Monthly Archives: November, 2011

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.


Commutative Algebra, Notes 1: Flat Modules

We will assume all rings are commutative and unital.

Definition 1.  Let R be a ring, E a right R-module, and M a left R-module.  E is M-flat if for every left R-module M' and monomorphism \phi:M'\to M, the homomorphism 1_E\otimes\phi:E\otimes_RM'\to E\otimes_RM is injective.

Some basic results can be shown:

Proposition 2.  If E is a right R-module and M is a left R-module such that E is M-flat, then if N\leq M, we have that E is N-flat and M/N-flat.  Also if \{M_i\}_{i\in I} is a family of left R-modules such that E is M_i-flat for all i\in I, then E is \oplus_{i\in I}E_i-flat.

Proposition 3.  The following are equivalent for a right R-module E.

  1. The canonical map f:E\otimes I\to E\otimes R_S is injective (where R_S=S^{-1}R for some multiplicative subset S).
  2. E is M-flat for every left R-module M.
  3. For every exact sequence of left R-modules and homomorphisms


   the sequence

E\otimes M'\stackrel{1\otimes f}{\longrightarrow}E\otimes M\stackrel{1\otimes g}{\longrightarrow}E\otimes M''

   is exact.

Definition 4.  A right R-module E satisfying the equivalent conditions above is flat.  (A dual definition exists for left modules)

Proposition 5.  Let E be a right R-module, then the following are equivalent.

1.  The sequence M'\stackrel{f}{\to}M\stackrel{g}{\to}M'' of left R-modules is exact iff

E\otimes M'\stackrel{1\otimes f}{\to}E\otimes M\stackrel{1\otimes g}{\to}E\otimes M''

is exact.

2.  E is flat and E\otimes N=0\Rightarrow N=0 for a left R-module N.

3.  E is flat and for every homomorphism f:N'\to N, 1_E\otimes f=0\Rightarrow f=0.

4.  E is flat and for every maximal left ideal \mathfrak{m} of R, E\neq E\mathfrak{m}.

Definition 6.  A right R-module E satisfying the above conditions is faithfully flat.

Recall that for a subset S of an R-module M, the annihilator of S is defined as

Ann_R(S)=\{r\in R:rs=0\mbox{~for all~}s\in S\}.

An R-module M is faithful if Ann_R(M)=0.

Definition 7.  Let E be a left R-module and suppose


is an exact sequence such that each L_i is free and finitely generated.  Then E is said to be finitely presented, and the exact sequence is called a finite presentation of E of length k.

Note that if f:R\to S is a ring homomorphism, then a finite presentation of a left R-module E induces a finite presentation of E\otimes_R S where the ith term of the first sequence, L_i, becomes L_i\otimes_R S with rs:=f(r)s.  (We will consider E to be the -1 term of the original sequence)

Proposition 8.  Let E be a left R-module.

  1. If E admits a finite presentation, then E is finitely generated.
  2. If R is left Noetherian, then if E is finitely generated, it admits a finite presentation.
  3. If E is projective and finitely generated, then it admits a finite presentation.

Definition 9.  Let E be a left R-module.  E is pseudo-coherent if every finitely generated submodule of E is admits a finite presentation.  E is coherent if it is pseudo-coherent and finitely generated.

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