# 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

$M'\stackrel{f}{\longrightarrow}M\stackrel{g}{\longrightarrow}M'',$

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

$L_k\stackrel{f_k}{\to}\cdots\stackrel{f_2}{\to}L_1\stackrel{f_1}{\to}L_0\stackrel{f_0}{\to}E$

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 $i$th 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.