Abstract Nonsense

Crushing one theorem at a time

Localization (Pt. III)

Point of Post: This post is a continuation of this one.

\text{ }

This tells us how to create a lot of ideals in D^{-1}R, sadly it doesn’t give us all the ideals! That said, there is a very nice correspondence between certain ideals (prime ideals) of R and those of D^{-1}R. But first, we need a small lemma

\text{ }

Lemma: Let R be a commutative unital ring, and D a unital multiplicative subset. Then, for any ideal \mathfrak{a} of R, one has that \langle\ell_D(\mathfrak{a})\rangle=D^{-1}R if and only if  \mathfrak{a}\cap D\ne\varnothing.

Proof: If d\in\mathfrak{a}\cap D then we obviously have, by the above characterization of \langle \ell_D(\mathfrak{a})\rangle that \frac{d}{d}=1\in\langle \ell_D(\mathfrak{a})\rangle and so evidently \langle\ell_D(\mathfrak{a})\rangle=D^{-1}R. Conversely, if \langle \ell_D(\mathfrak{a})\rangle=D^{-1}R then for some a\in\mathfrak{a} and d\in D we have that \frac{a}{d}=\frac{1}{1} and so d'a=dd'. But, clearly dd'\in D since D is multiplicative, but dd'\in\mathfrak{a} since \mathfrak{a} is an ideal and dd'=d'a. Thus, d'\in\mathfrak{a}\cap D. \blacksquare

\text{ }

With this we are able to prove the following correspondence between \text{Spec}(D^{-1}R) and \text{Spec}(R)

Theorem: Let R be a commutative unital ring and D a unitally multiplicatively closed subset. Then, for \mathfrak{p}\in\text{Spec}(R), one has that \ell_D(\mathfrak{p})\in\text{Spec}(D^{-1}R) if and only if \mathfrak{p}\cap D=\varnothing.

Proof: Evidently if \mathfrak{p}\cap D=\varnothing then by the above theorem we have that \langle \ell_D(\mathfrak{p})\rangle=D^{-1}R and so is not proper, and so by definition, not prime. Conversely, suppose that \mathfrak{a}\cap D=\varnothing. Let \displaystyle \frac{a}{d_1},\frac{a'}{d_2}\in D^{-1}R be such that \frac{aa'}{d_1d_2}=\frac{p}{d_3}\in\langle \ell_D(\mathfrak{p})\rangle. We see then that there exists d_4\in D such that d_4aa'd_3=d_1d_2pd_4. Note then that the right hand side of this formula is in \mathfrak{p} and so d_4d_3aa'\in\mathfrak{p}.  Now, we assumed that \mathfrak{p}\cap D=\varnothing, and so d_4d_3\notin\mathfrak{p} and thus since \mathfrak{p} is prime we have that a\in\mathfrak{p} or a'\in\mathfrak{p}. Thus, \frac{a}{d_1}\in\langle \ell_D(\mathfrak{p})\rangle or \frac{a'}{d_2}\in\langle \ell_D(\mathfrak{p})\rangle. The conclusion follows. \blacksquare

\text{ }

The last piece of the puzzle for how \text{Spec}(D^{-1}R) and \text{Spec}(R) are related is given in the following theorem:

\text{ }

Theorem: Let R be a commutative unital ring, D a unital multiplicative subset. Then, if \mathfrak{p} is a prime ideal in R disjoint from D then \ell_D^{-1}\left(\langle \ell_D(\mathfrak{p})\rangle\right)=\mathfrak{p}.

Proof: We trivially have that \mathfrak{p}\subseteq\ell_D^{-1}\left(\langle\ell_D(\mathfrak{p})\rangle\right) and so it suffices to prove the reverse inclusion. Indeed, let \frac{r}{1}\in\langle \ell_D(\mathfrak{p})\rangle, then there exists d,d'\in D and p\in\mathfrak{p} with dd'r=d'p. Now, the right hand side of this equation is in \mathfrak{p} and so obviously the left hand side is as well. But, since \mathfrak{p} is prime this tells us that either dd'\in\mathfrak{p} or r\in\mathfrak{p}, but since we assumed that the former can’t happen, we may conclude the latter. Since r was arbitrary the conclusion follows. \blacksquare

\text{ }

From this we get the conclusive relationship between the spectrum of a ring and the spectrum of its localization at a unital multiplicative subset:

\text{ }

Theorem: The map \left\{\mathfrak{p}\in\text{Spec}(R):\mathfrak{p}\cap D=\varnothing\right\}\to\text{Spec}(D^{-1}R) given by \mathfrak{p}\mapsto \langle \ell_D(\mathfrak{p})\rangle is a bijection.

\text{ }

\text{ }

Localizing at a Prime Ideal

\text{ }

We now discuss what is perhaps one of the most important cases of localizations, localization of a ring at a prime ideal, usually denoted R_\mathfrak{p}. This terminology is slightly confusing since it would seem to imply the multiplicative subset we are localizing at is the prime ideal itself. But, this is impossible since no prime ideal contains 1. In fact, we define for a commutative unital ring R and a prime ideal \mathfrak{p}\in\text{Spec}(R)  the localization at the prime ideal \mathfrak{p} denoted R_\mathfrak{p} to be equal to D^{-1}R where D=R-\mathfrak{p}. Perhaps one of the most instructive examples comes from taking R=\mathbb{Z} and \mathfrak{p}=(p) for some prime p. We see then by inspection that \mathbb{Z}_{(p)} is precisely the reduced elements of \mathbb{Q} which have denominators coprime to p.

\text{ }

For now all I want to say about localization at a prime ideal, is that they create what are called (unsurprisingly) local rings. Local rings are those with a unique maximal ideal. We know stupid examples, like simple rings such as fields or matrices over fields, but we have never seen an example of a local ring which has a non-trivial ideal theory. Well, the localization of a ring at a prime ideal provides such examples. Indeed:

\text{ }

Theorem: Let R be a commutative unital ring and D a unital multiplicative subset. Then, if \mathfrak{p} is a prime ideal of R the ring R_\mathfrak{p} has the unique maximal ideal \mathfrak{p}R_\mathfrak{p}.

Proof: We know from the above analysis of \text{Spec}\left(R_\mathfrak{p}\right) that any maximal ideal of R_\mathfrak{p} will be the image of a maximal-by-containment ideal in R disjoint from R-\mathfrak{p}, and so a maximal-by-containment ideal contained in \mathfrak{p}. Since clearly the only such maximal-by-containment ideal in the ideal sitting inside \mathfrak{p} itself, we may conclude that the only element of \text{MaxSpec}(R_\mathfrak{p}) is \langle \ell_D(\mathfrak{p})\rangle=\mathfrak{p}R_\mathfrak{p}. \blacksquare

\text{ }

\text{ }


[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.


October 7, 2011 - Posted by | Algebra, Ring Theory | , , , , , , , , ,

1 Comment »

  1. […] By the previous problem it suffices to prove that every prime ideal of is principal. That said, we know that every prime ideal of is of the form for some prime ideal . But, since is a PID we know […]

    Pingback by PIDs (Pt. I) « Abstract Nonsense | October 21, 2011 | Reply

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: