# Localization 2

Let’s figure out what “local” means and see if our construction somehow makes a local ring, i.e. is a “localization.”

Local: A ring is called local if there is a unique maximal ideal. This seems like a rather silly term, but it actually makes sense when you look at how rings arise in algebraic geometry or manifold theory. We won’t go there, though.

Sadly, it turns out that $S^{-1}R$ is not always a local ring. But this is where primality comes into play. If $P\subset R$ is a prime ideal then $S=R\setminus P$ is a multiplicative set. Suppose it weren’t, then there would be two elements $x,y\in S$ such that $xy\notin S$, i.e. $xy\in P$, but this is impossible, since by definition either $x\in P$ or $y\in P$. We now denote the localization of $R$ at $P$, to be $S^{-1}R=(R\setminus P)^{-1}R$ which we denote with the shorthand $R_P$. This does turn out to be local since by the property listed last time of the embedding $\phi^{-1}(S^{-1}P)=P$, so $S^{-1}P=\{r/s : r\in P, \ s\notin P\}$ is the unique maximal ideal in $R_P$.

Proof: Suppose $x\in R_P$, then $x=r/s$ with $r\in R$ and $s\notin P$. If $r\notin P$, then $r/s$ is a unit in $R_P$. So all nonunits are in $S^{-1}P$. Now if I is any ideal in $R_P$ that contains an element $r/s$ with $r\notin P$, then $I=R_P$. Thus every proper ideal in $R_P$ is contained in $S^{-1}P$. So $R_P$ is local with unique max ideal $S^{-1}P$. For notational purposes outside of this blog, people usually write the prime ideal as $\mathfrak{p}$ and the unique maximal ideal of $R_\mathfrak{p}$ as $\mathfrak{p}R_{\mathfrak{p}}$.

I guess I’ve been rather sparse on the examples. The first one that comes to mind is surely to take $\mathbb{Z}=R$. Then our prime ideals are just the principal ideals generated by the primes, so take $P=p\mathbb{Z}$ for some prime p. Then $\mathbb{Z}_P=\mathbb{Z}_{(p)}$.

I guess the importance of prime ideals leads us to explore some properties of prime ideals that could be useful.

Property 1: If $S\subset R$ is any multiplicative set (not containing 0) and if $P\subset R\setminus S$ is a maximal ideal, then $P$ is prime. Also, any ideal $I\subset R\setminus S$ is contained in such a $P$. I’ll omit proving this. The first part is fiddling with things until it works and the second statement uses Zorn’s Lemma.

OK, well I thought I had some other properties, but I can’t seem to find them/think of them now. I’m not sure where I’m going next. I’ll either move on to some related things to get at this better like the nilradical, or I’ll generalize this one more time to modules and do it using the categorical construction. If anyone has suggestions on which of these paths to take, just post. You probably have a few days as I’ll get busy again.

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### Author: hilbertthm90

I am a mathematics graduate student fascinated in how all my interests fit together.