Commutative Algebra, Notes 2: Localization

We will mostly adopt the Bourbaki definitions for the preliminary material.  An ideal \mathfrak{p} in a ring R is prime if A/\mathfrak{p} is an integral domain.  Two ideals \mathfrak{a},\mathfrak{b} are relatively prime if \mathfrak{a}+\mathfrak{b}=R.  A multiplicative subset of a set X is a submagma S of X.  In a ring, we assume it is a submagma with respect to multiplication of course.  We then have the canonical extension of R by formal inverses of S written S^{-1}R:=R[S^{-1}] (although I will use the notation R[S^{-1}] to reinforce the conceptualization of a ring extension).

Now suppose M is an R-module.  A natural question to ask is “can we induce some R[S^{-1}]-module”?  The natural choice will be the R[S^{-1}]-module M\otimes_R R[S^{-1}] where we define

r(m\otimes r')=m\otimes rr'

for r,r'\in R[S^{-1}].

Proposition 1.  Let R be a ring and S be a multiplicative subset.  Then there exists a ring \hat{R} and homomorphism f:R\to\hat{R} such that

  1. Elements of f(S) are invertible in \hat{R}.
  2. For every homomorphism g:R\to R' such that elements in g(S) are invertible in R', there exists a unique homomorphism h:\hat{R}\to R' such that g=h\circ f.

It naturally turns out that \hat{R}=R[S^{-1}] and f(r)=rs/s for some s\in S.  Similarly we have:

Proposition 2.  Let R be a ring, S be a multiplicative subset, and M be an R-module.  Then there exists an R[S^{-1}]-module \hat{M}  and homomorphism f:M\to\hat{M} such that

  1. For every s\in S, the map s:\hat{M}\to\hat{M} defined by \hat{m}\mapsto s\hat{m} is bijective.
  2. For every R-module N such that the map n\mapsto sn is bijective for every s\in S and if we have a homomorphism g:M\to N, there exists a unique homomorphism h:\hat{M}\to M such that g=h\circ f.

Here we can show that \hat{M}=M\otimes_R R[S^{-1}] and f(m)=m\otimes 1.  R[S^{-1}] is called the ring of fractions of R by S, while M\otimes_R R[S^{-1}], often denoted S^{-1}M for short, is called the module of fractions of M by S.

Definition 3.  Let \mathfrak{a} be an ideal in R.  We define the radical of \mathfrak{a} in R as

\sqrt[R]{\mathfrak{a}}=\sqrt{\mathfrak{a}}=\{r\in R:r^n\in\mathfrak{a}\mbox{~for some~}n\in\mathbb{N}\}.

We call \sqrt[R]{0}=\sqrt{0} the nilradical of R.  R is a reduced ring if \sqrt{0}=0.

Let A be an R-algebra and S be a multiplicative subset of R.  Then since A is an R-module, we can make A into an R[S^{-1}]-module

S^{-1}A=A\otimes_R R[S^{-1}].

But we can also turn it into an R[S^{-1}]-algebra.  We need a homomorphism

f:\left(A\otimes_R R[S^{-1}]\right)\otimes_{R[S^{-1}]}\left(A\otimes_R R[S^{-1}]\right)\to A\otimes_R R[S^{-1}].

But the domain collapses to (A\otimes_RA)\otimes_RR[S^{-1}] and then to A\otimes_RR[S^{-1}] under the product of A.  So we have the canonical product we need.

[1]  Bourbaki, N.  Commutative Algebra, Ch 1-7.  Springer-Verlag.  1989.

Advertisements

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 )

Twitter picture

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

Facebook photo

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

Google+ photo

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

Connecting to %s

%d bloggers like this: