting an element of $R$---i.e. passing to $S^{-1}R$ where$$S = \{1, f, f^2, f^3, \ldots\},$$where $f \in R$---on the algebra side corresponds to the geometric operation of removing from $\text{Spec}\,R$ the closed subset of all points where the "function" $f$ "vanishes", i.e. those primes to which $f$ belongs. Localization at a prime ideal $\mathfrak{p}$ of $R$ corresponds to taking a kind of intersection of all open subsets of $\text{Spec}\,R$ containing the point corresponding to $\mathfrak{p}$.

As for local versus global properties, yes, we pay attention to the distinction. But we usually gain an intuition for each property about why it is local or not, and in most cases it is not that hard to determine. I think the most subtle example, at least in the beginning, is the Hausdorff property. It seems that it should be local, but it is not---take two copies of the real line, and glue them via the identity map at all nonzero points, so that we get a real line with the origin "doubled". It is a union of two Hausdorff spaces, so it is locally Hausdorff, but it is not Hausdorff because we cannot separate the two origins into disjoint neighborhoods.