# General theory of localizations of rings and modules

V. V. Bavula

###### Abstract

The aim of the paper is to start to develop the most general theory of
localizations/inversion. Several new concepts are introduced and studied.

Key Words: Localizable set, localization of a ring at a localizable set,
Goldie’s Theorem, absolute quotient ring of a ring, maximal localizable set,
maximal left denominator set, the localization radical of a ring, maximal
localization of a ring, localizable element, non-localizable element,
completely localizable element, direct limit of rings, ultrafiler.

Mathematics subject classification 2020: 16S85, 16P50, 16U20, 12L10, 13B30,
16D30.

${\bf Contents}$

  1. 1.

Introduction.

  2. 2.

Localization of a ring at a set.

  3. 3.

Maximal localizable sets, the localization radical, the sets of localizable
and non-localizable elements.

  4. 4.

Localizations of commutative rings.

  5. 5.

Localizations of semiprime left Goldie rings and direct products of division
rings.

  6. 6.

Localization of a module at a localizable set.

##  1 Introduction

Throughout, all rings are associative, with a unit element 1, which is
inherited by subrings, preserved by homomorphisms and which acts unitally on
modules. Let $K$ be a commutative ring with 1. In the paper, a ring $R$ means
a $K$-algebra with fixed ring homomorphism $\nu_{R}:K\rightarrow R$ where the
image ${\rm im}(\nu_{R})$ belongs to the centre $Z(R)$ of $R$. A ring
homomorphism $f:R\rightarrow R^{\prime}$ is a $K$-homomorphism, that is
$f\nu_{R}=\nu_{R^{\prime}}$. Ring homomorphism means a $K$-homomorphism. For a
ring $R$, let $R^{\times}$ be its group of units and $\mathfrak{n}_{R}$ be its
prime radical.

Inversion of elements in a ring is an important and difficult operation that
‘simplifies’ the situation as a rule. Ore’s method of localization is an
example of a theory of one-sided fractions where by definition only the
elements of a denominator set can be inverted and the result is a ring of one-
sided fractions which always exsists, i.e. it is not equal to zero, [14, 9,
13]. In [1, Theorem 4.15], it is proven that the elements of an arbitrary
(left and right) Ore set can be inverted and the result is also a ring of one-
sided fractions (which always exists) but in general Ore set is not a
denominator set. This was the starting point in [7] and [8] where the most
general theory of one-sided fractions was presented. The goal of the paper is
to consider the general situation and to start to develop the most general
theory of localization/inversion where the result of localization/inversion of
elements does not necessarily yields one-sided fractions.

Let $R\langle X_{S}\rangle$ be a ring freely generated by the ring $R$ and a
set $X_{S}=\\{x_{s}\,|\,s\in S\\}$ of free noncommutative indeterminates
(indexed by the elements of the set $S$). Let $I_{S}$ be the ideal of
$R\langle X_{S}\rangle$ generated by the set $\\{sx_{s}-1,x_{s}s-1\,|\,s\in
S\\}$. The factor ring

$R\langle S^{-1}\rangle:=R\langle X_{S}\rangle/I_{S}.$ (1)

is called the localization of $R$ at $S$. Let ${\rm ass}(S)={\rm ass}_{R}(S)$
be the kernel of the ring homomorphism

$\sigma_{S}:R\rightarrow R\langle S^{-1}\rangle,\;\;r\mapsto r+I_{S}.$ (2)

The factor ring $\overline{R}:=R/{\rm ass}_{R}(S)$ is a subring of $R\langle
S^{-1}\rangle$ and the map $\pi_{S}:R\rightarrow\overline{R}$,
$r\mapsto\overline{r}:=r+{\rm ass}_{R}(S)$ is an epimorphism. The ideal ${\rm
ass}_{R}(S)$ of $R$ has a complex structure, its description is given in [7,
Proposition 2.12] when

$R\langle S^{-1}\rangle=\\{\overline{s}^{-1}\overline{r}\,|\,s\in S,r\in R\\}$

is a ring of left fractions. There is an example of a domain $R$ and a finite
set $S$ such that ${\rm ass}(S)=R$, i.e. $R\langle S^{-1}\rangle=\\{0\\}$ [12,
Exercises 9.5]. Proposition 2.10.(1) and its proof describe an explicit ideal
$\mathfrak{a}(S)$ of $R$ that is contained in ${\rm ass}_{R}(S)$. The ideal
$\mathfrak{a}(S)$ is the least ideal such that the elements of the set
$\\{s+\mathfrak{a}(S)\in R/\mathfrak{a}(S)\,|\,s\in S\\}$ are non-zero-
divisors in the factor ring $R/\mathfrak{a}(S)$, [7, Proposition 1.1]. The
proof of Proposition 2.10 contains an explicit description of the ideal
$\mathfrak{a}(S)$. If $S$ is a left denominator set then the set

${\rm ass}_{l}(S):=\\{r\in R\,|\,sr=0\;\;{\rm for\;some}\;\;s\in S\\}$

is an ideal of $R$ and ${\rm ass}_{l}(S)=\mathfrak{a}(S)={\rm ass}_{R}(S)$. In
general, for a multiplicative set $S$, ${\rm
ass}_{l}(S)\subseteq\mathfrak{a}(S)\subseteq{\rm ass}_{R}(S)$ (in general, the
inclusions are strict).

In Section 2, we recall Ore’s method of localization. The aim of this section
is to introduce several new concepts that are associated with localization of
a ring at a set and to prove some general properties for them (Corollary 2.2,
Lemma 2.6, Proposition 2.7, Proposition 2.9, Corollary 2.11): A subset $S$ of
$R$ is called a localizable set if $R\langle S^{-1}\rangle\neq\\{0\\}$;
$\mathrm{L}(R)$ is the set of all localizable sets of $R$ and
$\mathrm{L}(R,\mathfrak{a}):=\\{S\in\mathrm{L}(R)\,|\,{\rm
ass}_{R}(S)=\mathfrak{a}\\}$; $\mathrm{L}{\rm oc}(R)$ and $\mathrm{L}{\rm
oc}(R,\mathfrak{a})$ are the sets of $R$-isomorphism classes in the sets

$\\{R\langle S^{-1}\rangle\,|\,S\in\mathrm{L}(R)\\}\;\;{\rm
and}\;\;\\{R\langle S^{-1}\rangle\,|\,S\in\mathrm{L}(R,\mathfrak{a})\\},$

respectively;
$\mathrm{L}(R,\mathfrak{a},A):=\\{S\in\mathrm{L}(R,\mathfrak{a})\,|\,R\langle
S^{-1}\rangle\simeq A$, an $R$-isomorphism (necessarily unique)$\\}$. All the
above sets are partially ordered sets.

Proposition 2.1 is the universal property of localization. Proposition 2.4 is
a criterion for a subset of a ring to be a localizable set. Theorem 2.23 and
Theorem 2.24 demonstrates some of the results of this section for Ore sets.
Direct limits are essential in proving that the partially ordered sets above
admit maximal elements. Several results on direct limits of rings are proven
that are used later in the paper (Theorem 2.14, Corollary 2.15, and Lemma
2.16). For a ring $R$, the concept of the absolute quotient ring

$Q_{a}(R):=\varinjlim_{S\in\mathrm{L}(R)}R\langle S^{-1}\rangle$

is introduced. Theorem 2.18 is a description of the ring $Q_{a}(R)$ in terms
of maximals localization sets of $R$. Corollary 2.22 presents a sufficient
condition for a ring $R$ to have at least two maximal localization sets.

In Section 3, for a ring $R$, the following concepts are introduced: the
maximal localizable sets, the localization radical $\mathrm{L}{\rm rad}(R)$,
the set of localizable elements ${\cal L}\mathrm{L}(R)$, the set of non-
localizable elements ${\cal N}{\cal L}\mathrm{L}(R)$, the set of completely
localizable elements ${\cal C}\mathrm{L}(R)$, and the complete localization

$Q_{c}(R):=R\langle{\cal C}\mathrm{L}(R)^{-1}\rangle$

of $R$. These sets are invariant under the action of the automorphism group of
$R$. Proposition 3.10 shows that there are tight connections between the sets
${\cal L}\mathrm{L}(R)$, ${\cal N}{\cal L}\mathrm{L}(R)$, and $\mathrm{L}{\rm
rad}(R)$. For an arbitrary ring, Theorem 3.1 states that the set of maximal
localizable set is a non-empty set and every localizable set is a subset of a
maximal localizable set.

Proposition 3.5.(1), is an explicit description of the largest element ${\cal
S}(R,\mathfrak{a},A)$ of the poset $(\mathrm{L}(R,\mathfrak{a},A),\subseteq)$.
Corollary 3.6 is a criterion for a set $S\in\mathrm{L}(R,\mathfrak{a},A)$ to
be the largest element of the poset $\mathrm{L}(R,\mathfrak{a},A)$. Theorem
3.8 describes the sets $\max\mathrm{L}(R)$ and $\max\mathrm{L}{\rm oc}(R)$,
presents a bijection between them, and shows that $\max\mathrm{L}{\rm
oc}(R)\neq\emptyset$. Similarly, for each ideal $\mathfrak{a}\in{\rm
ass}\,\mathrm{L}(R)$, Theorem 3.9 describes the sets
$\max\mathrm{L}(R,\mathfrak{a})$ and $\max\mathrm{L}{\rm oc}(R,\mathfrak{a})$,
presents a bijection between them, and shows that they are nonempty sets. The
$\mathfrak{a}$-absolute quotient ring

$Q_{a}(R,\mathfrak{a}):=\varinjlim_{S\in\mathrm{L}(R,\mathfrak{a})}R\langle
S^{-1}\rangle$

is introduced for each ideal $\mathfrak{a}\in{\rm ass}\,\mathrm{L}(R)$.
Theorem 3.15 gives an explicit description of the ring
$Q_{a}(R,\mathfrak{a})$, a criterion for $Q_{a}(R,\mathfrak{a})\neq\\{0\\}$,
and a criterion for $\mathfrak{a}_{a,\mathfrak{a}}=\mathfrak{a}$.

Theorem 3.17.(1) is an $R$-isomorphism criterion for localizations of the ring
$R$ and Theorem 3.17.(2) is a criterion for existing of an $R$-homomorphism
between localizations of $R$. Proposition 3.14 contains some properties of the
set of completely localizable elements ${\cal C}\mathrm{L}(R)$ and the ideals
$\mathfrak{c}_{R}:={\rm ass}_{R}({\cal C}\mathrm{L}(R))$. Theorem 3.20 is a
criterion for the ring $R\langle S^{-1}\rangle$ to be an $R$-isomorphic to a
localization of $R$ at a denominator set and it also describes all such
denominator sets.

For a denominators set $T$ of $R$, Proposition 3.21 describes all localizable
sets $S$ of the ring $R$ such that the ring $R\langle S^{-1}\rangle$ is an
$R$-isomorphic to the localization of $R$ at $T$. For a localizable
multiplicative set $S$ of the ring $R$, Proposition 3.22 is a criterion for
the set $S$ to be a left denominator set of $R$.

In Section 4, for an arbitrary commutative ring $R$, descriptions of the
following sets are obtained: the sets of localizable and non-localizable
elements (Lemma 4.1), the maximal localizable sets, maximal localization
rings, the set of completely localizable elements, and the ideal
$\mathfrak{c}_{R}$ (Theorem 4.2). Theorem 4.2 also describes relations between
the ideals $\mathrm{L}{\rm rad}(R)$, $\mathfrak{n}_{R}$, and
$\mathfrak{c}_{R}$.

For a commutative ring $R$ with finitely many minimal primes and nilpotent
prime radical (eg, $R$ is a commutative Noetherian ring), Proposition 4.3
describes the spectrum ${\rm Spec}(Q_{c}(R))$ and the rings $Q_{c}(R)$ and
$Q_{c}(R)/\mathfrak{n}_{Q_{c}(R)}$.

In Section 5, Proposition 5.1.(4) describes the maximal localizable sets of
finite direct product of rings and their localizations. Proposition 5.1.(1)
describes localizations of the direct product of rings via localizations of
its components. For a direct product of simple rings

$A=\prod_{i=1}^{s}A_{i}\;\;{\rm such\;that}\;\;{\cal
L}\mathrm{L}(A_{i})=A_{i}^{\times}\;\;{\rm for}\;\;i=1,\ldots,s,$

Theorem 5.3 describes the sets $\max\mathrm{L}(A)$, ${\rm ass}_{A}(S)$ for all
$S\in\max\mathrm{L}(A)$, the sets of localizable, non-localizable, and
completely localizable elements of $A$. It also shows that $\mathrm{L}{\rm
rad}(A)=0$. Every semisimple Artinian ring satisfies the assumptions of
Theorem 5.3, see Corollary 5.4 for detail. For a semiprime left Goldie ring,
Lemma 5.8 describes all the maximal localizable sets that contain the set of
regular elements of the ring. Let

$D=\prod_{i\in I}D_{i}$

be a direct product of division rings $D_{i}$ where $I$ is an arbitrary set.
Proposition 5.9 describes all the localizable sets in the ring $D$ and all the
localizations of $D$. Theorem 5.11 describes the following sets:
$\max\mathrm{L}(D)$, $\max\mathrm{L}{\rm oc}(D)$, $\mathrm{L}{\rm rad}(D)$,
${\cal C}\mathrm{L}(D)$, $Q_{c}(D)$, ${\cal L}\mathrm{L}(D)$, and ${\cal
N}{\cal L}\mathrm{L}(D)$. Theorem 5.11.(2) shows that for every ${\cal
S}\in\max\mathrm{L}(D)$, the ring $D\langle{\cal S}^{-1}\rangle$ is a division
ring.

In Section 6, basic properties of the localization of a module at a
localizable set are considered. Proposition 6.1.(2) is a universal property of
localization of a module. Theorem 6.2 is a criterion for the localization
functor $M\rightarrow S^{-1}M$ to be an exact functor.

##  2 Localization of a ring at a set

At the beginning of the section we recall Ore’s method of localization, i.e.
the localization of a ring at a left or right denominator set. The aim of this
section is to introduce several new concepts that are associated with
localization of a ring at a set and to prove some general properties for them
(Corollary 2.2, Lemma 2.6, Proposition 2.7, Proposition 2.9, Corollary 2.11).
Proposition 2.1 is the universal property of localization. Proposition 2.4 is
a criterion for a subset of a ring to be a localizable set. Every Ore set is a
localizable set, [1, Theorem 4.15]. Theorem 2.23 and Theorem 2.24 demonstrates
some of the results of this section for Ore sets.

Direct limits play a prominent role in proving that certain natural posets
(partially ordered sets) that are associated with localizations admit maximal
elements. Several results on direct limits of rings are proven that are used
later in the paper (Theorem 2.14, Corollary 2.15, and Lemma 2.16).

For a ring $R$, the concept of the absolute quotient ring $Q_{a}(R)$ is
introduced. Theorem 2.18 is a description of the ring $Q_{a}(R)$ in terms of
maximals localization sets of $R$. Corollary 2.22 presents a sufficient
condition for a ring $R$ to have at least two maximal localization sets.

Ore and denominator sets, localization of a ring at a denominator set. Let $R$
be a ring. A subset $S$ of $R$ is called a multiplicative set if $SS\subseteq
S$, $1\in S$ and $0\not\in S$. A multiplicative subset $S$ of $R$ is called a
left Ore set if it satisfies the left Ore condition: for each $r\in R$ and
$s\in S$,

$Sr\bigcap Rs\neq\emptyset.$

Let ${\rm Ore}_{l}(R)$ be the set of all left Ore sets of $R$. For $S\in{\rm
Ore}_{l}(R)$, ${\rm ass}_{l}(S):=\\{r\in R\,|\,sr=0\;\;{\rm for\;some}\;\;s\in
S\\}$ is an ideal of the ring $R$.

A left Ore set $S$ is called a left denominator set of the ring $R$ if $rs=0$
for some elements $r\in R$ and $s\in S$ implies $tr=0$ for some element $t\in
S$, i.e., $r\in{\rm ass}_{l}(S)$. Let ${\rm Den}_{l}(R)$ (resp., ${\rm
Den}_{l}(R,\mathfrak{a})$) be the set of all left denominator sets of $R$
(resp., such that ${\rm ass}_{l}(S)=\mathfrak{a}$). For $S\in{\rm
Den}_{l}(R)$, let

$S^{-1}R=\\{s^{-1}r\,|\,s\in S,r\in R\\}$

be the left localization of the ring $R$ at $S$ (the left quotient ring of $R$
at $S$). Let us stress that in Ore’s method of localization one can localize
precisely at left denominator sets. In a similar way, right Ore and right
denominator sets are defined. Let ${\rm Ore}_{r}(R)$ and ${\rm Den}_{r}(R)$ be
the set of all right Ore and right denominator sets of $R$, respectively. For
$S\in{\rm Ore}_{r}(R)$, the set ${\rm ass}_{r}(S):=\\{r\in R\,|\,rs=0$ for
some $s\in S\\}$ is an ideal of $R$. For $S\in{\rm Den}_{r}(R)$,

$RS^{-1}=\\{rs^{-1}\,|\,s\in S,r\in R\\}$

is the right localization of the ring $R$ at $S$.

Given ring homomorphisms $\nu_{A}:R\rightarrow A$ and $\nu_{B}:R\rightarrow
B$. A ring homomorphism $f:A\rightarrow B$ is called an $R$-homomorphism if
$\nu_{B}=f\nu_{A}$. A left and right set is called an Ore set. Let ${\rm
Ore}(R)$ and ${\rm Den}(R)$ be the set of all Ore and denominator sets of $R$,
respectively. For $S\in{\rm Den}(R)$,

$S^{-1}R\simeq RS^{-1}$

(an $R$-isomorphism) is the localization of the ring $R$ at $S$, and ${\rm
ass}(S):={\rm ass}_{l}(S)={\rm ass}_{r}(S)$.

The ring $R\langle S^{-1}\rangle$ and the ideal ${\rm ass}_{R}(S)$. Let $R$ be
a ring and $S$ be a subset of $R$. Let $R\langle X_{S}\rangle$ be a ring
freely generated by the ring $R$ and a set $X_{S}=\\{x_{s}\,|\,s\in S\\}$ of
free noncommutative indeterminates (indexed by the elements of the set $S$).
Let $I_{S}$ be the ideal of $R\langle X_{S}\rangle$ generated by the set
$\\{sx_{s}-1,x_{s}s-1\,|\,s\in S\\}$ and

$R\langle S^{-1}\rangle:=R\langle X_{S}\rangle/I_{S}.$ (3)

The ring $R\langle S^{-1}\rangle$ is called the localization of $R$ at $S$.
Let ${\rm ass}(S)={\rm ass}_{R}(S)$ be the kernel of the ring homomorphism

$\sigma_{S}:R\rightarrow R\langle S^{-1}\rangle,\;\;r\mapsto r+I_{S}.$ (4)

The map $\pi_{S}:R\rightarrow\overline{R}:=R/{\rm ass}_{R}(S)$,
$r\mapsto\overline{r}:=r+{\rm ass}_{R}(S)$ is an epimorphism. The ideal ${\rm
ass}_{R}(S)$ of $R$ has a complex structure, its description is given in [7,
Proposition 2.12] when $R\langle
S^{-1}\rangle=\\{\overline{s}^{-1}\overline{r}\,|\,s\in S,r\in R\\}$ is a ring
of left fractions. We identify the factor ring $\overline{R}$ with its
isomorphic copy in the ring $R\langle S^{-1}\rangle$ via the monomorphism

$\overline{\sigma_{S}}:\overline{R}\rightarrow R\langle
S^{-1}\rangle,\;\;r+{\rm ass}_{R}(S)\mapsto r+I_{S}.$ (5)

Clearly, $\overline{S}:=(S+{\rm ass}_{R}(S))/{\rm ass}_{R}(S)\subseteq{\cal
C}_{R\langle S^{-1}\rangle}$. Corollary 2.2.(2) shows that the rings $R\langle
S^{-1}\rangle$ and $\overline{R}\langle\overline{S}^{-1}\rangle$ are
$R$-isomorphic. For $S=\emptyset$, $R\langle\emptyset^{-1}\rangle:=R$ and
${\rm ass}_{R}(\emptyset):=0$.

Definition. A subset $S$ of a ring $R$ is called a localizable set of $R$ if
$R\langle S^{-1}\rangle\neq\\{0\\}$. Let $\mathrm{L}(R)$ be the set of
localizable sets of $R$ and

${\rm ass}\,\mathrm{L}(R):=\\{{\rm ass}_{R}(S)\,|\,S\in\mathrm{L}(R)\\}.$ (6)

Proposition 2.1 is the universal property of localization.

###### Proposition 2.1

Let $R$ be a ring, $S\in\mathrm{L}(R)$, and $\sigma_{S}:R\rightarrow R\langle
S^{-1}\rangle$, $r\mapsto r+{\rm ass}_{R}(S)$. Let $f:R\rightarrow A$ be a
ring homomorphism such that $f(S)\subseteq A^{\times}$. The there is a unique
$R$-homomorphism $f^{\prime}:R\langle S^{-1}\rangle\rightarrow A$ such
$f=f^{\prime}\sigma_{S}$, i.e. the diagram below is commutative

$\begin{array}[c]{ccc}R&\stackrel{{\scriptstyle\sigma_{S}}}{{\rightarrow}}&R\langle
S^{-1}\rangle\\\ &\stackrel{{\scriptstyle
f}}{{\searrow}}&{\downarrow}^{\exists!\,f^{\prime}}\\\ &&A\end{array}$

Proof. The unique $R$-homomorphism $f^{\prime}$ is defined by the rule

$f^{\prime}:R\langle S^{-1}\rangle=R\langle X_{S}\rangle/I_{X}\rightarrow
A,\;\;r+I_{S}\mapsto f(r),\;\;x_{s}+I_{S}\mapsto f(s)^{-1}$

for all $r\in R$ and $s\in S$. $\Box$

###### Corollary 2.2

Suppose that $S\in\mathrm{L}(R,\mathfrak{a})$, $\mathfrak{b}$ is an ideal of
$R$ such that $\mathfrak{b}\subseteq\mathfrak{a}$,
$\overline{S}:=\pi_{\mathfrak{b}}(S)$ where $\pi_{\mathfrak{b}}:R\rightarrow
R/\mathfrak{b}$, $r\mapsto r+\mathfrak{b}$. Then

  1. 1.

$\overline{S}\in\mathrm{L}(R/\mathfrak{b},\mathfrak{a}/\mathfrak{b})$ and
$(R/\mathfrak{b})\langle\overline{S}^{-1}\rangle\simeq R\langle
S^{-1}\rangle$, an $R$-isomorphism.

  2. 2.

In particular, for $\mathfrak{a}=\mathfrak{b}$,
$\overline{S}\in\mathrm{L}(R/\mathfrak{a},0)$ and
$(R/\mathfrak{a})\langle\overline{S}^{-1}\rangle\simeq R\langle
S^{-1}\rangle$, an $R$-isomorphism.

Proof. 1. By the universal property of the localization, there is a unique
$R$-homomorphism $R\langle
S^{-1}\rangle\rightarrow(R/\mathfrak{b})\langle\overline{S}^{-1}\rangle$
(since $\mathfrak{b}\subseteq\mathfrak{a}$). Conversely, since
$\mathfrak{b}\subseteq\mathfrak{a}$, there is a unique $R$-homomorphism
$(R/\mathfrak{b})\langle\overline{S}^{-1}\rangle\rightarrow R\langle
S^{-1}\rangle$, by the universal property of the localization, and the lemma
follows.

2\. Statement 2 is a particular case of statement 1 where
$\mathfrak{b}=\mathfrak{a}$. $\Box$

Corollary 2.3 is a criterion for two localizations of a ring $R$ to be
$R$-isomorphic.

###### Corollary 2.3

Let $S\in\mathrm{L}(R,\mathfrak{a})$, $T\in\mathrm{L}(R,\mathfrak{b})$,
$\overline{S}=\pi(S)$ and $\overline{T}=\pi(T)$ where
$\pi:R\rightarrow\overline{R}=R/\mathfrak{a}$, $r\mapsto r+\mathfrak{a}$. Then
the rings $R\langle S^{-1}\rangle$ and $R\langle T^{-1}\rangle$ are
$R$-isomorphic iff $\mathfrak{a}=\mathfrak{b}$, $\overline{S}\subseteq
R\langle T^{-1}\rangle^{\times}$, and $\overline{T}\subseteq R\langle
S^{-1}\rangle^{\times}$.

Proof. $(\Rightarrow)$ The implication is obvious.

$(\Leftarrow)$ Suppose that $\mathfrak{a}=\mathfrak{b}$,
$\overline{S}\subseteq R\langle T^{-1}\rangle^{\times}$, and
$\overline{T}\subseteq R\langle S^{-1}\rangle^{\times}$. By Proposition 2.1,
we have commutative diagrams of $R$-homomorphisms,

$\begin{array}[c]{ccc}R&\stackrel{{\scriptstyle\sigma_{S}}}{{\rightarrow}}&R\langle
S^{-1}\rangle\\\
&\stackrel{{\scriptstyle\sigma_{T}}}{{\searrow}}&{\downarrow}^{\exists!\,f}\\\
&&R\langle T^{-1}\rangle\end{array}\;\;\;{\rm
and}\;\;\;\begin{array}[c]{ccc}R&\stackrel{{\scriptstyle\sigma_{T}}}{{\rightarrow}}&R\langle
T^{-1}\rangle\\\
&\stackrel{{\scriptstyle\sigma_{S}}}{{\searrow}}&{\downarrow}^{\exists!\,g}\\\
&&R\langle S^{-1}\rangle.\end{array}$

By Proposition 2.1, $fg={\rm id}_{R\langle T^{-1}\rangle}$ and $gt={\rm
id}_{R\langle S^{-1}\rangle}$, and so $f$ is an $R$-isomorphism. $\Box$

If $S\subseteq T\subseteq R$ then ${\rm ass}_{R}(S)\subseteq{\rm ass}_{R}(T)$
and, by Proposition 2.1, there is a unique $R$-homomorphism

$\phi_{TS}:R\langle S^{-1}\rangle\rightarrow R\langle T^{-1}\rangle,$ (7)

i.e. $\phi_{TS}(ra)=r\phi_{TS}(a)$ for all elements $a\in R\langle
S^{-1}\rangle$. Clearly, $\phi_{SS}={\rm id}_{R\langle S^{-1}\rangle}$ and

${\rm ker}(\phi_{TS})\supseteq R\langle S^{-1}\rangle\Big{(}{\rm
ass}_{R}(T)/{\rm ass}_{R}(S)\Big{)}R\langle S^{-1}\rangle.$ (8)

If $S\subseteq T\subseteq U\subseteq R$ then

$\phi_{US}=\phi_{UT}\phi_{TS}.$ (9)

Criterion for a set to be a localizable set. Proposition 2.4 is a criterion
for a set to be a localizable set.

###### Proposition 2.4

Let $S$ be a subset of $R$. The following statements are equivalent:

  1. 1.

$S\in\mathrm{L}(R)$.

  2. 2.

${\rm ass}_{R}(S)\neq R$.

  3. 3.

There is an ring homomorphism $f:R\rightarrow A$ such that $f(S)\subseteq
A^{\times}$ where $A^{\times}$ is a group of units of the ring $A$.

Proof. $(1\Leftrightarrow 2)$ $S\in\mathrm{L}(R)$ iff $R\langle
S^{-1}\rangle\neq\\{0\\}$ iff the image of the ring $R$ under the
$R$-homomorphism $R\rightarrow R\langle S^{-1}\rangle$ is not equal to $0$ iff
${\rm ass}_{R}(S)\neq R$.

$(1\Rightarrow 3)$ If $S\in\mathrm{L}(R)$ then the image of the set $S$ under
the homomorphism $\sigma_{S}:R\rightarrow R\langle S^{-1}\rangle$ consists of
units of the ring $R\langle S^{-1}\rangle$.

$(3\Rightarrow 2)$ Suppose that there is a ring homomorphism $f:R\rightarrow
A$ such that $f(S)\subseteq A^{\times}$. By the universal property of
localization, there is a unique $R$-homomorphism $R\langle
S^{-1}\rangle\rightarrow A$. Hence, $R\langle S^{-1}\rangle\neq\\{0\\}$ (since
$f(S)\subseteq A^{\times}$), and so ${\rm ass}_{R}(S)\neq R$. $\Box$

Proposition 2.5 gives a sufficient condition for a subset of a ring to be a
localizable set.

###### Proposition 2.5

Let $I$ be an ideal of a ring $R$, $\pi:R\rightarrow R/I$, $r\mapsto r+I$, and
$S$ be a subset of $R$ such that $\pi(S)\subseteq{\cal C}_{R/I}$ and ${\rm
ass}_{R}(S)=\mathfrak{a}(S)$. Then $S\in\mathrm{L}(R,\mathfrak{a}(S))$ and
$\mathfrak{a}(S)\subseteq I$.

Proof. By Corollary 2.11.(1), $\mathfrak{a}(S)\subseteq I$ since by induction
$\mathfrak{a}_{\lambda}\subseteq I$ for all $\lambda\in\mathbb{N}$. Since
$\mathfrak{a}(S)={\rm ass}_{R}(S)$, the set $S$ is a localizable set of $R$
since ${\rm ass}_{R}(S)\neq R$ (Proposition 2.4) as ${\rm ass}_{R}(S)\subseteq
I$. $\Box$

Let $S$ be a subset of the ring $R$ and $S_{mon}$ be the multiplicative
submonoid of $(R,\cdot)$ generated by the set $S$. Lemma 2.6 shows that the
localization of the ring $R$ at $S$ is the same as the localization of the
ring $R$ at $S_{mon}$.

###### Lemma 2.6

Let $S$ be a subset of the ring $R$. Then

  1. 1.

$S\in\mathrm{L}(R,\mathfrak{a})$ iff $S_{mon}\in\mathrm{L}(R,\mathfrak{a})$.

  2. 2.

If $S\in\mathrm{L}(R,\mathfrak{a})$ then $R\langle S^{-1}\rangle\simeq
R\langle S_{mon}^{-1}\rangle$, an $R$-isomorphism

Proof. 1. If $S\in\mathrm{L}(R)$ then the image of the monoid $S_{mon}$ under
the $R$-homomorphism $\sigma_{S}:R\rightarrow R\langle S^{-1}\rangle$ consists
of units, and so $S_{mon}\in\mathrm{L}(R)$, by Proposition 2.4. Similarly, if
$S_{mon}\in\mathrm{L}(R)$ then the image of the set $S$ under the
$R$-homomorphism $\sigma_{S_{mon}}:R\rightarrow R\langle S_{mon}^{-1}\rangle$
consists of units, and so $S\in\mathrm{L}(R)$, by Proposition 2.4.

2\. Since $S\in\mathrm{L}(R)$, $S_{mon}\in\mathrm{L}(R)$, by statement 1. By
the universal property of localization, there is a unique $R$-homomorphism
$R\langle S^{-1}\rangle\rightarrow R\langle S_{mon}^{-1}\rangle$. By the
universal property of localization, there is a unique $R$-homomorphism
$R\langle S_{mon}^{-1}\rangle\rightarrow R\langle S^{-1}\rangle$. So, the
rings $R\langle S^{-1}\rangle$ and $R\langle S_{mon}^{-1}\rangle$ are an
$R$-isomorphic, and statement 2 follows. $\Box$

The set ${\cal M}(S,R)$ of ideals of $R$ and its minimal element. For a set
$S\in\mathrm{L}(R)$, let ${\cal M}(S,R)$ be the set of ideal $\mathfrak{a}$ of
the ring $R$ such that there is a ring homomorphism
$\varphi:R/\mathfrak{a}\rightarrow A$ for some ring $A$ such that
$\varphi(S+\mathfrak{a})\subseteq A^{\times}$. Proposition 2.7 shows that
${\rm ass}_{R}(S)$ is the least element of the set ${\cal M}(S,R)$ w.r.t.
inclusion.

###### Proposition 2.7

For every set $S\in\mathrm{L}(R)$, $\min{\cal M}(S,R)=\\{{\rm ass}_{R}(S)\\}$.

Proof. Let $\mathfrak{a}\in{\cal M}(S,R)$ and

$f:R\rightarrow
R/\mathfrak{a}\stackrel{{\scriptstyle\varphi}}{{\rightarrow}}A,\;\;r\mapsto
r+\mathfrak{a}\mapsto\varphi(r+\mathfrak{a})$

for some ring $A$ such that $\varphi(S+\mathfrak{a})\subseteq A^{\times}$.
Clearly, ${\rm ker}(f)=\mathfrak{a}$. By Proposition 2.1, there is an
$R$-homomorphism $f^{\prime}:R\langle S^{-1}\rangle\rightarrow A$ such that
$f=f^{\prime}\sigma_{S}$. Therefore,

${\rm ass}_{R}(S)={\rm ker}(\sigma_{S})\subseteq{\rm
ker}(f^{\prime}\sigma_{S})={\rm ker}(f)=\mathfrak{a},$

and the proposition follows. $\Box$

General results on localizations. Some useful general results on localizations
are collected below that are used in the proofs of this paper.

###### Proposition 2.8

  1. 1.

If $S\subseteq R^{\times}$ then ${\rm ass}_{R}(S)=0$ and $R\langle
S^{-1}\rangle=R$.

  2. 2.

If $T\in\mathrm{L}(R)$ and $S\subseteq T$. Then $S\in\mathrm{L}(R)$, ${\rm
ass}_{R}(S)\subseteq{\rm ass}_{R}(T)$, $\sigma_{S}(T)\in\mathrm{L}(R\langle
S^{-1}\rangle)$,

$R\langle T^{-1}\rangle\simeq R\langle
S^{-1}\rangle\langle\sigma_{S}(T)^{-1}\rangle,$

an $R$-isomorphism, and ${\rm ass}_{R\langle
S^{-1}\rangle}(\sigma_{S}(T))\supseteq R\langle S^{-1}\rangle\Big{(}{\rm
ass}_{R}(T)/{\rm ass}_{R}(S)\Big{)}R\langle S^{-1}\rangle$.

  3. 3.

If $S,T\in\mathrm{L}(R)$ such that $S\cup T\in\mathrm{L}(R)$ then
$\sigma_{S}(T)\in\mathrm{L}(R\langle S^{-1}\rangle)$ and $R\langle(S\cup
T)^{-1}\rangle\simeq R\langle S^{-1}\rangle\langle\sigma_{S}(T)^{-1}\rangle$,
an $R$-isomorphism.

Proof. 1. Both rings $R\langle S^{-1}\rangle$ and $R$ satisfy the universal
property of localization, Proposition 2.1. Therefore, $R\langle
S^{-1}\rangle=R$. Hence, ${\rm ass}_{R}(S)=0$.

2\. Since $T\in\mathrm{L}(R)$ and $S\subseteq T$, there a unique
$R$-homomorphism $R\langle S^{-1}\rangle\rightarrow R\langle T^{-1}\rangle$.
Therefore, $S\in\mathrm{L}(R)$. Both rings $R\langle T^{-1}\rangle$ and
$R\langle S^{-1}\rangle\langle\sigma_{S}(T)^{-1}\rangle$ satisfy the universal
property of localization for the set $T$ (Proposition 2.1), and so they are
$R$-isomorphic. It follows that $\sigma_{S}(T)\in\mathrm{L}(R\langle
S^{-1}\rangle)$, by statement 2 (since $T\subseteq S\cup T$).

3\. Statement 3 is a particular case of statement 2 (since $S\subseteq S\cup
T$ and $S\cup T\in\mathrm{L}(R)$. $\Box$

###### Lemma 2.9

Suppose that $S\in\mathrm{L}(R,\mathfrak{a})$, $T\subseteq R$, and
$\sigma_{S}:R\rightarrow R\langle S^{-1}\rangle$.

  1. 1.

If $\sigma_{S}(T)\in\mathrm{L}(R\langle S^{-1}\rangle)$ then $S\cup
T\in\mathrm{L}(R)$ and $R\langle(S\cup T)^{-1}\rangle\simeq R\langle
S^{-1}\rangle\langle\sigma_{S}(T)^{-1}\rangle$, an $R$-isomorphism.

  2. 2.

If $\sigma_{S}(T)\subseteq R\langle S^{-1}\rangle^{\times}$ then $S\cup
T\in\mathrm{L}(R,\mathfrak{a})$ and $R\langle(S\cup T)^{-1}\rangle\simeq
R\langle S^{-1}\rangle$, an $R$-isomorphism.

  3. 3.

Let ${\cal T}=\sigma^{-1}(R\langle S^{-1}\rangle^{\times})$. Then ${\cal
T}\in\mathrm{L}(R,\mathfrak{a})$ and $R\langle{\cal T}^{-1}\rangle\simeq
R\langle S^{-1}\rangle$, an $R$-isomorphism.

Proof. 1. The image of the set $S\cup T$ in $R\langle
S^{-1}\rangle\langle\sigma_{S}(T)^{-1}\rangle$ consists of units, hence $S\cup
T\in\mathrm{L}(R)$, by Proposition 2.4. By Proposition 2.8.(3), the rings
$R\langle(S\cup T)^{-1}\rangle$ and $R\langle
S^{-1}\rangle\langle\sigma_{S}(T)^{-1}\rangle$ are $R$-isomorphic.

2\. Since $\sigma_{S}(T)\subseteq R\langle S^{-1}\rangle^{\times}$,
$\sigma_{S}(T)\in\mathrm{L}(R\langle S^{-1}\rangle)$. Now, by statement 1,
$S\cup T\in\mathrm{L}(R)$ and

$R\langle(S\cup T)^{-1}\rangle\simeq R\langle
S^{-1}\rangle\langle\sigma_{S}(T)^{-1}\rangle\simeq R\langle S^{-1}\rangle,$

$R$-isomorphisms where the second $R$-isomorphism is due to Proposition
2.8.(1). Hence, ${\rm ass}_{R}(S\cup T)=\mathfrak{a}$.

3\. Statement 3 follows from statement 2 where $T={\cal T}$ since $S\cup{\cal
T}={\cal T}$. $\Box$

The ideals $\mathfrak{a}(S)$, ${}^{\prime}\mathfrak{a}(S)$ and
$\mathfrak{a}^{\prime}(S)$. The ideals $\mathfrak{a}(S)$,
${}^{\prime}\mathfrak{a}(S)$ and $\mathfrak{a}^{\prime}(S)$ are contained in
the ideal ${\rm ass}_{R}(S)$. They are defined iteratively and it is difficult
to compute them in general. They reveal a complex structure of the ideal ${\rm
ass}_{R}(S)$.

For each element $r\in R$, let $r\cdot:R\rightarrow R$, $x\mapsto rx$ and
$\cdot r:R\rightarrow R$, $x\mapsto xr$. The sets

${}^{\prime}\mathcal{C}_{R}:=\\{r\in R\,|\,{\rm ker}(\cdot r)=0\\}\;\;{\rm
and}\;\;\mathcal{C}_{R}^{\prime}:=\\{r\in R\,|\,{\rm ker}(r\cdot)=0\\}$

are called the sets of left and right regular elements of $R$, respectively.
Their intersection

$\mathcal{C}_{R}={}^{\prime}\mathcal{C}_{R}\cap\mathcal{C}_{R}^{\prime}$

is the set of regular elements of $R$. The rings

$Q_{l,cl}(R):=\mathcal{C}_{R}^{-1}R\;\;{\rm
and}\;\;Q_{r,cl}(R):=R\mathcal{C}_{R}^{-1}$

are called the classical left and right quotient rings of $R$, respectively.
If both rings exist then they are isomorphic and the ring

$Q_{cl}(R):=Q_{l,cl}(R)\simeq Q_{r,cl}(R)$

is called the classical quotient ring of $R$. Goldie’s Theorem states that the
ring $Q_{l,cl}(R)$ is a semisimple Artinian ring iff the ring $R$ is a
semiprime left Goldie ring (the ring $R$ is called a left Goldie ring if ${\rm
udim}(R)<\infty$ and the ring $R$ satisfies the a.c.c. on left annihilators
where ${\rm udim}$ stands for the left uniform dimension). In a similar way a
right Goldie ring is defined. A left and right Goldie ring is called a Goldie
ring.

###### Proposition 2.10

([7, Proposition 1.1]) Let $R$ be a ring and $S$ be a non-empty subset of $R$.

  1. 1.

There is the least ideal, $\mathfrak{a}=\mathfrak{a}(S)$, such that
$(S+\mathfrak{a})/\mathfrak{a}\subseteq\mathcal{C}_{R/\mathfrak{b}}$.

  2. 2.

There is the least ideal,
${}^{\prime}\mathfrak{a}={}^{\prime}\mathfrak{a}(S)$, such that
$(S+{}^{\prime}\mathfrak{a})/{}^{\prime}\mathfrak{a}\subseteq{}^{\prime}\mathcal{C}_{R/\mathfrak{b}}$.

  3. 3.

There is the least ideal, $\mathfrak{a}^{\prime}=\mathfrak{a}^{\prime}(S)$,
such that
$(S+\mathfrak{a}^{\prime})/\mathfrak{a}^{\prime}\subseteq\mathcal{C}_{R/\mathfrak{b}}^{\prime}$.

Notice that ${}^{\prime}\mathfrak{a}(S)\subseteq\mathfrak{a}(S)$ and
$\mathfrak{a}^{\prime}(S)\subseteq\mathfrak{a}(S)$.

Recall that $S_{mon}$ is the multiplicative submonoid of $(R,\cdot)$ which is
generated by the set $S$. If $0\in S_{mon}$ then
$\mathfrak{a}(S)={}^{\prime}\mathfrak{a}(S)=\mathfrak{a}^{\prime}(S)=R$ and
$R\langle S^{-1}\rangle=0$. So, we can assume that $0\not\in S_{mon}$, i.e.
the set $S$ is a multiplicative set. Notice that (Lemma 2.6)

$\mathfrak{a}(S)=\mathfrak{a}(S_{mon}),\;\;{\rm ass}(S)={\rm
ass}(S_{mon}),\;\;{\rm and}\;\;R\langle S^{-1}\rangle=R\langle
S_{mon}^{-1}\rangle.$ (10)

The proof of Proposition 2.10 ([7, Proposition 1.1]) contains an explicit
description of the ideals $\mathfrak{a}=\mathfrak{a}(S)$,
${}^{\prime}\mathfrak{a}={}^{\prime}\mathfrak{a}(S)$, and
$\mathfrak{a}^{\prime}=\mathfrak{a}^{\prime}(S)$. In more detail, let $\Gamma$
be the set of ordinals. The ideal $\mathfrak{a}$ (resp.,
${}^{\prime}\mathfrak{a}$, $\mathfrak{a}^{\prime}$) is the union

$\mathfrak{a}=\bigcup_{\lambda\in\Gamma}\mathfrak{a}_{\lambda}\;\;\;({\rm
resp.,}\;\;{}^{\prime}\mathfrak{a}=\bigcup_{\lambda\in\Gamma}{}^{\prime}\mathfrak{a}_{\lambda},\;\;\mathfrak{a}^{\prime}=\bigcup_{\lambda\in\Gamma}\mathfrak{a}_{\lambda}^{\prime})$
(11)

of ascending chain of ideals $\\{\mathfrak{a}_{\lambda}\\}_{\lambda\in\Gamma}$
(resp., $\\{{}^{\prime}\mathfrak{a}_{\lambda}\\}_{\lambda\in\Gamma}$,
$\\{\mathfrak{a}^{\prime}_{\lambda}\\}_{\lambda\in\Gamma}$), where
$\lambda\leq\mu$ in $\Gamma$ implies
$\mathfrak{a}_{\lambda}\subseteq\mathfrak{a}_{\mu}$ (resp.,
${}^{\prime}\mathfrak{a}_{\lambda}\subseteq{}^{\prime}\mathfrak{a}_{\mu}$,
$\mathfrak{a}_{\lambda}^{\prime}\subseteq\mathfrak{a}_{\mu}^{\prime}$). The
ideals $\mathfrak{a}_{\lambda}$ (resp., ${}^{\prime}\mathfrak{a}_{\lambda}$,
$\mathfrak{a}_{\lambda}^{\prime}$) are defined inductively as follows: the
ideal $\mathfrak{a}_{0}$ (resp., ${}^{\prime}\mathfrak{a}_{0}$,
$\mathfrak{a}_{0}^{\prime}$) is generated by the set $\\{r\in R\,|\,sr=0$ or
$rt=0$ for some elements $s,t\in S\\}$ (resp., $\\{r\in R\,|\,rt=0$ for some
element $t\in S\\}$, $\\{r\in R\,|\,sr=0$ for some element $s\in S\\}$), and
for $\lambda\in\Gamma$ such that $\lambda>0$ (where below
$\Big{(}\\{\ldots\\}\Big{)}$ means the ideal of $R$ generated by the set
$\\{\ldots\\}$),

$\mathfrak{a}_{\lambda}=\begin{cases}\bigcup_{\mu<\lambda\in\Gamma}\mathfrak{a}_{\mu}&\text{if
$\lambda$ is a limit ordinal},\\\ \Big{(}\\{r\in
R\,|\,sr\in\mathfrak{a}_{\lambda-1}\;{\rm
or}\;rt\in\mathfrak{a}_{\lambda-1}\;{\rm for\;some}\;s,t\in
S\\}\Big{)}&\text{if if $\lambda$ is not a limit ordinal}.\\\ \end{cases}$
(12)

(resp.,

${}^{\prime}\mathfrak{a}_{\lambda}=\begin{cases}\bigcup_{\mu<\lambda\in\Gamma}{}^{\prime}\mathfrak{a}_{\mu}&\text{if
$\lambda$ is a limit ordinal},\\\ \Big{(}\\{r\in
R\,|\,rt\in{}^{\prime}\mathfrak{a}_{\lambda-1}\;{\rm for\;some}\;t\in
S\\}\Big{)}&\text{if if $\lambda$ is not a limit ordinal}.\\\ \end{cases}$
(13)

$\mathfrak{a}_{\lambda}^{\prime}=\begin{cases}\bigcup_{\mu<\lambda\in\Gamma}\mathfrak{a}_{\mu}^{\prime}&\text{if
$\lambda$ is a limit ordinal},\\\ \Big{(}\\{r\in
R\,|\,sr\in\mathfrak{a}_{\lambda-1}^{\prime}\;{\rm for\;some}\;s\in
S\\}\Big{)}&\text{if if $\lambda$ is not a limit ordinal}).\end{cases}$ (14)

Let us define an ideal $\mathfrak{b}$ of $R$ which is the union
$\mathfrak{b}=\bigcup_{\lambda\in\Gamma}\mathfrak{b}_{\lambda}$ of an
ascending chain of ideals $\mathfrak{b}_{\lambda}$ that are defined
inductively

$\mathfrak{b}_{\lambda}=\begin{cases}\bigcup_{\mu<\lambda\in\Gamma}\mathfrak{b}_{\mu}&\text{if
$\lambda$ is a limit ordinal},\\\ \Big{(}\\{r\in
R\,|\,sr\in\mathfrak{b}_{\lambda-1}\;{\rm
or}\;rt\in\mathfrak{b}_{\lambda-1}\;{\rm
or}\;srt\in\mathfrak{b}_{\lambda-1}\;{\rm for\;some}\;s,t\in
S\\}\Big{)}&\text{if if $\lambda$ is not a limit ordinal}.\\\ \end{cases}$
(15)

where the ideal $\mathfrak{b}_{0}$ is generated by the set $\\{r\in
R\,|\,sr=0$ or $rt=0$ or $srt=0$ for some elements $s,t\in S\\}$.

###### Corollary 2.11

Let $R$ be a ring and $S$ be a non-empty subset of $R$. Then

  1. 1.

$\mathfrak{a}(S)=\bigcup_{\lambda\in\mathbb{N}}\mathfrak{a}_{\lambda}=\bigcup_{\lambda\in\mathbb{N}}\mathfrak{b}_{\lambda}$.

  2. 2.

${}^{\prime}\mathfrak{a}(S)=\bigcup_{\lambda\in\mathbb{N}}{}^{\prime}\mathfrak{a}_{\lambda}$.

  3. 3.

$\mathfrak{a}^{\prime}(S)=\bigcup_{\lambda\in\mathbb{N}}\mathfrak{a}^{\prime}_{\lambda}$.

Proof. The corollary follows from the definitions of the ideals
$\mathfrak{a}_{\mathbb{N}}$, $\mathfrak{b}_{\mathbb{N}}$,
${}^{\prime}\mathfrak{a}_{\mathbb{N}}$, and
$\mathfrak{a}_{\mathbb{N}}^{\prime}$. In more detail, let us show that
$\mathfrak{a}(S)=\mathfrak{b}_{\mathbb{N}}:=\bigcup_{\lambda\in\mathbb{N}}\mathfrak{b}_{\lambda}$.
The other cases can be treated in a similar way. The inclusion
$\mathfrak{a}(S)\supseteq\mathfrak{b}_{\mathbb{N}}$ is obvious. To prove that
the reverse inclusion holds it suffices to show that the maps

$s\cdot:R/\mathfrak{b}_{\mathbb{N}}\rightarrow
R/\mathfrak{b}_{\mathbb{N}},\;\;r\mapsto sr+\mathfrak{b}_{\mathbb{N}}\;\;{\rm
and}\;\;\cdot s:R/\mathfrak{b}_{\mathbb{N}}\rightarrow
R/\mathfrak{b}_{\mathbb{N}},\;\;r\mapsto rs+\mathfrak{b}_{\mathbb{N}}$

are injections for all $s\in S$. Suppose that
$r+\mathfrak{b}_{\mathbb{N}}\in{\rm ker}(s\cdot)$ or
$r+\mathfrak{b}_{\mathbb{N}}\in{\rm ker}(\cdot s)$, that is
$sr\in\mathfrak{b}_{\mathbb{N}}$ or $rs\in\mathfrak{b}_{\mathbb{N}}$. Then
$sr\in\mathfrak{b}_{\lambda}$ or $rs\in\mathfrak{b}_{\lambda}$ for some
$\lambda\in\mathbb{N}$, and so $r\in\mathfrak{b}_{\lambda+1}$ in both cases.
This means that $r\in\mathfrak{b}_{\mathbb{N}}$, as required. $\Box$

Definition. The first natural number such that the ascending chain of ideals
in (15) (resp., (11)) stabilizes is called the $\Gamma$-length of the ideal
denoted $l_{\Gamma}(\mathfrak{a})$ (resp.,
$l_{\Gamma}({}^{\prime}\mathfrak{a})$, $l_{\Gamma}(\mathfrak{a}^{\prime})$).
If there is no such a number then $l_{\Gamma}(\mathfrak{a}):=\infty$ (resp.,
$l_{\Gamma}({}^{\prime}\mathfrak{a}):=\infty$,
$l_{\Gamma}(\mathfrak{a}^{\prime}):=\infty$).

Lemma 2.12 shows that the $\Gamma$-length $l_{\Gamma}(\mathfrak{a})$ can be
any natural number or $\infty$.

Let $K$ be a field. Consider the following rings

  1. 1.

$R_{0}=K\langle x,y_{0}\,|\,xy_{0}=0\rangle$,

  2. 2.

$R_{n}=K\langle
x,y_{i},u_{j},v_{j}\,|\,xy_{0}=0,xy_{j}=u_{j}y_{j-1}v_{j},i=0,\ldots,n,j=1,\ldots,n\rangle$,

  3. 3.

$R_{\infty}=K\langle
x,y_{i},u_{j},v_{j}\,|\,xy_{0}=0,xy_{j}=u_{j}y_{j-1}v_{j},i\geq 0,j\geq
1\rangle$.

Notice that $R_{0}/(y_{0})\simeq K[x]$ and $R_{n}/(y_{0},\ldots,y_{n})\simeq
K\langle x,u_{j},v_{j}\,|\,j=1,\ldots,n\rangle$ is a free algebra for all
$n\geq 1$ and $n=\infty$.

###### Lemma 2.12

Let $R_{n}$, $n\in\mathbb{N}\cup\\{\infty\\}$, be the rings above,
$S=\\{x^{i}\,|\,i\geq 0\\}$, and $\mathfrak{a}(S,R_{n}):=\mathfrak{a}(S)$
where $S\subseteq R_{n}$. Then $\mathfrak{a}(S,R_{n})={\rm
ass}_{R_{n}}(S)=(y_{0},\ldots,y_{n})$, $l_{\Gamma}(\mathfrak{a}(S,R_{n}))=n$,
and

$R_{n}\langle S^{-1}\rangle\simeq\begin{cases}K[x,x^{-1}]&\text{if }n=0,\\\
K\langle x^{\pm 1},u_{j},v_{j}\,|\,j=1,\ldots,n\rangle&\text{if
}n=1,\ldots,\infty.\\\ \end{cases}$

Proof. Suppose that $n=0$. Then
$y_{0}\in\mathfrak{b}_{0}\subseteq\mathfrak{b}\subseteq\mathfrak{a}(S,R_{0})\subseteq{\rm
ass}_{R_{0}}(S)$, $R_{0}/(y_{0})=K[x]$, and so $R_{0}\langle
S^{-1}\rangle\simeq K[x,x^{-1}]$ and $\mathfrak{a}(S,R_{0})={\rm
ass}_{R_{0}}(S)=(y_{0})=\mathfrak{b}_{0}$. Hence,
$l_{\Gamma}(\mathfrak{a}(S,R_{0}))=0$.

Suppose that $n\geq 1$. Then

$R_{n}=\bigoplus_{w\in W_{n}}Kw$

where $W_{n}$ is the set of all words in the alphabet
$\\{x,y_{i},u_{j},v_{j}\,|\,i=0,\ldots,n,j=1,\ldots,n\\}$ that do not
contained the subwords $xy_{i}$, where $i=0,\ldots,n$. Then
$\mathfrak{b}_{0}=(y_{0})$ since ${\rm ker}_{R_{n}}(x\cdot)=y_{0}R_{n}$ and
${\rm ker}_{R_{n}}(\cdot x)=0$. Using the same argument in the case of the
factor ring

$R_{n}/(\mathfrak{b}_{0})=K\langle
x,y_{i},u_{j},v_{j}\,|\,xy_{1}=0,xy_{k}=u_{k}y_{k-1}v_{k},k=2,\ldots,n\rangle\;\;{\rm
where}\;\;i=1,\ldots,n,j=1,\ldots,n,$

we conclude that $\mathfrak{b}_{1}=(y_{0},y_{1})$ and

$R_{n}/(\mathfrak{b}_{1})=K\langle
x,y_{i},u_{j},v_{j}\,|\,xy_{2}=0,xy_{k}=u_{k}y_{k-1}v_{k},k=3,\ldots,n\rangle\;\;{\rm
where}\;\;i=2,\ldots,n,j=1,\ldots,n.$

Repeating the same argument again and again (or use induction), we prove that
for all $m=1,\ldots,n$, $\mathfrak{b}_{m}=(y_{0},\ldots,y_{m})$ and

$R_{n}/(\mathfrak{b}_{m})=K\langle
x,y_{i},u_{j},v_{j}\,|\,xy_{m+1}=0,xy_{k}=u_{k}y_{k-1}v_{k},k=m+2,\ldots,n\rangle,i=m+1,\ldots,n,j=1,\ldots,n.$

Hence,

$R_{n}\langle S^{-1}\rangle\simeq(R_{n}/\mathfrak{b}_{n})\langle
S^{-1}\rangle\simeq K\langle x^{\pm 1},u_{j},v_{j}\,|\,j=1,\ldots,n\rangle$

and $\mathfrak{a}(S,R_{n})={\rm ass}_{R_{n}}(S)=\mathfrak{b}_{n}$. It follows
that $l_{\Gamma}(\mathfrak{a}(S,R_{n}))=n$. $\Box$

Perfect localization set and perfect localizations. By [7, Lemma 1.2],

$\mathfrak{a}(S)\subseteq{\rm ass}_{R}(S).$ (16)

Definition, [7]. If the equality holds, $\mathfrak{a}(S)={\rm ass}_{R}(S)$,
then set $S$ is called a perfect localization set and the localization
$R\langle S^{-1}\rangle$ is called a perfect localization.

Theorem 2.13 shows that there are plenty of perfect localizations (especially
in applications).

###### Theorem 2.13

Let $R$ be a $K$-algebra over a field $K$, $\mathfrak{a}:=\mathfrak{a}(S)$,
$\pi_{\mathfrak{a}}:R\rightarrow\overline{R}:=R/\mathfrak{a}$,
$r\mapsto\overline{r}:=r+\mathfrak{a}$, $S$ be a multiplicative set in $R$.
Suppose that $\overline{R}$ is a domain and for each pair of elements $s\in S$
and $r\in R$, the subalgebra $K\langle\overline{s},\overline{r}\rangle$ of
$\overline{R}$ is not a free algebra. Then

  1. 1.

${\rm ass}_{R}(S)=\mathfrak{a}$.

  2. 2.

$\overline{S}:=\pi_{\mathfrak{a}}(S)\in{\rm Den}(\overline{R},0)$ and
$R\langle
S^{-1}\rangle\simeq\overline{S}^{-1}\overline{R}\simeq\overline{R}\overline{S}^{-1}$.

Proof. By the assumption, for each pair of elements $s\in S$ and $r\in R$, the
subalgebra $K\langle\overline{s},\overline{r}\rangle$ of $\overline{R}$ is not
a free algebra. Then $\overline{S}:=\pi_{\mathfrak{a}}(S)\in{\rm
Den}(\overline{R},0)$ (by Jategaonkar’s Lemma, see [11, Lemma 9.21]).
Therefore, ${\rm ass}_{R}(S)=\mathfrak{a}$ and $R\langle
S^{-1}\rangle\simeq\overline{S}^{-1}\overline{R}\simeq\overline{R}\overline{S}^{-1}$.
$\Box$

The localizable sets of a ring and the posets $(\mathrm{L}(R),\subseteq)$ and
$(\mathrm{L}{\rm oc}(R),\rightarrow)$. For an ideal $\mathfrak{a}$ of $R$, let
$\mathrm{L}(R,\mathfrak{a}):=\\{S\in\mathrm{L}(R)\,|\,{\rm
ass}_{R}(S)=\mathfrak{a}\\}$. Then

$\mathrm{L}(R)=\coprod_{\mathfrak{a}\in{\rm
ass}\,\mathrm{L}(R)}\mathrm{L}(R,\mathfrak{a})$ (17)

is a disjoint union of non-empty sets. A set with binary operation,
$(I,\leq)$, is a partially ordered set (a poset, for short) if $i\leq i$ for
all $i\in I$, and if $i\leq j$ and $j\leq k$ for some elements $i,j,k\in I$
then $i\leq k$. The set $(\mathrm{L}(R),\subseteq)$ is a poset (w.r.t.
inclusion $\subseteq$), and $(\mathrm{L}(R,\mathfrak{a}),\subseteq)$ is a sub-
poset of $(\mathrm{L}(R),\subseteq)$ for every $\mathfrak{a}\in{\rm
ass}\mathrm{L}(R)$.

Let $\mathrm{L}{\rm oc}(R)$ be the set of $R$-isomorphism classes of all the
localizations of the ring $R$, i.e. $R$-isomorphism classes in the set
$\\{R\langle S^{-1}\rangle\,|\,S\in\mathrm{L}(R)\\}$. If the rings $R\langle
S^{-1}\rangle$ and $R\langle T^{-1}\rangle$ are $R$-isomorphic for some
$S,T\in\mathrm{L}(R)$ then ${\rm ass}_{R}(S)={\rm ass}_{R}(T)$. Therefore,

$\mathrm{L}{\rm oc}(R)=\coprod_{\mathfrak{a}\in{\rm
ass}\,\mathrm{L}(R)}\mathrm{L}{\rm oc}(R,\mathfrak{a})$ (18)

is a disjoint union of non-empty sets where $\mathrm{L}{\rm
oc}(R,\mathfrak{a})$ is the set of $R$-isomorphism classes in the set
$\\{R\langle S^{-1}\rangle\,|\,S\in\mathrm{L}(R,\mathfrak{a})\\}$.

The set $(\mathrm{L}{\rm oc}(R,\mathfrak{a}),\rightarrow)$ is a poset where
$A\rightarrow B$ if there is a necessarily unique $R$-homomorphism from $A$ to
$B$ which is denoted by

$\phi_{BA}:A\rightarrow B.$

In particular, the unique $R$-homomorphism $\phi_{TS}:R\langle
S^{-1}\rangle\rightarrow R\langle T^{-1}\rangle$ is also denoted by
$\phi_{R\langle S^{-1}\rangle R\langle T^{-1}\rangle}$. If $A\rightarrow B$
for some rings $A\in\mathrm{L}{\rm oc}(R,\mathfrak{a})$ and
$B\in\mathrm{L}{\rm oc}(R,\mathfrak{b})$ then

$\mathfrak{a}\subseteq\mathfrak{b}.$ (19)

If, in addition, $A=R\langle S^{-1}\rangle$ and $B=R\langle T^{-1}\rangle$ for
some localizable sets $S,T\in\mathrm{L}(R)$ then if necessary we can assume
that $S\subseteq T$ (for example, by replacing $T$ by
$\sigma_{T}^{-1}(B^{\times})$ where $\sigma_{T}:R\rightarrow B$, see
Proposition 3.18.(1)).

The map

$R\langle(\cdot)^{-1}\rangle:\,\mathrm{L}(R)\rightarrow\mathrm{L}{\rm
oc}(R),\;\;S\mapsto R\langle S^{-1}\rangle,$ (20)

is an epimorphism of posets. For each ideal $\mathfrak{a}\in{\rm
ass}\,\mathrm{L}(R)$, it induces the epimorphism of posets,

$R\langle(\cdot)^{-1}\rangle:\,\mathrm{L}(R,\mathfrak{a})\rightarrow\mathrm{L}{\rm
oc}(R,\mathfrak{a}),\;\;S\mapsto R\langle S^{-1}\rangle.$ (21)

For each ideal $\mathfrak{a}\in{\rm ass}\,\mathrm{L}(R)$,

$\mathrm{L}(R,\mathfrak{a})=\bigsqcup_{A\in\mathrm{L}{\rm
oc}(R,\mathfrak{a})}\mathrm{L}(R,\mathfrak{a},A)$ (22)

where
$\mathrm{L}(R,\mathfrak{a},A):=\\{S\in\mathrm{L}(R,\mathfrak{a})\,|\,R\langle
S^{-1}\rangle\simeq A$, an $R$-isomorphism (necessarily unique)$\\}$.

The poset of localizable multiplicative sets $(\mathrm{L}(R)_{m},\subseteq)$.
An Ore set or a denominator set of a ring is a multiplicative set. This fact
suggests the following definitions. Let $\mathrm{L}(R)_{m}$ be the set of all
localizable multiplicative sets of $R$. By Lemma 2.6, the localization of a
ring at a localizable set is the same as the localization of the ring at the
localizable multiplicative set that the set generates. For an ideal
$\mathfrak{a}$ of $R$, let
$\mathrm{L}(R,\mathfrak{a})_{m}:=\\{S\in\mathrm{L}(R)_{m}\,|\,{\rm
ass}_{R}(S)=\mathfrak{a}\\}$. By Lemma 2.6, ${\rm ass}\,\mathrm{L}(R)={\rm
ass}\,\mathrm{L}(R)_{m}:=\\{{\rm ass}_{R}(S)\,|\,S\in\mathrm{L}(R)_{m}\\}$,
and so

$\mathrm{L}(R)_{m}=\coprod_{\mathfrak{a}\in{\rm
ass}\,\mathrm{L}(R)}\mathrm{L}(R,\mathfrak{a})_{m}$ (23)

is a disjoint union of non-empty sets. The map

$\mathrm{L}(R)\rightarrow\mathrm{L}(R)_{m},\;\;S\mapsto S_{mon}$ (24)

is an epimorphism of posets with section

$\mathrm{L}(R)_{m}\rightarrow\mathrm{L}(R),\;\;S\mapsto S.$ (25)

Similarly, for each $\mathfrak{a}\in{\rm ass}\,\mathrm{L}(R)$, the map

$\mathrm{L}(R,\mathfrak{a})\rightarrow\mathrm{L}(R,\mathfrak{a})_{m},\;\;S\mapsto
S_{mon}$ (26)

is an epimorphism of posets with section

$\mathrm{L}(R,\mathfrak{a})_{m}\rightarrow\mathrm{L}(R,\mathfrak{a}),\;\;S\mapsto
S.$ (27)

The map

$R\langle(\cdot)^{-1}\rangle:\,\mathrm{L}(R)_{m}\rightarrow\mathrm{L}{\rm
oc}(R),\;\;S\mapsto R\langle S^{-1}\rangle,$ (28)

is an epimorphism of posets. For each ideal $\mathfrak{a}\in{\rm
ass}\,\mathrm{L}(R)$, it induces the epimorphism of posets,

$R\langle(\cdot)^{-1}\rangle:\,\mathrm{L}(R,\mathfrak{a})_{m}\rightarrow\mathrm{L}{\rm
oc}(R,\mathfrak{a}),\;\;S\mapsto R\langle S^{-1}\rangle.$ (29)

For each ideal $\mathfrak{a}\in{\rm ass}\,\mathrm{L}(R)$,

$\mathrm{L}(R,\mathfrak{a})_{m}=\bigsqcup_{A\in\mathrm{L}{\rm
oc}(R,\mathfrak{a})}\mathrm{L}(R,\mathfrak{a},A)_{m}$ (30)

where
$\mathrm{L}(R,\mathfrak{a},A)_{m}:=\\{S\in\mathrm{L}(R,\mathfrak{a})_{m}\,|\,R\langle
S^{-1}\rangle\simeq A$, an $R$-isomorphism (necessarily unique)$\\}$.

Direct limits of rings. Let $(I,\leq)$ be a poset, $\\{R_{i}\\}_{i\in I}$ be
rings, and for each pair of elements $i$ and $j$ of $I$ such that $i\leq j$
there is a ring homomorphism $f_{ji}:R_{i}\rightarrow R_{j}$ such that
$f_{ii}$ is the identity map for all $i\in I$, and

$f_{ki}=f_{kj}f_{ji}$

for all elements $i,j,k\in I$ such that $i\leq j\leq k$. The system
$(R_{i},f_{ij})$ is called a direct system of rings on $I$. A ring
$\varinjlim_{i\in I}R_{i}$ together with ring homomorphisms
$g_{i}:R_{i}\rightarrow\varinjlim_{i\in I}R_{i}$ such that $g_{i}=g_{j}f_{ji}$
for all $i\leq j$ is called the direct limit of the direct system
$(R_{i},f_{ji})$ if it satisfies the following universal property: For each
system of ring homomorphisms $g_{i}^{\prime}:R_{i}\rightarrow A$ such that
$g_{i}^{\prime}=g_{j}^{\prime}f_{ji}$ for all $i\leq j$, there is a unique
homomorphism

$\alpha:\varinjlim_{i\in I}R_{i}\rightarrow A$

such that $g_{i}^{\prime}=\alpha g_{i}$ for all $i\in I$. It follows from the
universal property that the direct limit is unique (up to isomorphism).

Suppose that the poset $(I,\leq)$ is a directed set, that is for each pair of
elements $i,j\in I$ there is an element $k\in I$ such that $i\leq k$ and
$j\leq k$. On the disjoint union $\coprod_{i\in I}R_{i}$, we say that elements
$a_{i}\in R_{i}$ and $a_{j}\in R_{j}$ are equivalent, $a_{i}\sim a_{k}$, if

$f_{ki}(a_{i})=f_{kj}(a_{j})$

for some element $k\in I$ such that $i,j\leq k$. Then the set of equivalence
classes, $\coprod_{i\in I}R_{i}/\sim,$ is isomorphic to $\varinjlim_{i\in
I}R_{i}$ where for each $i\in I$, the map

$g_{i}:R_{i}\rightarrow\varinjlim_{i\in I}R_{i},\;\;r\mapsto\widetilde{r}$

is a ring homomorphism and $\widetilde{r}$ is the equivalence class of the
element $r$, such that for all $i\leq j$, $g_{i}=g_{j}f_{ji}$.

Let $(I,\leq)$ be a poset, $(R_{i},f_{ij})$ be a direct system of rings on
$I$. Let ${\cal I}$ be an additive subgroup of the the direct sum of
$K$-modules $\bigoplus_{i\in I}R_{i}$ generated by all the elements

$\\{a_{i}-a_{j}\,|\,f_{ki}(a_{i})=f_{kj}(a_{j}),a_{i}\in R_{i},a_{j}\in
R_{j},i\leq k,j\leq k\\}.$ (31)

By the definition, ${\cal I}$ is a $K$-submodule of $\bigoplus_{i\in I}R_{i}$.
Let $*_{i\in I}R_{i}$ be the free product of $K$-algebras $R_{i}$. For each
$i\in I$, there is a natural ring homomorphism
$\theta_{i}:R_{i}\rightarrow*_{i\in I}R_{i}$, and the ring $*_{i\in I}R_{i}$
is generated by the images of the rings $R_{i}$. Let

$\theta:\bigoplus_{i\in I}R_{i}\rightarrow*_{i\in
I}R_{i},\;\;r_{i}\mapsto\theta_{i}(r_{i})\;\;{\rm for}\;\;r_{i}\in R_{i}.$
(32)

An element $i\in I$ is the largest element of the poset $I$ if for all
elements $j\in I$, $j\leq i$.

###### Theorem 2.14

Let $(I,\leq)$ be a poset and $(R_{i},f_{ij})$ be a direct system of rings on
$I$. Then

  1. 1.

$\varinjlim_{i\in I}R_{i}\simeq*_{i\in I}R_{i}/(\theta({\cal I}))$ where
$(\theta({\cal I}))$ is the ideal of the free product of $K$-algebras $*_{i\in
I}R_{i}$ generated by the set $\theta({\cal I})$.

  2. 2.

Suppose that for each element $i\in I$ there is an element $j\in\max(I)$ such
that $i\leq j$. Then

$\varinjlim_{i\in I}R_{i}\simeq*_{\mu\in\max(I)}R_{\mu}/\mathfrak{f}$

where $*_{\mu\in\max(I)}R_{\mu}$ is a free product of the set of $K$-algebras
$\\{R_{\mu}\,|\,\mu\in\max(I)\\}$ and $\mathfrak{f}$ is an ideal of
$*_{\mu\in\max(I)}R_{\mu}$ which is generated by the set
$\theta\bigg{(}\bigg{(}\bigoplus_{\mu\in\max(I)}R_{\mu}\bigg{)}\bigcap{\cal
I}\bigg{)}$.

  3. 3.

Suppose that an element $i\in I$ is the largest element of the poset $I$. Then
$\varinjlim_{j\in I}R_{j}\simeq R_{i}$.

Proof. 1. To prove statement 1 it suffices to show that the ring $*_{i\in
I}R_{i}/(\theta({\cal I}))$ satisfies the universal property of the ring
$\varinjlim_{i\in I}R_{i}$. Let $g_{i}^{\prime}:R_{i}\rightarrow A$, $i\in I$,
be a set of ring homomorphisms such that $g_{i}^{\prime}=g_{j}^{\prime}f_{ji}$
for all $i\leq j$. Then there is a unique ring homomorphism $\rho:*_{i\in
I}R_{i}\rightarrow A$ such that $g_{i}^{\prime}=\rho\theta_{i}$ for all $i\in
I$. Then

$\rho\theta({\cal I})=0$

since for all elements $a_{i}\in R_{i}$ and $a_{j}\in R_{j}$ such that
$f_{ki}(a_{i})=f_{kj}(a_{j})$ for some $k\in I$ such that $i,j\leq k$,

$\rho(\theta_{i}(a_{i})-\theta_{j}(a_{j}))=g_{i}^{\prime}(a_{i})-g_{j}^{\prime}(a_{j})=g_{k}^{\prime}f_{ki}(a_{i})-g_{k}^{\prime}f_{kj}(a_{j})=g_{k}^{\prime}(f_{ki}(a_{i})-f_{kj}(a_{j}))=g_{k}^{\prime}(0)=0.$

2\. Let $\varinjlim^{K}_{i\in I}R_{i}$ be the direct limit of the $K$-modules
$R_{i}$.

(i) $\varinjlim^{K}_{i\in I}R_{i}=\bigg{(}\bigoplus_{i\in
I}R_{i}\bigg{)}/{\cal I}$: By the definition, the $K$-module
$\bigg{(}\bigoplus_{i\in I}R_{i}\bigg{)}/{\cal I}$ satisfies the universal
property of the direct limit of $K$-modules $R_{i}$ and the statement (i)
follows from the uniqueness of the direct limit of $K$-modules.

(ii) $\bigg{(}\bigoplus_{i\in I}R_{i}\bigg{)}/{\cal
I}\simeq\bigg{(}\bigoplus_{\mu\in\max(I)}R_{\mu}\bigg{)}/\bigg{(}\bigoplus_{\mu\in\max(I)}R_{\mu}\bigg{)}\bigcap{\cal
I}$: By the assumption, for each element $i\in I$ there is an element
$j\in\max(I)$ such that $i\leq j$. Then it follows from the inclusion
$\bigoplus_{\mu\in\max(I)}R_{\mu}\subseteq\bigoplus_{i\in I}R_{i}$ that

$\bigg{(}\bigoplus_{i\in I}R_{i}\bigg{)}/{\cal
I}=\bigg{(}\bigoplus_{\mu\in\max(I)}R_{i}+{\cal I}\bigg{)}/{\cal
I}\simeq\bigg{(}\bigoplus_{\mu\in\max(I)}R_{\mu}\bigg{)}/\bigg{(}\bigoplus_{\mu\in\max(I)}R_{\mu}\bigg{)}\bigcap{\cal
I}.$

(iii) $\varinjlim{}^{K}_{i\in I}R_{i}=\bigg{(}\bigoplus_{i\in
I}R_{i}\bigg{)}/{\cal
I}\simeq\bigg{(}\bigoplus_{\mu\in\max(I)}R_{\mu}\bigg{)}/\bigg{(}\bigoplus_{\mu\in\max(I)}R_{\mu}\bigg{)}\bigcap{\cal
I}$: The statement (iii) follows from the statements (i) and (ii).

(iv) The ring $*_{\mu\in\max(I)}R_{\mu}/\mathfrak{f}$ satisfies the universal
property of the ring $\varinjlim_{i\in I}R_{i}$: Let
$g_{i}^{\prime}:R_{i}\rightarrow A$, $i\in I$, be a set of ring homomorphisms
such that $g_{i}^{\prime}=g_{j}^{\prime}f_{ji}$ for all $i\leq j$. By
statement 1 and the statement (iii), the homomorphisms $g_{i}^{\prime}$ induce
a unique ring homomorphism,
$\xi:*_{\mu\in\max(I)}R_{\mu}/\mathfrak{f}\rightarrow A$ such that
$g_{\mu}^{\prime}=\xi\theta_{\mu}$ for all $\mu\in\max(I)$. Now, statement 2
follows from the statement (iv).

3\. Statement 3 follows from statement 2. $\Box$

Let $T\in\mathrm{L}(R)$. Then the power set of $T$, i.e. the set of all
subsets of the set $T$, $({\cal P}(T),\subseteq)$, is a poset. The system
$\\{R\langle S^{-1}\rangle,\phi_{SS^{\prime}}\\}$ is a direct system where
$S,S^{\prime}\in{\cal P}(T)$. Corollary 2.15.(1) describe the direct limit
$\varinjlim_{\\{S\subseteq T\\}}R\langle S^{-1}\rangle$.

###### Corollary 2.15

Let $R$ be a ring and $T\in\mathrm{L}(R)$. Then

  1. 1.

$R\langle T^{-1}\rangle\simeq\varinjlim_{\\{S\subseteq T\\}}R\langle
S^{-1}\rangle$.

  2. 2.

${\rm ass}_{R}(T)=\bigcup_{S\subseteq T}{\rm ass}_{R}(S)$.

Proof. 1. Statement 1 follows from Proposition 2.8.(2).

2\. Statement 2 follows from statement 1. $\Box$

We collect two results on direct limits, Lemma 2.16 and Lemma 5.8, that are
used later in the paper and are interesting on their own.

###### Lemma 2.16

Let $(I,\leq)$ be a directed set and $(R_{i},f_{ij})$ be a direct system of
rings on $I$. Suppose that all rings $R_{i}$, where $i\in I$, are isomorphic
to a ring $R$ and all the maps $f_{ij}$ are isomorphisms. Then
$\varinjlim_{i\in I}R_{i}\simeq R$.

Proof. Fix an element $i\in I$ and an isomorphism $g_{i}:R_{i}\rightarrow R$.
For each element $j\in I$ such that $i\leq j$, let
$g_{j}:=g_{i}f_{ij}:R_{j}\rightarrow R$ where $f_{ij}:=f_{ji}^{-1}$. For each
element $k\in I$ such that $i\not\leq k$, fix an element $j\in I$ such that
$i\leq j$ and $k\leq j$. Let

$g_{k}:=g_{i}f_{ij}f_{jk}:R_{k}\rightarrow R\;\;{\rm
where}\;\;f_{jk}:=f_{kj}.$

Then it is easy to verify that the map $g_{k}$ does not depend on the choice
of the element $j$ and $(R,g_{i})_{i\in I}$ satisfies the universal property
of the direct limit for the direct system $(R_{i},f_{ij})$, and so
$R=\varinjlim_{i\in I}R_{i}$. $\Box$

Lemma 5.8 gives an example of a direct system of rings with zero direct limit.

###### Lemma 2.17

Let $J=\\{1\\}\coprod I$ be a disjoint union of two sets where the set $I$
consists at least of two elements. The set $J$ is a poset where $1<i$ for all
$i\in I$. Let $\\{R_{i}\,|\,i\in I\\}$ be a set of rings and
$A_{1}=\prod_{i\in I}R_{i}$. For each $i\in I$, let the map
$f_{1i}:R_{1}\rightarrow R_{i}$ be the projection onto $R_{i}$, $f_{11}={\rm
id}_{R_{1}}$ and $f_{ii}={\rm id}_{R_{i}}$ (the identity maps). Then
$\varinjlim_{j\in J}R_{i}\simeq\\{0\\}$.

Proof. For each $i\in I$, let $1_{i}$ be the identity element of the ring
$R_{i}$, and $1_{i}^{\prime}$ be an element of $R_{1}$ where on the $i$’th
place is $1_{i}$ and zeros are elsewhere. Clearly, $\max(J)=I$. Now, by
Theorem 2.14.(2), $\varinjlim_{j\in J}R_{j}\simeq*_{i\in
I}R_{i}/\mathfrak{f}$. Since for all $i\in I$,

$1_{i}=1_{i}-0=(1_{i}-1_{i}^{\prime})+(1_{i}^{\prime}-p_{j}(1_{i}^{\prime}))\in\mathfrak{f}$

where $j\neq i$, $\varinjlim_{j\in J}R_{i}\simeq\\{0\\}$ (see Theorem 2.14.(2)
for the definition of the ideal $\mathfrak{f}$). $\Box$

The absolute quotient ring $Q_{a}(R)$ of a ring $R$. The set
$(\mathrm{L}(R),\subseteq)$ is a poset and the system $\\{R\langle
S^{-1}\rangle,\phi_{ST}\\}$ is a direct system of rings where
$S,T\in\mathrm{L}(R)$.

Definition. The direct limit $Q_{a}(R):=\varinjlim_{S\in\mathrm{L}(R)}R\langle
S^{-1}\rangle$ is called the absolute quotient ring of the ring $R$. Let
$\sigma_{a}:R\rightarrow Q_{a}(R)$ be a unique $R$-homomorphism,
$\mathfrak{a}_{a}(R):={\rm ker}(\sigma_{a})$, $R_{a}:=R/\mathfrak{a}_{a}(R)$,
$\pi_{a}:R\rightarrow R_{a}$, $r\mapsto r+\mathfrak{a}_{a}$, and
$\overline{\sigma_{a}}:R_{a}\rightarrow Q_{a}(R)$ is a unique
$R$-homomorphism. Clearly, $\sigma_{a}=\overline{\sigma_{a}}\pi_{a}$.

###### Theorem 2.18

$Q_{a}(R)=\begin{cases}R\langle S^{-1}\rangle&\text{if
}\max\mathrm{L}(R)=\\{S\\},\\\ \\{0\\}&\text{otherwise},\\\ \end{cases}$ and
$\mathfrak{a}_{a}(R)=\begin{cases}{\rm ass}_{R}(S)&\text{if
}\max\mathrm{L}(R)=\\{S\\},\\\ R&\text{otherwise}.\\\ \end{cases}$

Proof. Suppose that $\max\mathrm{L}(R)=\\{S\\}$. Then, by Theorem 3.1.(2), the
element $S$ is the largest element of the poset $\mathrm{L}(R)$, and the
statement follows from Theorem 2.14.(3).

Suppose that the set $\max\mathrm{L}(R)$ contains at least two elements, say
$S$ and $T$. The set ${\cal S}:=\sigma_{a}^{-1}(Q_{a}(R)^{\times})$ is a
localizable set in $R$ since $\sigma_{a}({\cal S})\subseteq
Q_{a}(R)^{\times}$. The diagram below is commutative

$\begin{array}[c]{ccc}R&\stackrel{{\scriptstyle\sigma_{S}}}{{\rightarrow}}&R\langle
S^{-1}\rangle\\\
&\stackrel{{\scriptstyle\sigma_{a}}}{{\searrow}}&{\downarrow}^{\phi_{S}}\\\
&&Q_{a}(R)\end{array}$

where $\phi_{S}$ is a unique $R$-homomorphism. Then it follows from the
inclusion $\phi_{S}(R\langle S^{-1}\rangle^{\times})\subseteq
Q_{a}(R)^{\times}$ that

${\cal
S}=\sigma_{a}^{-1}(Q_{a}(R)^{\times})=(\phi_{S}\sigma_{S})^{-1}(Q_{a}(R)^{\times})=\sigma_{S}^{-1}\phi_{S}^{-1}(Q_{a}(R)^{\times})\supseteq\sigma_{S}^{-1}(R\langle
S^{-1}\rangle^{\times})=S,$

by Corollary 3.7.(1). Similarly, ${\cal S}\supseteq T$. Therefore, ${\cal
S}\supseteq S\cup T$, and so the set $S\cup T$ is a localizable set of $R$ as
a subset of the localizable set ${\cal S}$, a contradiction (since
$S,T\in\max\mathrm{L}(R)$ and $S\neq T$). The proof of the first equality of
the theorem is complete. The second equality of the theorem follows from the
first. $\Box$

Example. If $R$ is a commutative domain then the set ${\cal
C}_{R}=R\backslash\\{0\\}$ is the largest element of $\mathrm{L}(R)$. By
Theorem 2.18, $Q_{a}(R)=Q_{cl}(R)$ is the classical quotient ring of $R$, i.e.
the field of fractions of $R$, and $\mathfrak{a}_{a}(R)=0$.

###### Lemma 2.19

If $R$ is a simple Artinian rings then $\max\mathrm{L}(R)=\\{R^{\times}\\}$,
$Q_{a}(R)=R$ and $\mathfrak{a}_{a}(R)=0$.

Proof. Every non-unit of the simple Artinian $R$ ring is a zero divisor. Since
the ring $R$ is simple every non-unit $r\in R$ is a non-localizable element
since ${\rm ass}_{R}(r)=R$. Therefore, ${\cal L}\mathrm{L}(R)=R^{\times}$ and
${\cal N}{\cal L}\mathrm{L}(R)=R\backslash R^{\times}$. Hence,
$\max\mathrm{L}(R)=\\{R^{\times}\\}$. Now, by Theorem 2.18,
$Q_{a}(R)=R\langle(R^{\times})^{-1}\rangle=R$ and $\mathfrak{a}_{a}(R)=0$.
$\Box$

We need the lemma below in order to prove Proposition 2.21 which gives a
practical sufficient condition for $Q_{a}(R)=\\{0\\}$ and
$|\max\mathrm{L}(R)|\geq 2$.

###### Lemma 2.20

  1. 1.

$Q_{a}(R)\simeq\Big{(}*_{S\in\max\mathrm{L}(R)}R\langle
S^{-1}\rangle\Big{)}/\mathfrak{f}_{a}$ where $\mathfrak{f}_{a}$ is an ideal of
the $K$-algebra $*_{S\in\max\mathrm{L}(R)}R\langle S^{-1}\rangle$ which is
generated by the set $\theta\Big{(}\bigoplus_{S\in\max\mathrm{L}(R)}R\langle
S^{-1}\rangle\cap{\cal I}\Big{)}$, see Theorem 2.14.(2).

  2. 2.

$\mathfrak{a}_{a}(R)\supseteq\sum_{S\in\max\mathrm{L}(R)}{\rm ass}_{R}(S)$.

  3. 3.

$Q_{a}(R)=\\{0\\}$ iff $\mathfrak{a}_{a}(R)=R$. If
$\sum_{S\in\max\mathrm{L}(R)}{\rm ass}_{R}(S)=R$ then $Q_{a}(R)=\\{0\\}$.

Proof. 1. Statement 1 is a particular case of Theorem 2.14.(2).

2\. Let $S\in\max\mathrm{L}(R)$. Then for each element $r\in{\rm ass}_{R}(S)$
and each $T\in\max\mathrm{L}(R)$,

$\sigma_{T}(r)=(\sigma_{T}(r)-r)-(0-r)=(\sigma_{T}(r)-r)-(\sigma_{S}(r)-r)\in\theta\bigg{(}\bigoplus_{S\in\max\mathrm{L}(R)}R\langle
S^{-1}\rangle\cap{\cal I}\bigg{)},$

and so ${\rm ass}_{R}(S)\subseteq\mathfrak{a}_{a}(R)$, and statement 2
follows.

3\. $Q_{a}(R)=\\{0\\}$ iff $1=0$ in $Q_{a}(R)$ iff $\mathfrak{a}_{a}(R)=R$. If
$\sum_{S\in\max\mathrm{L}(R)}{\rm ass}_{R}(S)=R$ then, by statement 2,
$\mathfrak{a}_{a}(R)=R$, and so $Q_{a}(R)=\\{0\\}$. $\Box$

###### Proposition 2.21

Given localizable sets $\\{S_{i}\,|\,i\in I\\}$ of a ring $R$ such that
$\sum_{i\in I}{\rm ass}_{R}(S_{i})=R$. Then $Q_{a}(R)=\\{0\\}$ and
$|\max\mathrm{L}(R)|\geq 2$.

Proof. By Theorem 3.1.(2), every localizable set $S_{i}$ is contained in a
maximal localizable set $S_{i}^{\prime}$. Since ${\rm
ass}_{R}(S_{I})\subseteq{\rm ass}_{R}(S_{I}^{\prime})$ and $\sum_{i\in I}{\rm
ass}_{R}(S_{i})=R$, we have that $\sum_{i\in I}{\rm
ass}_{R}(S_{i}^{\prime})=R$, and so $\sum_{S\in\max\mathrm{L}(R)}{\rm
ass}_{R}(S)=R$. By Lemma 2.20.(2), $\mathfrak{a}_{a}(R)=R$, and so
$Q_{a}(R)=0$, by Lemma 2.20.(3). By Theorem 2.18, $|\max\mathrm{L}(R)|\geq 2$.
$\Box$

###### Corollary 2.22

Let $A=\prod_{i\in I}A_{i}$ be a product of rings where $I$ is an arbitrary
set that contains at least two elements. Then $Q_{a}(A)=0$ and
$|\max\mathrm{L}(R)|\geq 2$.

Proof. By the assumption, the set $I$ contains at least two elements. For each
$i\in I$, the set $S_{i}=A_{i}^{\times}\times\prod_{j\neq i}A_{j}$ is a
localizable set of $A$ with ${\rm ass}_{R}(S_{i})=\\{0\\}\times\prod_{j\neq
i}A_{j}$ since $A\langle S_{i}^{-1}\rangle\simeq A_{i}$. Let $i$ and $j$ be
distinct elements of the set $I$. Since $A={\rm ass}_{A}(S_{i})+{\rm
ass}_{A}(S_{j})$, we have that $Q_{a}(A)=\\{0\\}$ and $|\max\mathrm{L}(R)|\geq
2$, by Proposition 2.21. $\Box$

Every Ore set is a localizable set. Let $S$ be an Ore set of the ring $R$.
Theorem 2.23 states that every Ore set is localizable, gives an explicit
description of this ideal and the ring $R\langle S^{-1}\rangle$, and
$\mathfrak{a}(S)={\rm ass}_{R}(S)$. So, the localization an any Ore set is an
example of the perfect localization. Furthermore, Theorem 2.23 also states
that the ring $R\langle S^{-1}\rangle$ is $R$-isomorphic to the localization
$\overline{S}^{-1}\overline{R}$ of the ring $\overline{R}$ at the denominator
set $\overline{S}$ of $\overline{R}$.

###### Theorem 2.23

Let $R$ be a ring and $S\in{\rm Ore}(R)$.

  1. 1.

[1, Theorem 4.15] Every Ore set is a localizable set.

  2. 2.

[7, Theorem 1.6.(1)] $\mathfrak{a}:=\\{r\in R\,|\,srt=0$ for some elements
$s,t\in S\\}$ is an ideal of $R$ such that $\mathfrak{a}\neq R$.

  3. 3.

[7, Theorem 1.6.(2)] Let $\pi:R\rightarrow\overline{R}:=R/\mathfrak{a}$,
$r\mapsto\overline{r}=r+\mathfrak{a}$. Then $\overline{S}:=\pi(S)\in{\rm
Den}(\overline{R},0)$, $\mathfrak{a}=\mathfrak{a}(S)={\rm ass}_{R}(S)$,
$S\in\mathrm{L}(R,\mathfrak{a})$, and
$S^{-1}R\simeq\overline{S}^{-1}\overline{R}$, an $R$-isomorphism. In
particular, every Ore set is localizable.

Criterion for a left Ore set to be a left localizable set. For a ring $R$ and
its ideal $\mathfrak{a}$, let

$\displaystyle{}^{\prime}{\rm Den}_{l}(R,\mathfrak{a})$ $\displaystyle:=$
$\displaystyle\\{S\in{\rm Den}_{l}(R)\,|\,{\rm
ass}_{l}(S)=\mathfrak{a},S\subseteq{}^{\prime}{\cal C}_{R}\\},$
$\displaystyle{\rm Den}_{r}^{\prime}(R,\mathfrak{a})$ $\displaystyle:=$
$\displaystyle\\{S\in{\rm Den}_{r}(R)\,|\,{\rm
ass}_{r}(S)=\mathfrak{a},S\subseteq{\cal C}_{R}^{\prime}\\}.$

Theorem 2.24.(1) is a criterion for a left Ore set to be a left localizable
set and Theorem 2.24.(2) describes the structure of the localization of a ring
at a localizable left Ore set.

###### Theorem 2.24

([7, Theorem 1.5]) Let $R$ be a ring, $S\in{\rm Ore}_{l}(R)$, and
$\mathfrak{a}={\rm ass}_{R}(S)$. Then

  1. 1.

$S\in\mathrm{L}(R)$ iff ${}^{\prime}\mathfrak{a}\neq R$ where the ideal
${}^{\prime}\mathfrak{a}={}^{\prime}\mathfrak{a}(S)$ of $R$ is as in
Proposition 2.10.(2) and (13).

  2. 2.

Suppose that ${}^{\prime}\mathfrak{a}\neq R$. Let
${}^{\prime}\pi:R\rightarrow{}^{\prime}R:=R/{}^{\prime}\mathfrak{a}$,
$r\mapsto{}^{\prime}r=r+{}^{\prime}\mathfrak{a}$ and
${}^{\prime}S={}^{\prime}\pi(S)$. Then

    1. (a)

${}^{\prime}S\in{}^{\prime}{\rm Den}_{l}({}^{\prime}R)$.

    2. (b)

$\mathfrak{a}={}^{\prime}\pi^{-1}({\rm ass}_{l}({}^{\prime}S))$.

    3. (c)

$S^{-1}R\simeq{}^{\prime}S^{-1}{}^{\prime}R$, an $R$-isomorphism.

[7, Theorem 2.11] is a criterion for a right Ore set to be a localizable set.

###### Definition 2.25

([7]) A multiplicative set $S$ of a ring $R$ is called a left localizable set
of $R$ if

$R\langle
S^{-1}\rangle=\\{\overline{s}^{-1}\overline{r}\,|\,\overline{s}\in\overline{S},\overline{r}\in\overline{R}\\}\neq\\{0\\}$

where $\overline{R}=R/\mathfrak{a}$, $\mathfrak{a}={\rm ass}_{R}(S)$ and
$\overline{S}=(S+\mathfrak{a})/\mathfrak{a}$, i.e., every element of the ring
$R\langle S^{-1}\rangle$ is a left fraction $\overline{s}^{-1}\overline{r}$
for some elements $\overline{s}\in\overline{S}$ and
$\overline{r}\in\overline{R}$. Similarly, a multiplicative set $S$ of a ring
$R$ is called a right localizable set of $R$ if

$R\langle
S^{-1}\rangle=\\{\overline{r}\overline{s}^{-1}\,|\,\overline{s}\in\overline{S},\overline{r}\in\overline{R}\\}\neq\\{0\\},$

i.e., every element of the ring $R\langle S^{-1}\rangle$ is a right fraction
$\overline{r}\overline{s}^{-1}$ for some elements
$\overline{s}\in\overline{S}$ and $\overline{r}\in\overline{R}$. A right and
left localizable set of $R$ is called a localizable set of $R$.

##  3 Maximal localizable sets, the localization radical, the sets of
localizable and non-localizable elements

For a ring $R$, the following concepts are introduced: the maximal localizable
sets, the localization radical $\mathrm{L}{\rm rad}(R)$, the set of
localizable elements ${\cal L}\mathrm{L}(R)$, the set of non-localizable
elements ${\cal N}{\cal L}\mathrm{L}(R)$, the set of completely localizable
elements ${\cal C}\mathrm{L}(R)$. All these objects are characteristic subsets
of $R$, i.e. they are invariant under the action of the automorphism group of
$R$. Proposition 3.10 shows that there are tight connections between the sets
${\cal L}\mathrm{L}(R)$, ${\cal N}{\cal L}\mathrm{L}(R)$, and $\mathrm{L}{\rm
rad}(R)$. For an arbitrary ring, Theorem 3.1 states that the set of maximal
localizable set is a non-empty set and every localizable set is a subset of a
maximal localizable set.

Proposition 3.5.(1), is an explicit description of the largest element ${\cal
S}(R,\mathfrak{a},A)$ of the poset $(\mathrm{L}(R,\mathfrak{a},A),\subseteq)$.
Corollary 3.6 is a criterion for a set $S\in\mathrm{L}(R,\mathfrak{a},A)$ to
be the largest element of the poset $\mathrm{L}(R,\mathfrak{a},A)$. Theorem
3.8 describes the sets $\max\mathrm{L}(R)$ and $\max\mathrm{L}{\rm oc}(R)$,
presents a bijection between them, and shows that $\max\mathrm{L}{\rm
oc}(R)\neq\emptyset$. Similarly, for each ideal $\mathfrak{a}\in{\rm
ass}\mathrm{L}(R)$, Theorem 3.9 describes the sets
$\max\mathrm{L}(R,\mathfrak{a})$ and $\max\mathrm{L}{\rm oc}(R,\mathfrak{a})$,
presents a bijection between them, and shows that they are nonempty sets.

The $\mathfrak{a}$-absolute quotient ring $Q_{a}(R,\mathfrak{a})$ is
introduced for each ideal $\mathfrak{a}\in{\rm ass}\,\mathrm{L}(R)$. Theorem
3.15 gives an explicit description of the ring $Q_{a}(R,\mathfrak{a})$, a
criterion for $Q_{a}(R,\mathfrak{a})\neq\\{0\\}$, and a criterion for
$\mathfrak{a}_{a,\mathfrak{a}}=\mathfrak{a}$.

Theorem 3.17.(1) is an $R$-isomorphism criterion for localizations of the ring
$R$ and Theorem 3.17.(2) is a criterion for existing of an $R$-homomorphism
between localizations of $R$. Proposition 3.14 contains some properties of
${\cal C}\mathrm{L}(R)$ and $\mathfrak{c}_{R}$. Theorem 3.20 is a criterion
for the ring $R\langle S^{-1}\rangle$ to be an $R$-isomorphic to a
localization of $R$ at a denominator set and it also describes all such
denominator sets.

For a denominators set $T\in{\rm Den}_{*}(R,\mathfrak{a})$, Proposition 3.21
describes all localizable sets $S$ of the ring $R$ such that the ring
$R\langle S^{-1}\rangle$ is an $R$-isomorphic to the localization of $R$ at
$T$. For a localizable multiplicative set $S$ of the ring $R$, Proposition
3.22 is a criterion for the set $S$ to be a left denominator set of $R$.

The set $\max\mathrm{L}(R)$ of maximal elements of $\mathrm{L}(R)$ is a non-
empty set. For a ring $R$, the set ${\rm max.Den}_{l}(R)$ of maximal left
denominator sets (w.r.t. $\subseteq$) is a non-empty set, [3, Lemma 3.7.(2)]
and the sets of maximal localizable left or right or two-sided Ore sets are
also non-empty sets [7, Theorem 1.9]. Let $\max\,\mathrm{L}(R)$ be the set of
maximal (w.r.t. $\subseteq$) elements of the set $\mathrm{L}(R)$. An element
of the set $\max\,\mathrm{L}(R)$ is called a maximal localizable set of the
ring $R$.

###### Theorem 3.1

Let $R$ be a ring. Then

  1. 1.

$\max\mathrm{L}(R)\neq\emptyset$.

  2. 2.

Each set $S\in\mathrm{L}(R)$ is contained in a set ${\cal
S}\in\max\mathrm{L}(R)$.

Proof. 1. The theorem follows from Zorn’s Lemma. Let
$\\{S_{i}\\}_{i\in\Gamma}\subseteq\mathrm{L}(R)$ be an ordered by inclusion
set of localizable sets in $R$ indexed by an ordinal $\Gamma$, i.e. for each
pair $i\leq j$ in $\Gamma$, $S_{i}\subseteq S_{j}$ and if $i\in\Gamma$ is a
limit ordinal then $S_{i}=\bigcup_{j<i}S_{j}$. In view of Zorn’s Lemma, it
suffices to show that $\bigcup_{i\in\Gamma}S_{i}\in\mathrm{L}(R)$. By (7) and
(9), we have the direct system of ring homomorphisms, $(R\langle
S_{i}^{-1}\rangle,\phi_{S_{j}S_{i}})$, where for each pair $i\leq j$, the map
$\phi_{S_{j}S_{i}}:R\langle S_{i}^{-1}\rangle\rightarrow R\langle
S_{j}^{-1}\rangle$ is defined in (7). Since the rings $R\langle
S_{i}^{-1}\rangle$ are unital and $\phi_{S_{j}S_{i}}(1)=1$ for all $i\leq j$,
the direct limit $\varinjlim R\langle S_{i}^{-1}\rangle$ is a nonzero ring and
and is isomorphic to
$R\langle\Big{(}\bigcup_{i\in\Gamma}S_{i}\Big{)}^{-1}\rangle$. Therefore,
$\bigcup_{i\in\Gamma}S_{i}\in\mathrm{L}(R)$.

2\. Repeat the proof of statement 1 for the set of all localizable sets that
contain the set $S$. $\Box$

###### Corollary 3.2

Let $R$ be a ring and $T\in\max\,\mathrm{L}(R)$. Then

  1. 1.

$R\langle T^{-1}\rangle\simeq\varinjlim_{\\{S\subseteq T\\}}R\langle
S^{-1}\rangle$.

  2. 2.

${\rm ass}_{R}(T)=\bigcup_{S\subseteq T}{\rm ass}_{R}(S)$.

Proof. The corollary is a particular case of Corollary 2.15. $\Box$

Let $\max{\rm ass}\,\mathrm{L}(R)$ be the set of maximal elements of the set
${\rm ass}\,\mathrm{L}(R)=\\{{\rm ass}_{R}(S)\,|\,S\in\mathrm{L}(R)\\}$ and
${\rm ass}\max\mathrm{L}(R):=\\{{\rm ass}_{R}({\cal S})\,|\,{\cal
S}\in\max\mathrm{L}(R)\\}$.

###### Corollary 3.3

$\max{\rm ass}\,\mathrm{L}(R)\subseteq{\rm ass}\max\mathrm{L}(R)$.

Proof. The corollary follows from Theorem 3.1. $\Box$

A set of subsets is called an antichain if the sets are incomparable, that is
none of the sets is a subset of another one.

Question. Is $\max{\rm ass}\,\mathrm{L}(R)={\rm ass}\max\mathrm{L}(R)$? I.e.
the set ${\rm ass}\max\mathrm{L}(R)$ is an antichain.

The largest element ${\cal S}(R,\mathfrak{a},A)$ in
$(\mathrm{L}(R,\mathfrak{a},A)\subseteq)$ and its characterizations where
$A\in\mathrm{L}{\rm oc}(R,\mathfrak{a})$.

###### Lemma 3.4

Given sets $S,{\cal S}\in\mathrm{L}(R,\mathfrak{a},A)$ such that
$S\subseteq{\cal S}$. Then $T\in\mathrm{L}(R,\mathfrak{a},A)$ for all subsets
$T$ of $R$ such $S\subseteq T\subseteq{\cal S}$.

Proof. The inclusions $S\subseteq T\subseteq{\cal S}$, imply that
$\mathfrak{a}={\rm ass}_{R}(S)\subseteq{\rm ass}_{R}(T)\subseteq{\rm
ass}_{R}({\cal S})=\mathfrak{a}$, and so ${\rm ass}_{R}(T)=\mathfrak{a}$.

It suffices to show that the rings $R\langle T^{-1}\rangle$ and $A$ are
$R$-isomorphic. Since $S\subseteq{\cal S}$ and $S,{\cal
S}\in\mathrm{L}(R,\mathfrak{a},A)$, the unique $R$-homomorphism $A\simeq
R\langle S^{-1}\rangle\rightarrow R\langle{\cal S}^{-1}\rangle\simeq A$ is the
identity map ${\rm id}_{A}$. In particular,
$\sigma_{S}(T)\subseteq\sigma_{S}({\cal S})\subseteq A^{\times}$ since
$T\subseteq{\cal S}$ where $\sigma_{S}:R\rightarrow R\langle
S^{-1}\rangle\simeq A$. Now,

$R\langle T^{-1}\rangle\simeq R\langle
S^{-1}\rangle\langle\sigma_{S}(T)^{-1}\rangle\simeq
A\langle\sigma_{S}(T)^{-1}\rangle=A$

since $\sigma_{S}(T)\subseteq A^{\times}$, as required. $\Box$

Proposition 3.5.(1), is an explicit description of the largest element ${\cal
S}(R,\mathfrak{a},A)$ of the poset $(\mathrm{L}(R,\mathfrak{a},A),\subseteq)$.

###### Proposition 3.5

Let $A\in\mathrm{L}{\rm oc}(R,\mathfrak{a})$, $A^{\times}$ be the group of
units of the ring $A$, and $\sigma:R\rightarrow A$, $r\mapsto r+\mathfrak{a}$.
Then

  1. 1.

The set

${\cal
S}(R,\mathfrak{a},A):=\sigma^{-1}(A^{\times})=\bigcup_{S\in\mathrm{L}(R,\mathfrak{a},A)}S$

is the largest element of the poset
$(\mathrm{L}(R,\mathfrak{a},A),\subseteq)$. In particular, the poset
$(\mathrm{L}(R,\mathfrak{a},A),\subseteq)$ is a directed set such that if
$S\in\mathrm{L}(R,\mathfrak{a},A)$ then $T\in\mathrm{L}(R,\mathfrak{a},A)$ for
all subsets $T$ of $R$ such that $S\subseteq T\subseteq{\cal
S}(R,\mathfrak{a},A)$.

  2. 2.

The set ${\cal S}(R,\mathfrak{a},A)$ is a multiplicative set in $R$ such that
${\cal S}(R,\mathfrak{a},A)+\mathfrak{a}={\cal S}(R,\mathfrak{a},A)$.

  3. 3.

$A\simeq R\langle{\cal
S}(R,\mathfrak{a},A)^{-1}\rangle\simeq\varinjlim_{S\in\mathrm{L}(R,\mathfrak{a},A)}R\langle
S^{-1}\rangle$, $R$-isomorphisms.

Proof. 1. Let ${\cal S}:=\sigma^{-1}(A^{\times})$.

(i) ${\cal S}\in\mathrm{L}(R,\mathfrak{a},A)$: Fix an element
$S\in\mathrm{L}(R,\mathfrak{a},A)$. Then the statement (i) follows from Lemma
2.9.(3).

(ii) ${\cal S}=\bigcup_{S\in\mathrm{L}(R,\mathfrak{a},A)}S$: By the statement
(i), ${\cal S}\in\mathrm{L}(R,\mathfrak{a},A)$. For all
$S\in\mathrm{L}(R,\mathfrak{a},A)$, $\sigma(S)\subseteq A^{\times}$, hence
$S\subseteq{\cal S}$, and the statement (ii) follows.

By the statement (ii), the set ${\cal S}$ is the largest element of the poset
$(\mathrm{L}(R,\mathfrak{a},A),\subseteq)$, and so the poset
$(\mathrm{L}(R,\mathfrak{a},A),\subseteq)$ is a directed set. By Lemma 3.4, if
$S\in\mathrm{L}(R,\mathfrak{a},A)$ then $T\in\mathrm{L}(R,\mathfrak{a},A)$ for
all subsets $T$ of $R$ such that $S\subseteq T\subseteq{\cal
S}(R,\mathfrak{a},A)$.

2\. By statement 1, ${\cal S}(R,\mathfrak{a},A)=\sigma^{-1}(A^{\times})$, and
so the set ${\cal S}(R,\mathfrak{a},A)$ is a multiplicative set in $R$ such
that ${\cal S}(R,\mathfrak{a},A)+\mathfrak{a}={\cal S}(R,\mathfrak{a},A)$.

3\. Since ${\cal S}\in\mathrm{L}(R,\mathfrak{a},A)$, there is an
$R$-isomorphism $A\simeq R\langle{\cal S}(R,\mathfrak{a},A)^{-1}\rangle$. By
statement 1, the set $\mathrm{L}(R,\mathfrak{a},A)$ is a directed set w.r.t.
inclusion such that for all sets $S,T\in\mathrm{L}(R,\mathfrak{a},A)$ such
that $S\subseteq T$, $R\langle S^{-1}\rangle\simeq R\langle
T^{-1}\rangle\simeq A$, $R$-isomorphisms. By Lemma 2.16,
$\varinjlim_{S\in\mathrm{L}(R,\mathfrak{a},A)}R\langle S^{-1}\rangle\simeq A$,
an $R$-isomorphism. $\Box$

Corollary 3.6 is a criterion for a set $S\in\mathrm{L}(R,\mathfrak{a},A)$ to
be the largest element of the poset $\mathrm{L}(R,\mathfrak{a},A)$.

###### Corollary 3.6

Let $S\in\mathrm{L}(R,\mathfrak{a},A)$. Then $S={\cal S}(R,\mathfrak{a},A)$
iff for every element $r\in R\backslash S$ the rings $A$ and
$R\langle(S\cup\\{r\\})^{-1}\rangle$ are not $R$-isomorphic.

Proof. (i) For all elements $x\in{\cal S}\backslash S$,
$S\cup\\{x\\}\in\mathrm{L}(R,\mathfrak{a},A)$: Since
$S\in\mathrm{L}(R,\mathfrak{a},A)$, we must have that $S\subseteq{\cal
S}:={\cal S}(R,\mathfrak{a},A)$, by the maximality of the element ${\cal S}$.
Let $x\in{\cal S}\backslash S$ and $T=S\cup\\{x\\}$. Then $S\subset
T\subseteq{\cal S}$, and so $\mathfrak{a}={\rm ass}_{R}(S)\subseteq{\rm
ass}_{R}(T)\subseteq{\rm ass}_{R}({\cal S})=\mathfrak{a}$, i.e. ${\rm
ass}_{R}(T)=\mathfrak{a}$ and there are unique $R$-homomorphisms

$A\simeq R\langle S^{-1}\rangle\rightarrow R\langle T^{-1}\rangle\rightarrow
R\langle{\cal S}^{-1}\rangle\simeq A.$

Hence, $T\in\mathrm{L}(R,\mathfrak{a},A)$.

(ii) For all elements $x\in R\backslash{\cal S}$, ${\cal
S}\cup\\{x\\}\not\in\mathrm{L}(R,\mathfrak{a},A)$: The statement (ii) follows
from the maximality of the element ${\cal S}$.

Now, the corollary follows from the statements (i) and (ii). $\Box$

By Proposition 3.5, for rings $A_{1},A_{2}\in\mathrm{L}{\rm oc}(R)$,

$A_{1}\rightarrow A_{2}\;\;{\rm iff}\;\;{\cal
S}_{l}(R,\mathfrak{a},A_{1})\subseteq{\cal S}_{l}(R,\mathfrak{a},A_{2}).$

Corollary 3.7 is a description of the largest element of the directed set
$(\mathrm{L}(R,\mathfrak{a},A),\subseteq)$.

###### Corollary 3.7

Let ${\cal S}\in\max\mathrm{L}(R)$, $\mathfrak{a}={\rm ass}_{R}({\cal S})$,
$A=R\langle{\cal S}^{-1}\rangle$, and $\sigma=\sigma_{\cal S}:R\rightarrow A$.
Then

  1. 1.

The set

${\cal S}(A):={\cal
S}(R,\mathfrak{a},A)=\sigma^{-1}(A^{\times})=\bigcup_{S\in\mathrm{L}(R,\mathfrak{a},A)}$

is the largest element of the directed set
$(\mathrm{L}(R,\mathfrak{a},A),\subseteq)$, and if
$S\in\mathrm{L}(R,\mathfrak{a},A)$ then $T\in\mathrm{L}(R,\mathfrak{a},A)$ for
all subsets $T$ of $R$ such that $S\subseteq T\subseteq{\cal S}$.

  2. 2.

The set ${\cal S}(A)$ is a multiplicative set in $R$ such that ${\cal
S}(A)+\mathfrak{a}={\cal S}(A)$.

  3. 3.

$A\simeq R\langle{\cal
S}(A)^{-1}\rangle\simeq\varinjlim_{S\in\mathrm{L}(R,\mathfrak{a},A)}R\langle
S^{-1}\rangle$, $R$-isomorphisms.

Proof. The corollary is a particular case of Proposition 3.5. $\Box$

By (18), for each ring $A\in\mathrm{L}{\rm oc}(R)$ there is a unique ideal
$\mathfrak{a}\in{\rm ass}\,\mathrm{L}(R)$ such that $A\in\mathrm{L}{\rm
oc}(R,\mathfrak{a})$. By Proposition 3.5.(1), for each ring
$A\in\mathrm{L}{\rm oc}(R,\mathfrak{a})$, there is the largest element ${\cal
S}(R,\mathfrak{a},A)$ of the set $\mathrm{L}(R,\mathfrak{a},A)$. By
Proposition 3.5.(1,2), the set

${\cal S}(R,\mathfrak{a},A))=\sigma^{-1}(A^{\times})$

is a multiplicative submonoid of $R$ where $\sigma:R\rightarrow A$, $r\mapsto
r+\mathfrak{a}$. Hence, the map

$\mathrm{L}{\rm oc}(R)=\coprod_{\mathfrak{a}\in{\rm
ass}\,\mathrm{L}(R)}\mathrm{L}{\rm
oc}(R,\mathfrak{a})\rightarrow\mathrm{L}(R),\;\;\mathrm{L}{\rm
oc}(R,\mathfrak{a})\ni A\mapsto{\cal S}(R,\mathfrak{a},A),$ (33)

is a section of the epimorphism of the posets (20). The set

${\cal M}(R)=\\{{\cal S}(R,\mathfrak{a},A)\,|\,\mathfrak{a}\in{\rm
ass}\,\mathrm{L}(R),A\in\mathrm{L}{\rm oc}(R,\mathfrak{a})\\}$ (34)

is the image of the section above. In particular, the map

$\mathrm{L}{\rm oc}(R)=\coprod_{\mathfrak{a}\in{\rm
ass}\,\mathrm{L}(R)}\mathrm{L}{\rm oc}(R,\mathfrak{a})\rightarrow{\cal
M}(R),\;\;\mathrm{L}{\rm oc}(R,\mathfrak{a})\ni A\mapsto{\cal
S}(R,\mathfrak{a},A),$ (35)

is an isomorphism of the posets $(\mathrm{L}{\rm oc}(R),\rightarrow)$ and
$({\cal M}(R),\subseteq)$ with inverse ${\cal S}\mapsto R\langle{\cal
S}^{-1}\rangle$.

Theorem 3.8 describes the sets $\max\mathrm{L}(R)$ and $\max\mathrm{L}{\rm
oc}(R)$, presents a bijection between them, and shows that $\max\mathrm{L}{\rm
oc}(R)\neq\emptyset$.

###### Theorem 3.8

  1. 1.

The map $\max\mathrm{L}(R)\rightarrow\max\mathrm{L}{\rm oc}(R)$, ${\cal
S}\mapsto R\langle{\cal S}^{-1}\rangle$ is a bijection with inverse
$A\mapsto\sigma_{A}^{-1}(A^{\times})$ where $\sigma_{A}:R\rightarrow A$.

  2. 2.

$\max\mathrm{L}(R)=\\{\sigma_{A}^{-1}(A^{\times})\,|\,A\in\max\mathrm{L}{\rm
oc}(R)\\}$.

  3. 3.

$\max\mathrm{L}{\rm oc}(R)=\\{R\langle{\cal S}^{-1}\rangle\,|\,{\cal
S}\in\max\mathrm{L}(R)\\}\neq\emptyset$.

Proof. 1. Statement 1 follows from the isomorphisms of posets given in (35)
and Corollary 3.7.

2 and 3. Statements 2 and 3 follow from statement 1 and Corollary 3.7
($\max\mathrm{L}{\rm oc}(R)\neq\emptyset$ since
$\max\mathrm{L}(R)\neq\emptyset$, by Theorem 3.1).(1)). $\Box$

Similarly, for each ideal $\mathfrak{a}\in{\rm ass}\mathrm{L}(R)$, Theorem 3.9
describes the sets $\max\mathrm{L}(R,\mathfrak{a})$ and $\max\mathrm{L}{\rm
oc}(R,\mathfrak{a})$, presents a bijection between them, and shows that they
are nonempty sets.

###### Theorem 3.9

Let $\mathfrak{a}\in{\rm ass}\,\mathrm{L}(R)$.

  1. 1.

The map $\max\mathrm{L}(R,\mathfrak{a})\rightarrow\max\mathrm{L}{\rm
oc}(R,\mathfrak{a})$, ${\cal S}\mapsto R\langle{\cal S}^{-1}\rangle$ is a
bijection with inverse $A\mapsto\sigma_{A}^{-1}(A^{\times})$ where
$\sigma_{A}:R\rightarrow A$.

  2. 2.

$\max\mathrm{L}(R,\mathfrak{a})=\\{\sigma_{A}^{-1}(A^{\times})\,|\,A\in\max\mathrm{L}{\rm
oc}(R,\mathfrak{a})\\}\neq\emptyset$.

  3. 3.

$\max\mathrm{L}{\rm oc}(R,\mathfrak{a})=\\{R\langle{\cal
S}^{-1}\rangle\,|\,{\cal S}\in\max\mathrm{L}(R,\mathfrak{a})\\}\neq\emptyset$.

Proof. 1. Statement 1 follows from the isomorphisms of posets given in (35)
and Corollary 3.7.

2\. (i) $\max\mathrm{L}(R,\mathfrak{a})\neq\emptyset$: The statement (i)
follows from Zorn’s Lemma. In more detail, let
$\\{S_{i}\\}_{i\in\Gamma}\subseteq\mathrm{L}(R,\mathfrak{a})$ be an ordered by
inclusion set of localizable sets in $R$ indexed by an ordinal $\Gamma$, i.e.
for each pair $i\leq j$ in $\Gamma$, $S_{i}\subseteq S_{j}$ and if
$i\in\Gamma$ is a limit ordinal then $S_{i}=\bigcup_{j<i}S_{j}$. In view of
Zorn’s Lemma, it suffices to show that
$\bigcup_{i\in\Gamma}S_{i}\in\mathrm{L}(R,\mathfrak{a})$. By (7) and (9), we
have the direct system of ring homomorphisms, $(R\langle
S_{i}^{-1}\rangle,\phi_{S_{j}S_{i}})$, where for each pair $i\leq j$, the map
$\phi_{S_{j}S_{i}}:R\langle S_{i}^{-1}\rangle\rightarrow R\langle
S_{j}^{-1}\rangle$ is defined in (7). Since the rings $R\langle
S_{i}^{-1}\rangle$ are unital and $\phi_{S_{j}S_{i}}(1)=1$ for all $i\leq j$,
the direct limit $\varinjlim R\langle S_{i}^{-1}\rangle$ is a nonzero ring and
and is isomorphic to
$R\langle\Big{(}\bigcup_{i\in\Gamma}S_{i}\Big{)}^{-1}\rangle$. Therefore,
$\bigcup_{i\in\Gamma}S_{i}\in\mathrm{L}(R)$. Finally,

${\rm ann}_{R}(\bigcup_{i\in\Gamma}S_{i})=\bigcup_{i\in\Gamma}{\rm
ann}_{R}(S_{i})=\bigcup_{i\in\Gamma}\mathfrak{a}=\mathfrak{a},$

and the statement (ii) follows.

(ii)
$\max\mathrm{L}(R,\mathfrak{a})=\\{\sigma_{A}^{-1}(A^{\times})\,|\,A\in\max\mathrm{L}{\rm
oc}(R,\mathfrak{a})\\}$: The statement (ii) follows from statement 1 and
Corollary 3.7.

3\. Statement 3 follows from statement 1 and Corollary 3.7 (by statements 1
and 2, $\max\mathrm{L}{\rm oc}(R,\mathfrak{a})\neq\emptyset$). $\Box$

The $\mathrm{L}$-radical $\mathrm{L}{\rm rad}(R)$ of the ring $R$, the sets of
localizable, non-localizable and completely localizable elements of $R$.

Definition. An element $S$ of the set $\max\mathrm{L}(R)$ is called a maximal
localizable set, and the rings $R\langle S^{-1}\rangle$ is called the maximal
localization of $R$. The set of maximal localizations of the ring $R$ is
denoted my $\max\mathrm{L}{\rm oc}(R)$. The intersection and the sum,

$\mathfrak{l}(R):=\mathrm{L}{\rm rad}(R):=\bigcap_{S\in\max\mathrm{L}(R)}{\rm
ass}_{R}(S)\;\;{\rm and}\;\;\mathfrak{c}\mathfrak{l}(R):={\rm
co-}\mathrm{L}{\rm rad}(R):=\sum_{S\in\max\mathrm{L}(R)}{\rm ass}_{R}(S),$
(36)

are called the localization radical and the localization coradical,
respectively, or the $\mathrm{L}$-radical and the $\mathrm{L}$-coradical of
$R$, for short. By Theorem 3.1.(2),

$\mathfrak{c}\mathfrak{l}(R)={\rm co-}\mathrm{L}{\rm
rad}(R)=\sum_{S\in\mathrm{L}(R)}{\rm ass}_{R}(S).$

For the ring $R$, there is the canonical exact sequence,

$0\rightarrow\mathrm{L}{\rm rad}(R)\rightarrow
R\stackrel{{\scriptstyle\sigma}}{{\rightarrow}}\prod_{S\in\max\mathrm{L}(R)}R\langle
S^{-1}\rangle,\;\;\sigma:=\prod_{S\in\max\mathrm{L}(R)}\,\sigma_{S},$ (37)

where $\sigma_{S}:R\rightarrow R\langle S^{-1}\rangle$.

Definition. The sets ${\cal
L}\mathrm{L}(R):=\bigcup_{S\in\mathrm{L}(R)}S=\bigcup_{S\in\max\mathrm{L}(R)}S$
and ${\cal N}{\cal L}\mathrm{L}(R):=R\backslash{\cal L}\mathrm{L}(R)$ are
called the set of localizable and non-localizable elements of $R$, resp., and
the intersection

${\cal C}\mathrm{L}(R):=\bigcap_{S\in\max\mathrm{L}(R)}S$

is called the set of completely localizable elements of the ring $R$. The ring
$Q_{c}(R):=R\langle{\cal C}\mathrm{L}(R)^{-1}\rangle$ is called the complete
localization or the ${\cal C}$-localization of $R$. Let
$\mathfrak{c}_{R}:={\rm ass}_{R}({\cal C}\mathrm{L}(R))$,
$R_{c}:=R/\mathfrak{c}_{R}$, $\pi_{\mathfrak{c}_{R}}:R\rightarrow R_{c}$,
$r\mapsto r+\mathfrak{c}_{R}$, and ${\cal C}_{c}:={\cal
C}\mathrm{L}(R)_{c}:=\pi_{\mathfrak{c}_{R}}({\cal C}\mathrm{L}(R))$.

By Corollary 2.2.(2), $R_{c}\subseteq Q_{c}$, ${\cal
C}_{c}\in\mathrm{L}(R_{c})$, and

$Q_{c}(R)\simeq R_{c}\langle{\cal C}_{c}^{-1}\rangle.$ (38)

By the definition, the sets ${\cal L}\mathrm{L}(R)$, ${\cal N}{\cal
L}\mathrm{L}(R)$, and ${\cal C}\mathrm{L}(R)$ are invariant under the action
of the automorphism group of the ring $R$, i.e., they are characteristic sets.

Proposition 3.10 shows that there are tight connections between the sets
${\cal L}\mathrm{L}(R)$, ${\cal N}{\cal L}\mathrm{L}(R)$, and $\mathrm{L}{\rm
rad}(R)$.

###### Proposition 3.10

  1. 1.

${\cal L}\mathrm{L}(R)\cap\mathrm{L}{\rm rad}(R)=\emptyset$.

  2. 2.

$\mathrm{L}{\rm rad}(R)\subseteq{\cal N}{\cal L}\mathrm{L}(R)$.

  3. 3.

${\cal L}\mathrm{L}(R)+\mathrm{L}{\rm rad}(R)={\cal L}\mathrm{L}(R)$.

  4. 4.

${\cal N}{\cal L}\mathrm{L}(R)+\mathrm{L}{\rm rad}(R)={\cal N}{\cal
L}\mathrm{L}(R)$.

Proof. 1. For each element $s\in{\cal L}\mathrm{L}(R)$, there is a maximal
localizable set, say $S\in\max\mathrm{L}(R)$, such that $s\in S$. Then
$s\not\in{\rm ass}_{R}(S)$ since the element $s$ is a unit in the ring
$R\langle S^{-1}\rangle$. Hence, $s\not\in\mathrm{L}{\rm rad}(R)$, and so
${\cal L}\mathrm{L}(R)\cap\mathrm{L}{\rm rad}(R)=\emptyset$.

2\. Statement 2 follows from statement 1: $\mathrm{L}{\rm rad}(R)\subseteq
R\backslash{\cal L}\mathrm{L}(R)={\cal N}{\cal L}\mathrm{L}(R)$.

3\. Since $0\in\mathrm{L}{\rm rad}(R)$, to prove statement 3 it suffices to
show that ${\cal L}\mathrm{L}(R)+\mathrm{L}{\rm rad}(R)\subseteq{\cal
L}\mathrm{L}(R)$. For each element $s\in{\cal L}\mathrm{L}(R)$, there is a
maximal localizable set, say $S\in\max\mathrm{L}(R)$, that contains the
element $s$. Then the image of the elements $s+\mathrm{L}{\rm rad}(R)$ in the
ring $R\langle S^{-1}\rangle$ are units (since $\mathrm{L}{\rm
rad}(R)\subseteq{\rm ass}_{R}(S)$). Therefore, $s+\mathrm{L}{\rm
rad}(R)\subseteq{\cal L}\mathrm{L}(R)$, and statement 3 follows.

4\. Recall that $R={\cal L}\mathrm{L}(R)\coprod{\cal N}{\cal L}\mathrm{L}(R)$,
a disjoint union. Suppose that statement 4 does not hold, i.e. there are
elements $n\in{\cal N}{\cal L}\mathrm{L}(R)$ and $r\in\mathrm{L}{\rm rad}(R)$
such $n+r\not\in{\cal N}{\cal L}\mathrm{L}(R)$. Then $n+r\in{\cal
L}\mathrm{L}(R)$, and so

${\cal L}\mathrm{L}(R)\not\ni n\in-r+{\cal
L}\mathrm{L}(R)\subseteq\mathrm{L}{\rm rad}(R)+{\cal
L}\mathrm{L}(R)\stackrel{{\scriptstyle{\rm st.3}}}{{=}}{\cal L}\mathrm{L}(R),$

a contradiction. $\Box$

Corollary 3.11 provides a useful sufficient condition for the localization
radical to be zero. Corollary 3.11 is used in the description of the
localization radical of an arbitrary direct product of division rings (Theorem
5.11.(3)).

###### Corollary 3.11

If ${\cal L}\mathrm{L}(R)=R\backslash\\{0\\}$ then $\mathrm{L}{\rm rad}(R)=0$.

Proof. By Proposition 3.10.(2), $\mathrm{L}{\rm rad}(R)\subseteq{\cal N}{\cal
L}\mathrm{L}(R)=R\backslash{\cal L}\mathrm{L}(R)=\\{0\\}$ since ${\cal
L}\mathrm{L}(R)=R\backslash\\{0\\}$. Hence $\mathrm{L}{\rm rad}(R)=0$. $\Box$

Lemma 3.12 is a criterion for an element of $R$ to be a localizable element.

###### Lemma 3.12

Let $r\in R$. The following statements are equivalent:

  1. 1.

$r\in{\cal L}\mathrm{L}(R)$.

  2. 2.

$\\{r\\}\in\mathrm{L}(R)$.

  3. 3.

${\rm ass}_{R}(r)\neq R$.

  4. 4.

There is an $R$-homomorphism of rings $f:R\rightarrow A$ such that $f(r)\in
A^{\times}$.

Proof. $(1\Rightarrow 2)$ If $r\in S$ for some $S\in\mathrm{L}(R)$, then
$\\{r\\}\in\mathrm{L}(R)$, by Proposition 2.8.(2).

$(2\Rightarrow 1)$ The implication follows from the inclusions $r\in\\{r\\}$
and $\\{r\\}\in\mathrm{L}(R)$.

$(2\Leftrightarrow 3\Leftrightarrow 4)$ Proposition 2.4. $\Box$

###### Corollary 3.13

  1. 1.

Every localizable set consists of localizable elements.

  2. 2.

Every subset of a ring that consists of localizable elements is a localizable
set iff there is the largest localizable set.

Proof. 1. Statement 1 follows from Proposition 2.8.(2) or Lemma 3.12.(1,2).

2\. If ${\cal S}$ is the largest localizable set of a ring $R$ then ${\cal
L}\mathrm{L}(R)={\cal S}$, and so every subset of $R$ that consists of
localizable elements is a subset of ${\cal S}$, hence localizable, by
Proposition 2.8.(2).

Suppose that a ring $R$ has two maximal localizable sets, ${\cal S}$ and
${\cal T}$. Then their union consists of localizable elements but it is not a
localizable set otherwise, by Theorem 3.1.(2), it would have contained in a
maximal localizable set, which is not possible (since ${\cal S}$ and ${\cal
T}$ are maximal localizable sets). $\Box$

The set ${\cal C}\mathrm{L}(R)$ of completely localizable elements of $R$. Let
$\mathfrak{c}_{R}:={\rm ass}_{R}({\cal C}\mathrm{L}(R))$. Consider the maps

$0\rightarrow\mathrm{L}{\rm rad}(R)\rightarrow
R\stackrel{{\scriptstyle\pi_{\mathfrak{c}_{R}}}}{{\rightarrow}}R/\mathfrak{c}_{R}\stackrel{{\scriptstyle\overline{\kappa}}}{{\rightarrow}}Q_{c}(R)\rightarrow\prod_{S\in\max\mathrm{L}(R)}R\langle
S^{-1}\rangle\;\;{\rm
and}\;\;\kappa:=\overline{\kappa}\pi_{\mathfrak{c}_{R}}:R\rightarrow Q_{c}(R)$
(39)

is the unique $R$-homomorphism from $R$ to its localization $Q_{c}(R)$,
$\pi_{\mathfrak{c}_{R}}(r):=r+\mathfrak{c}_{R}$, and $\overline{\kappa}$ is
the unique $R$-homomorphism from $R/\mathfrak{c}_{R}$ to $Q_{c}(R)$ (induced
by $\kappa$).

Proposition 3.14 contains some properties of ${\cal C}\mathrm{L}(R)$ and
$\mathfrak{c}_{R}$.

###### Proposition 3.14

Let ${\cal C}:={\cal C}\mathrm{L}(R)$.

  1. 1.

${\cal C}=\sigma^{-1}\Big{(}\prod_{S\in\max\mathrm{L}(R)}R\langle
S^{-1}\rangle^{\times}\Big{)}$ where
$\sigma:R\rightarrow\prod_{S\in\max\mathrm{L}(R)}R\langle S^{-1}\rangle$.

  2. 2.

${\cal C}\mathrm{L}(R)+\mathrm{L}{\rm rad}(R)={\cal C}\mathrm{L}(R)$.

  3. 3.

$\mathfrak{c}_{R}\subseteq\mathrm{L}{\rm rad}(R)$.

  4. 4.

If the map $\overline{\kappa}:R/\mathfrak{c}_{R}\rightarrow Q_{c}(R)$ is an
injection then $\mathfrak{c}_{R}=\mathrm{L}{\rm rad}(R)$.

  5. 5.

${\cal C}+\mathfrak{c}_{R}={\cal C}$.

Proof. 1. Statement 1 follows from the definition of the set ${\cal C}$ and
Corollary 3.7.(1):

${\cal
C}=\bigcap_{S\in\max\mathrm{L}(R)}S=\sigma^{-1}\bigg{(}\prod_{S\in\max\mathrm{L}(R)}R\langle
S^{-1}\rangle^{\times}\bigg{)}.$

2\. By Corollary 3.7.(3), $S+{\rm ass}_{R}(S)=S$ for all
$S\in\max\mathrm{L}(R)$. Now, $S+\mathrm{L}{\rm rad}(R)=S$ for all
$S\in\max\mathrm{L}(R)$ since $\mathrm{L}{\rm
rad}(R)=\bigcap_{S\in\max\mathrm{L}(R)}{\rm ass}_{R}(S)$. It follows that
${\cal C}\mathrm{L}(R)+\mathrm{L}{\rm rad}(R)={\cal C}\mathrm{L}(R)$ since
${\cal C}\mathrm{L}(R)=\bigcap_{S\in\max\mathrm{L}(R)}S$.

3\. The inclusion $\mathfrak{c}_{R}\subseteq\mathrm{L}{\rm rad}(R)$ follows
from (39).

4\. If the map $\overline{\kappa}$ is an injection then
$\mathfrak{c}_{R}\supseteq\mathrm{L}{\rm rad}(R)$ (by (39)), and statement 4
follows from statement 3.

5\. Statement 5 follows from statements 2 and 3. $\Box$

The $\mathfrak{a}$-absolute quotient ring $Q_{a}(R,\mathfrak{a}))$ of a ring
$R$ where $\mathfrak{a}\in{\rm ass}\,\mathrm{L}(R)$. Let $\mathfrak{a}\in{\rm
ass}\,\mathrm{L}(R)$ and

$U(R,\mathfrak{a}):=\bigcup_{S\in\max\mathrm{L}(R,\mathfrak{a})}S.$ (40)

The set $(\mathrm{L}(R,\mathfrak{a}),\subseteq)$ is a poset and the system
$\\{R\langle S^{-1}\rangle,\phi_{ST}\\}$ is a direct system of rings where
$S,T\in\mathrm{L}(R,\mathfrak{a})$.

Definition. The direct limit
$Q_{a}(R,\mathfrak{a}):=\varinjlim_{S\in\mathrm{L}(R,\mathfrak{a})}R\langle
S^{-1}\rangle$ is called the $\mathfrak{a}$-absolute quotient ring of the ring
$R$. Let $\sigma_{a,\mathfrak{a}}:R\rightarrow Q_{a}(R,\mathfrak{a})$ be a
unique $R$-homomorphism and $\mathfrak{a}_{a,\mathfrak{a}}(R):={\rm
ker}(\sigma_{a,\mathfrak{a}})$,
$R_{a,\mathfrak{a}}:=R/\mathfrak{a}_{a,\mathfrak{a}}(R)$,
$\pi_{a,\mathfrak{a}}:R\rightarrow R_{a,\mathfrak{a}}$, $r\mapsto
r+\mathfrak{a}_{a,\mathfrak{a}}$, and
$\overline{\sigma_{a,\mathfrak{a}}}:R_{a,\mathfrak{a}}\rightarrow
Q_{a}(R,\mathfrak{a})$ is a unique $R$-homomorphism. Clearly,
$\sigma_{a,\mathfrak{a}}=\overline{\sigma_{a,\mathfrak{a}}}\pi_{a,\mathfrak{a}}$
and $\mathfrak{a}_{a,\mathfrak{a}}(R)\supseteq\mathfrak{a}$.

Theorem 3.15 gives an explicit description of the ring
$Q_{a}(R,\mathfrak{a})$, a criterion for $Q_{a}(R,\mathfrak{a})\neq\\{0\\}$,
and a criterion for $\mathfrak{a}_{a,\mathfrak{a}}=\mathfrak{a}$.

###### Theorem 3.15

Let $\mathfrak{a}\in{\rm ass}\,\mathrm{L}(R)$ and $U=U(R,\mathfrak{a})$.

  1. 1.

$Q_{a}(R,\mathfrak{a})\neq\\{0\\}$ iff $U\in\mathrm{L}(R)$.

  2. 2.

Suppose that $U\in\mathrm{L}(R,\mathfrak{b})$. Then

    1. (a)

The rings $Q_{a}(R,\mathfrak{a})$ and $R\langle U^{-1}\rangle$ are
$R$-isomorphic and $\mathfrak{a}_{a,\mathfrak{a}}=\mathfrak{b}$.

    2. (b)

$\mathfrak{a}_{a,\mathfrak{a}}=\mathfrak{a}$ iff
$\max\mathrm{L}(R,\mathfrak{a})=\\{S\\}$ iff the rings $Q_{a}(R,\mathfrak{a})$
and $R\langle S^{-1}\rangle$ are $R$-isomorphic for some
$S\in\max\mathrm{L}(R,\mathfrak{a})$.

Proof. 1. $(\Rightarrow)$ Suppose that $Q_{a}(R,\mathfrak{a})\neq\\{0\\}$.
Then the set ${\cal
S}:=\sigma_{a,\mathfrak{a}}^{-1}(Q_{a}(R,\mathfrak{a})^{\times})$ is a
localizable set in $R$ since $\sigma_{a,\mathfrak{a}}({\cal S})\subseteq
Q_{a}(R,\mathfrak{a})^{\times}$. Then for each
$S\in\max\mathrm{L}(R,\mathfrak{a})$, there is a commutative diagram

$\begin{array}[c]{ccc}R&\stackrel{{\scriptstyle\sigma_{S}}}{{\rightarrow}}&R\langle
S^{-1}\rangle\\\
&\stackrel{{\scriptstyle\sigma_{a,\mathfrak{a}}}}{{\searrow}}&{\downarrow}^{\phi_{S}}\\\
&&Q_{a}(R,\mathfrak{a})\end{array}$

where $\phi_{S}$ is a unique $R$-homomorphism. Then it follows from the
inclusion $\phi_{S}(R\langle S^{-1}\rangle^{\times})\subseteq
Q_{a}(R,\mathfrak{a})^{\times}$ that

${\cal
S}=\sigma_{a,\mathfrak{a}}^{-1}(Q_{a}(R,\mathfrak{a})^{\times})=(\phi_{S}\sigma_{S})^{-1}(Q_{a}(R,\mathfrak{a})^{\times})=\sigma_{S}^{-1}\phi_{S}^{-1}(Q_{a}(R,\mathfrak{a})^{\times})\supseteq\sigma_{S}^{-1}(R\langle
S^{-1}\rangle^{\times})=S,$

by Corollary 3.7.(1). Therefore, ${\cal
S}\supseteq\bigcup_{S\in\max\mathrm{L}(R,\mathfrak{a})}S=U$, and so
$U\in\mathrm{L}(R)$ since ${\cal S}\in\mathrm{L}(R)$.

$(\Leftarrow)$ Suppose that $U\in\mathrm{L}(R)$. Let us show that the rings
$Q_{a}(R,\mathfrak{a})$ and $R\langle U^{-1}\rangle$ are $R$-isomorphic. By
the definition, the ring $Q_{a}(R,\mathfrak{a})$ is the direct limit of the
direct system ${\cal D}:=\\{R\langle S^{-1}\rangle,\phi_{ST}\\}$ where
$S,T\in\mathrm{L}(R,\mathfrak{a})$. Since for all
$S\in\mathrm{L}(R,\mathfrak{a})$, $S\subseteq U$, we can extend this direct
system to a direct system ${\cal D}^{\prime}$ by adding the $R$-homomorphisms
$R\langle S^{-1}\rangle\rightarrow R\langle U^{-1}\rangle$ for all
$S\in\mathrm{L}(R,\mathfrak{a})$. The element $U$ is the largest element of
the poset $\mathrm{L}^{\prime}:=\mathrm{L}(R,\mathfrak{a})\cup\\{U\\}$. By
Theorem 2.14.(3),

$\varinjlim_{S\in\mathrm{L}^{\prime}}R\langle S^{-1}\rangle\simeq R\langle
U^{-1}\rangle,$

an $R$-isomorphism. By the definitions of the two direct systems, ${\cal D}$
and ${\cal D}^{\prime}$, there is necessarily a unique $R$-homomorphism
$\alpha:Q_{a}(R,\mathfrak{a})\rightarrow R\langle U^{-1}\rangle$. Since

$\sigma_{a,\mathfrak{a}}(U)\subseteq Q_{a}(R,\mathfrak{a})^{\times},$

where $\sigma_{a,\mathfrak{a}}:R\rightarrow Q_{a}(R,\mathfrak{a})$, there is
necessarily a unique $R$-homomorphism

$\beta:R\langle U^{-1}\rangle\rightarrow Q_{a}(R,\mathfrak{a}).$

Hence, the rings $Q_{a}(R,\mathfrak{a})$ and $R\langle U^{-1}\rangle$ are
$R$-isomorphic, by the universal properties of localization and the direct
limit. In particular, $\mathfrak{a}_{a,\mathfrak{a}}=\mathfrak{b}$. This
finishes the proof of statements 1 and 2(a).

2(b). (i)
$\mathfrak{a}_{a,\mathfrak{a}}=\mathfrak{a}\Rightarrow\max\mathrm{L}(R,\mathfrak{a})=\\{S\\}$:
Suppose that the set $\max\mathrm{L}(R,\mathfrak{a})$ contains at least two
elements, say $S$ and $T$. Then

$S\subset S\cup T\subseteq U\;\;{\rm
and}\;\;S,U\in\mathrm{L}(R,\mathfrak{a}).$

This contradicts to the fact that $S\in\max\mathrm{L}(R,\mathfrak{a})$, and
the statement (i) follows.

(ii) $\max\mathrm{L}(R,\mathfrak{a})=\\{S\\}\Rightarrow$ the rings
$Q_{a}(R,\mathfrak{a})$ and $R\langle S^{-1}\rangle$ are $R$-isomorphic: The
implication follows from Theorem 2.14.(3).

(iii) The rings $Q_{a}(R,\mathfrak{a})$ and $R\langle S^{-1}\rangle$ are
$R$-isomorphic for some $S\in\max\mathrm{L}(R,\mathfrak{a})$ then
$\mathfrak{a}_{a,\mathfrak{a}}=\mathfrak{a}$: The implication is obvious.
$\Box$

Example. If $R$ is a commutative domain then the set ${\cal
C}_{R}=R\backslash\\{0\\}$ is the largest element of $\mathrm{L}(R,0)$. By
Theorem 3.15, $Q_{a}(R,0)=Q_{cl}(R)$ is the classical quotient ring of $R$,
i.e. the field of fractions of $R$, and $\mathfrak{a}_{a,0}(R)=0$.

###### Lemma 3.16

If $R$ is a simple Artinian rings then $\max\mathrm{L}(R,0)=\\{R^{\times}\\}$,
$Q_{a}(R,0)=R$ and $\mathfrak{a}_{a,0}(R)=0$.

Proof. Lemma follows from Lemma 2.19. $\Box$

Isomorphism criterion for localizations of a ring. Theorem 3.17.(1) is an
$R$-isomorphism criterion for localizations of the ring $R$ and Theorem
3.17.(2) is a criterion for existing of an $R$-homomorphism between
localizations of $R$.

###### Theorem 3.17

Let $S,T\in\mathrm{L}(R)$.

  1. 1.

The rings $R\langle S^{-1}\rangle$ and $R\langle T^{-1}\rangle$ are
$R$-isomorphic iff

$\sigma_{S}^{-1}(R\langle S^{-1}\rangle^{\times})=\sigma_{T}^{-1}(R\langle
T^{-1}\rangle^{\times})$

where $\sigma_{S}:R\rightarrow R\langle S^{-1}\rangle$ and
$\sigma_{T}:R\rightarrow R\langle T^{-1}\rangle$.

  2. 2.

There is an $R$-homomorphism $R\langle S^{-1}\rangle\rightarrow R\langle
T^{-1}\rangle$ iff $\sigma_{S}^{-1}(R\langle
S^{-1}\rangle^{\times})\subseteq\sigma_{T}^{-1}(R\langle
T^{-1}\rangle^{\times}).$

Proof. 1. $(\Rightarrow)$ The implication is obvious.

$(\Leftarrow)$ The implication follows from the fact that the rings $R\langle
S^{-1}\rangle$ and $R\langle\sigma_{S}^{-1}(R\langle
S^{-1}\rangle^{\times})\rangle$ are isomorphic for all $S\in\mathrm{L}(R)$, by
Proposition 3.5.(1).

2\. $(\Rightarrow)$ The implication is obvious.

$(\Leftarrow)$ The implication follows from the fact that the rings $R\langle
S^{-1}\rangle$ and $R\langle\sigma_{S}^{-1}(R\langle
S^{-1}\rangle^{\times})\rangle$ are isomorphic for all $S\in\mathrm{L}(R)$
(Proposition 3.5.(1)) and the fact that if $S_{1},S_{2}\in\mathrm{L}(R)$ and
$S_{1}\subseteq S_{2}$ then there is a unique $R$-homomorphism $R\langle
S_{1}^{-1}\rangle\rightarrow R\langle S_{2}^{-1}\rangle$. $\Box$

Proposition 3.18 is a practical criterion for a subset $S$ of a ring $R$ to be
an element of the set $\mathrm{L}(R,\mathfrak{a})$.

###### Proposition 3.18

Let $S$ be a subset of a ring $R$.

  1. 1.

Suppose that there exists an ideal $\mathfrak{a}$ of $R$ such that
$\mathfrak{a}\subseteq{\rm ass}_{R}(S)$ and
$\overline{S}:=\pi(S)\in\mathrm{L}(\overline{R},0)$ where
$\pi:R\rightarrow\overline{R}:=R/\mathfrak{a}$, $r\mapsto r+\mathfrak{a}$.
Then $S\in\mathrm{L}(R,\mathfrak{a})$.

  2. 2.

$S\in\mathrm{L}(R,\mathfrak{b})$ iff there is an ideal $\mathfrak{a}$ of $R$
such that $\mathfrak{a}\subseteq\mathfrak{b}$ and
$\overline{S}:=\pi(S)\in\mathrm{L}(\overline{R},0)$ where
$\pi:R\rightarrow\overline{R}:=R/\mathfrak{a}$, $r\mapsto r+\mathfrak{a}$.

Proof. 1. Since the elements of the set $\overline{S}$ are invertible in the
ring $\overline{R}\langle\overline{S}^{-1}\rangle$, ${\rm
ass}_{R}(S)\subseteq\mathfrak{a}$, by Proposition 2.7. By the assumption,
${\rm ass}_{R}(S)\supseteq\mathfrak{a}$. Therefore, ${\rm
ass}_{R}(S)=\mathfrak{a}$ and $S\in\mathrm{L}(R,\mathfrak{a})$.

2\. $(\Rightarrow)$ If $S\in\mathrm{L}(R,\mathfrak{b})$ then it suffices to
take $\mathfrak{a}=\mathfrak{b}$, by Corollary 2.2.(2).

$(\Leftarrow)$ This implication follows from statement 1. $\Box$

###### Proposition 3.19

Suppose that $S\in\mathrm{L}(R,\mathfrak{a})$, $T\in\mathrm{L}(R\langle
S^{-1}\rangle,0)$ and $\overline{S}\subseteq T$ where
$\overline{S}=\sigma_{S}(S)$ and $\sigma_{S}:R\rightarrow R\langle
S^{-1}\rangle$. Then

  1. 1.

$\overline{T}:=T\cap\overline{R}\in\mathrm{L}(\overline{R},0)$,
$\overline{T}\in\mathrm{L}(R\langle S^{-1}\rangle,0)$,

$\overline{R}\langle\overline{S}^{-1}\rangle\subseteq\overline{R}\langle\overline{T}^{-1}\rangle=R\langle
S^{-1}\rangle\langle\overline{T}^{-1}\rangle\subseteq R\langle
S^{-1}\rangle\langle T^{-1}\rangle,$

and
$T^{\prime}:=\sigma_{S}^{-1}(T)\in\mathrm{L}(R,\mathfrak{a},\overline{R}\langle\overline{T}^{-1}\rangle)$.

  2. 2.

Let $\sigma:R\rightarrow A:=R\langle S^{-1}\rangle\langle T^{-1}\rangle$ and
${\cal T}:=\sigma^{-1}(A^{\times})$. Then ${\cal
T}\in\mathrm{L}(R,\mathfrak{a},A)$.

Proof. 1. Recall that the rings $R\langle S^{-1}\rangle$ and
$\overline{R}\langle\overline{S}^{-1}\rangle$ are $R$-isomorphic.

(i) $\overline{T}\in\mathrm{L}(\overline{R},0)$ and
$\overline{T}\in\mathrm{L}(R\langle S^{-1}\rangle,0)$: The chain of inclusions
below follows from the inclusions
$\overline{R}\subseteq\overline{R}\langle\overline{S}^{-1}\rangle$ and
$\overline{T}\subseteq T$,

${\rm ass}_{\overline{R}}(\overline{T})\subseteq{\rm
ass}_{\overline{R}\langle\overline{S}^{-1}\rangle}(\overline{T})={\rm
ass}_{R\langle S^{-1}\rangle}(\overline{T})\subseteq{\rm ass}_{R\langle
S^{-1}\rangle}(T)=0,$

and the statement (i) follows.

(ii) $\overline{S}\subseteq\overline{T}$ and
$\overline{R}\langle\overline{S}^{-1}\rangle\subseteq\overline{R}\langle\overline{T}^{-1}\rangle=R\langle
S^{-1}\rangle\langle\overline{T}^{-1}\rangle\subseteq R\langle
S^{-1}\rangle\langle T^{-1}\rangle$: Since $\overline{S}\subseteq T$ and
$\overline{S}\subseteq\overline{R}$, we have that $\overline{S}\subseteq
T\cap\overline{R}=\overline{T}$. Hence

$\overline{R}\langle\overline{S}^{-1}\rangle\subseteq\overline{R}\langle\overline{T}^{-1}\rangle$

since $\overline{T}\in\mathrm{L}(R\langle
S^{-1}\rangle,0)=\mathrm{L}(\overline{R}\langle\overline{S}^{-1}\rangle,0)$
(the statement (i)). Now,

$\overline{R}\langle\overline{T}^{-1}\rangle=\overline{R}\langle(\overline{S}\cup\overline{T})^{-1}\rangle=\overline{R}\langle\overline{S}^{-1}\rangle\langle\overline{T}^{-1}\rangle=R\langle
S^{-1}\rangle\langle\overline{T}^{-1}\rangle\subseteq R\langle
S^{-1}\rangle\langle T^{-1}\rangle$

since $\overline{S}\subseteq\overline{T}\subseteq T$ and
$\overline{T},T\in\mathrm{L}(R\langle S^{-1}\rangle,0)$.

Recall that $\pi_{S}:R\rightarrow\overline{R}$, $r\mapsto r+\mathfrak{a}$.

(iii) $T^{\prime}=\pi_{S}^{-1}(\overline{T})\supseteq S$ and
$\pi_{S}(T^{\prime})=\overline{T}$: The map $\sigma_{S}:R\rightarrow R\langle
S^{-1}\rangle$ is the composition of two maps

$\sigma_{S}:R\stackrel{{\scriptstyle\pi_{S}}}{{\rightarrow}}\overline{R}\rightarrow
R\langle S^{-1}\rangle$

where the second map is a natural inclusion. Hence,
$T^{\prime}=\pi_{S}^{-1}(\overline{T})$. Clearly,
$\pi_{S}^{-1}(\overline{T})\supseteq S$ since $\overline{S}\subseteq T$ and
$\pi_{S}(S)=\overline{S}$. The map $\pi_{S}:R\rightarrow\overline{R}$ is an
epimorphism, and so
$\pi_{S}(T^{\prime})=\pi_{S}(\pi_{S}^{-1}(\overline{T}))=\overline{T}$.

(iv)
$T^{\prime}\in\mathrm{L}(R,\mathfrak{a},\overline{R}\langle\overline{T}^{-1}\rangle)$:
It follows from the inclusion $S\subseteq T^{\prime}$ that

$\mathfrak{a}={\rm ass}_{R}(S)\subseteq{\rm ass}_{R}(T^{\prime}).$

By the statements (i) and (iii),
$\pi_{S}(T^{\prime})=\overline{T}\in\mathrm{L}(\overline{R},0)$. Therefore,
${\rm ass}_{R}(T^{\prime})=\mathfrak{a}$, by Proposition 3.18.(1). Now,

$R\langle T^{\prime}{}^{-1}\rangle=R\langle(S\cup
T)^{\prime}{}^{-1}\rangle=R\langle
S^{-1}\rangle\langle\pi_{S}(T^{\prime})^{-1}\rangle=R\langle
S^{-1}\rangle\langle\overline{T}^{-1}\rangle=\overline{R}\langle\overline{T}^{-1}\rangle,$

by the statement (ii), and the statement (iv) follows.

2\. Since $T\in\mathrm{L}(R\langle S^{-1}\rangle,0)$, ${\rm
ker}(\sigma)=\mathfrak{a}$. Now, statement 2 follows from Proposition 3.5.(1).
$\Box$

Localizable sets and denominator sets. Theorem 3.20 is a criterion for the
ring $R\langle S^{-1}\rangle$ to be an $R$-isomorphic to a localization of $R$
at a denominator set and it also describes all such denominator sets.

###### Theorem 3.20

Let $S\in\mathrm{L}(R,\mathfrak{a})$, $\sigma_{S}:R\rightarrow R\langle
S^{-1}\rangle$, and ${\cal S}:=\sigma_{S}^{-1}(R\langle
S^{-1}\rangle^{\times})$. Then

  1. 1.

The ring $R\langle S^{-1}\rangle$ is $R$-isomorphic to a localization of the
ring $R$ at a *-denominator set where $*\in\\{l,r,\emptyset\\}$ iff ${\cal
S}\in{\rm Den}_{*}(R,\mathfrak{a})$.

  2. 2.

The set of all *-denominator sets of $R$ with localization $R$-isomorphic to
the ring $R\langle S^{-1}\rangle$ is equal to $\\{T\in{\rm
Den}_{*}(R,\mathfrak{a})\,|\,T\subseteq{\cal S},{\cal
S}+\mathfrak{a}\subseteq(T^{-1}R)^{\times}\\}$.

Proof. 1. Let us prove statement 1 for left denominator sets. The other two
case can be proven in a similar way (or reduced to this case using an opposite
ring construction).

$(\Rightarrow)$ Suppose that the ring $R\langle S^{-1}\rangle$ is
$R$-isomorphic to a localization of the ring $R$ at a left denominator set $T$
of $R$. Then $T\in{\rm Den}_{l}(R,\mathfrak{a})$. By [3, Proposition 3.1.(1)],
${\cal S}\in{\rm Den}_{l}(R,\mathfrak{a})$ and the rings ${\cal S}^{-1}R$ and
$T^{-1}R$ are $R$-isomorphic.

$(\Leftarrow)$ Suppose that ${\cal S}\in{\rm Den}_{l}(R,\mathfrak{a})$. Then
the rings ${\cal S}^{-1}R$ and $R\langle{\cal S}^{-1}\rangle$ are
$R$-isomorphic. By Proposition 3.5.(1), the rings $R\langle S^{-1}\rangle$ and
$R\langle{\cal S}^{-1}\rangle$ are $R$-isomorphic, and the implication
follows.

2\. Statement follows from Corollary 2.3. $\Box$

For a denominators set $T\in{\rm Den}_{*}(R,\mathfrak{a})$, Proposition 3.21
describes all localizable sets $S$ of the ring $R$ such that the ring
$R\langle S^{-1}\rangle$ is an $R$-isomorphic to the localization of $R$ at
$T$.

###### Proposition 3.21

Let $T\in{\rm Den}_{*}(R,\mathfrak{a})$ where $*\in\\{l,r,\emptyset\\}$, $A$
be the localization of $R$ at $T$, and
$\pi_{T}:R\rightarrow\overline{R}=R/\mathfrak{a}$, $r\mapsto r+\mathfrak{a}$.
Then the set of all localizable sets $S$ of $R$ such that the rings $R\langle
S^{-1}\rangle$ and $A$ are $R$-isomorphic is equal to
$\\{S\in\mathrm{L}(R,\mathfrak{a})\,|\,\pi_{S}(S)\subseteq
A^{\times},\pi_{S}(T)\subseteq R\langle S^{-1}\rangle^{\times}\\}$.

Proof. The statement follows from Corollary 2.3. $\Box$

For a localizable multiplicative set $S$ of the ring $R$, Proposition 3.22 is
a criterion for the set $S$ to be a left denominator set of $R$.

###### Proposition 3.22

A localizable multiplicative set $S\in\mathrm{L}(R,\mathfrak{a})$ is a left
denominator set of $R$ iff

  1. 1.

$\mathfrak{a}=\sum_{s\in S}{\rm ker}_{R}(s\cdot)\supseteq\sum_{s\in S}{\rm
ker}_{R}(\cdot s)$, and

  2. 2.

the left $R$-module $R\langle S^{-1}\rangle/\overline{R}$ is $S$-torsion where
$\overline{R}=R\mathfrak{a}$.

Proof. $(\Rightarrow)$ The implication is obvious.

$(\Leftarrow)$ Recall that ${\rm ass}_{l}(S)=\sum_{s\in S}{\rm
ker}_{R}(s\cdot)$. By statement 1,

$\mathfrak{a}={\rm ass}_{l}(S).$

Let $a\in R\langle S^{-1}\rangle$. By statement 2,
$sa=\overline{r}\in\overline{R}$ for some elements $s\in S$ and
$r\in\overline{R}$, and so the element $a=s^{-1}\overline{r}$ is a left
fraction. Therefore, the set $S$ is a left Ore set since $\mathfrak{a}={\rm
ass}_{l}(S)$. In fact, $S\in{\rm Den}_{l}(R,\mathfrak{a})$ since ${\rm
ass}_{l}(S)\supseteq\sum_{s\in S}{\rm ker}_{R}(\cdot s)$, see statement 1.
$\Box$

Corollary 3.23 (resp., Corollary 3.24) is a criterion for the set $S$ to be a
right (resp., left and right) denominator set of $R$.

###### Corollary 3.23

A localizable multiplicative set $S\in\mathrm{L}(R,\mathfrak{a})$ is a right
denominator set of $R$ iff

  1. 1.

$\mathfrak{a}=\sum_{s\in S}{\rm ker}_{R}(\cdot s)\supseteq\sum_{s\in S}{\rm
ker}_{R}(s\cdot)$, and

  2. 2.

the right $R$-module $R\langle S^{-1}\rangle/\overline{R}$ is $S$-torsion
where $\overline{R}=R\mathfrak{a}$.

Proof. Repeat the proof of Proposition 3.22 replacing ‘left’ to ‘right’
everywhere. $\Box$

###### Corollary 3.24

A localizable multiplicative set $S\in\mathrm{L}(R,\mathfrak{a})$ is a
denominator set of $R$ iff

  1. 1.

$\mathfrak{a}=\sum_{s\in S}{\rm ker}_{R}(\cdot s)=\sum_{s\in S}{\rm
ker}_{R}(s\cdot)$, and

  2. 2.

$R\langle S^{-1}\rangle/\overline{R}$ is an $S$-torsion left and right
$R$-module where $\overline{R}=R\mathfrak{a}$.

Proof. The corollary follows from Proposition 3.21 and Corollary 3.23. $\Box$

##  4 Localizations of commutative rings

In this section, let $R$ be a commutative ring and
$\mathfrak{n}=\mathfrak{n}_{R}$ be its prime radical.

For an arbitrary commutative ring $R$, descriptions of the following sets are
obtained: the sets of localizable and non-localizable elements (Lemma 4.1),
the maximal localizable sets, maximal localization rings, the set of
completely localizable elements, and the ideal $\mathfrak{c}_{R}$ (Theorem
4.2). Theorem 4.2 also describes relations between the ideals $\mathrm{L}{\rm
rad}(R)$, $\mathfrak{n}_{R}$, and $\mathfrak{c}_{R}$.

For a commutative ring $R$ such that $|\min(R)|<\infty$ and $\mathfrak{n}_{R}$
is a nilpotent ideal (eg, $R$ is a commutative Noetherian ring), Proposition
4.3 describes ${\rm Spec}(Q_{c}(R))$ and the rings $Q_{c}(R)$ and
$Q_{c}(R)/\mathfrak{n}_{Q_{c}(R)}$.

For a commutative ring $R$ such that $|\min(R)|<\infty$ and
$\mathfrak{n}_{R}=0$ (eg, $R$ is a commutative semiprime Noetherian ring),
Proposition 4.3 shows that ${\cal C}\mathrm{L}(R)={\cal C}_{R}$,
$Q_{c}(R)=Q_{cl}(R)$, and $\mathrm{L}{\rm
rad}(R)=\mathfrak{c}_{R}=\mathfrak{n}_{Q_{c}(R)}=0$. A characterization of
commutative semiprime Goldie rings is given (Corollary 4.5).

Notice that the prime radical $\mathfrak{n}_{R}$ of a commutative ring $R$
contains precisely all the nilpotent elements of the ring $R$. Lemma 4.1
describes the sets of localizable and non-localizable elements of an arbitrary
commutative ring.

###### Lemma 4.1

Let $R$ be a commutative ring.

  1. 1.

${\cal L}\mathrm{L}(R)=R\backslash\mathfrak{n}_{R}$.

  2. 2.

${\cal N}{\cal L}\mathrm{L}(R)=\mathfrak{n}_{R}$.

  3. 3.

Let $S\subseteq R$. Then $S\in\mathrm{L}(R)$ iff $S_{mon}$ is a multiplicative
subset of $R$. If $S\in\mathrm{L}(R)$ then $R\langle
S^{-1}\rangle=S_{mon}^{-1}R$ and ${\rm ass}_{R}(S)={\rm
ass}_{R}(S_{mon})=\\{r\in R\,|\,tr=0$ for some $t\in S_{mon}\\}$.

Proof. 1 and 2. The prime radical $\mathfrak{n}_{R}$ contains precisely all
the nilpotent elements of the ring $R$. Hence, $\mathfrak{n}_{R}\subseteq{\cal
N}{\cal L}\mathrm{L}(R)$. The ring $R$ is a commutative. So, for each element
$t\in R\backslash\mathfrak{n}_{R}$, the set
$S_{t}:=\\{t^{i}\,|\,i\in\mathbb{N}\\}$ is a multiplicative set of $R$, and so
$s\in{\cal L}\mathrm{L}(R)$, i.e. $R\backslash\mathfrak{n}_{R}\subseteq{\cal
L}\mathrm{L}(R)$. Since,

$R={\cal L}\mathrm{L}(R)\coprod{\cal N}{\cal
L}\mathrm{L}(R)=R\backslash\mathfrak{n}_{R}\coprod\mathfrak{n}_{R},$

we must have that ${\cal L}\mathrm{L}(R)=R\backslash\mathfrak{n}_{R}$ and
${\cal N}{\cal L}\mathrm{L}(R)=\mathfrak{n}_{R}$.

3\. Statement 3 follows from statement 1. $\Box$

For an arbitrary commutative ring $R$, Theorem 4.2 describes its maximal
localizable sets, maximal localization rings, the set of completely
localizable elements, and the ideal $\mathfrak{c}_{R}$. It also describes
relations between the ideals $\mathrm{L}{\rm rad}(R)$, $\mathfrak{n}_{R}$, and
$\mathfrak{c}_{R}$.

###### Theorem 4.2

Let $R$ be a commutative ring and ${\cal C}:={\cal C}\mathrm{L}(R)$. Then

  1. 1.

$\max\mathrm{L}(R)=\\{S_{\mathfrak{p}}\,|\,\mathfrak{p}\in\min(R)\\}$ where
$S_{\mathfrak{p}}=R\backslash\mathfrak{p}$ is a multiplicative subset of $R$.

  2. 2.

$R\langle
S_{\mathfrak{p}}^{-1}\rangle=S_{\mathfrak{p}}^{-1}R:=R_{\mathfrak{p}}$ is the
localization of the ring $R$ at the prime ideal $\mathfrak{p}$.

  3. 3.

${\rm ass}_{R}(S_{\mathfrak{p}})=\\{r\in R\,|\,sr=0$ for some $s\in
S_{\mathfrak{p}}\\}\subseteq\mathfrak{p}$.

  4. 4.

$\mathrm{L}{\rm rad}(R)\subseteq\mathfrak{n}_{R}$.

  5. 5.

${\cal
C}=R\backslash\Big{(}\bigcup_{\mathfrak{p}\in\min(R)}\mathfrak{p}\Big{)}$ is a
multiplicative subset of $R$ and
$Q_{c}(R)=\bigg{(}R\backslash\Big{(}\bigcup_{\mathfrak{p}\in\min(R)}\mathfrak{p}\Big{)}\bigg{)}^{-1}R$.

  6. 6.

$\mathfrak{c}_{R}=\\{r\in R\,|\,cr=0$ for some $c\in{\cal
C}\\}\subseteq\mathfrak{n}_{R}$.

  7. 7.

$\mathfrak{n}_{Q_{c}(R)}={\cal C}^{-1}\mathfrak{n}_{R}$.

Proof. 1. Let $S\in\mathrm{L}(R)$. By Lemma 4.1.(3), the monoid $S_{mon}$ is a
multiplicative subset of $R$ and $R\langle S^{-1}\rangle=S_{mon}^{-1}R$.
Choose a prime ideal $P\in{\rm Spec}(S_{mon}^{-1}R)$. Then
$\mathfrak{p}^{\prime}:=\sigma_{S}^{-1}(P)\in{\rm Spec}(R)$ where
$\sigma_{S}:R\rightarrow S_{mon}^{-1}R$, $r\mapsto\frac{r}{1}$, and so

$S\subseteq
S_{\mathfrak{p}^{\prime}}=R\backslash\mathfrak{p}^{\prime}\subseteq
S_{\mathfrak{p}}$

for each $\mathfrak{p}\in\min(R)$ such that
$\mathfrak{p}\subseteq\mathfrak{p}^{\prime}$, and statement 1 follows.

2\. Statement 2 follows from statement 1.

3\. Clearly, ${\rm ass}_{R}(S_{\mathfrak{p}})=\\{r\in R\,|\,sr=0$ for some
$s\in S_{\mathfrak{p}}\\}$. If $r\in{\rm ass}_{R}(S_{\mathfrak{p}})$ then
$sr=0$ for some element $s\in S_{\mathfrak{p}}=R\backslash\mathfrak{p}$, and
so $r\in\mathfrak{p}$ (since $s\not\in\mathfrak{p}$ and the ideal
$\mathfrak{p}$ is a prime ideal). Hence, ${\rm
ass}_{R}(S_{\mathfrak{p}})\subseteq\mathfrak{p}$.

4\. By statements 1 and 3, $\mathrm{L}{\rm
rad}(R)=\bigcap_{\mathfrak{p}\in\min(R)}{\rm
ass}_{R}(S_{\mathfrak{p}})\subseteq\bigcap_{\mathfrak{p}\in\min(R)}\mathfrak{p}=\mathfrak{n}_{R}$.

5\. By statement 1 and Lemma 4.1.(1), the set

${\cal
C}=\bigcap_{\mathfrak{p}\in\min(R)}S_{\mathfrak{p}}=R\backslash\Big{(}\bigcup_{\mathfrak{p}\in\min(R)}\mathfrak{p}\Big{)}\subseteq
R\backslash\Big{(}\bigcap_{\mathfrak{p}\in\min(R)}\mathfrak{p}\Big{)}=R\backslash\mathfrak{n}_{R}={\cal
L}\mathrm{L}(R)$

is a localizable set. Hence, the set ${\cal C}$ is a multiplicative subset of
$R$.

6\. By statement 5, $\mathfrak{c}_{R}=\\{r\in R\,|\,cr=0$ for some $c\in{\cal
C}\\}$. Since $c\not\in\mathfrak{p}$ for all $\mathfrak{p}\in\min(R)$, we must
have $r\in\bigcap_{\mathfrak{p}\in\min(R)}\mathfrak{p}=\mathfrak{n}_{R}$, and
so $\mathfrak{c}_{R}\subseteq\mathfrak{n}_{R}$.

7\. An element $c^{-1}r\in Q_{c}(R)$, where $c\in{\cal C}$ and $r\in R$,
belongs to the prime radical $\mathfrak{n}_{Q_{c}(R)}$ iff it is a nilpotent
element iff $\frac{r}{1}$ is a nilpotent element of $Q_{c}(R)$ iff
$c^{\prime}r^{n}=0$ for some element $c^{\prime}\in{\cal C}$ and a natural
number $n\geq 1$ iff $(c^{\prime}r)^{n}=0$ iff
$c^{\prime}r\in\mathfrak{n}_{R}$ iff $c^{-1}r\in\mathfrak{n}_{Q_{c}(R)}$.
$\Box$

For a commutative ring $R$ such that $|\min(R)|<\infty$ and $\mathfrak{n}_{R}$
is a nilpotent ideal (eg, $R$ is a commutative Noetherian ring), Proposition
4.3 describes ${\rm Spec}(Q_{c}(R))$ and the rings $Q_{c}(R)$ and
$Q_{c}(R)/\mathfrak{n}_{Q_{c}(R)}$.

###### Proposition 4.3

Let $R$ be a commutative ring and ${\cal C}:={\cal C}\mathrm{L}(R)$. Then

  1. 1.

Suppose that $|\min(R)|<\infty$. Then $K{\rm dim}(Q_{c}(R))=0$ and ${\rm
Spec}(Q_{c}(R))=\\{{\cal
C}^{-1}\mathfrak{p}\,|\,\mathfrak{p}\in\min(R)\\}=\min(Q_{c}(R))$.

  2. 2.

Suppose that $|\min(R)|<\infty$ and $\mathfrak{n}_{R}$ is a nilpotent ideal,
eg, $R$ is a commutative Noetherian ring. Then

    1. (a)

$Q_{c}(R)\simeq\prod_{\mathfrak{p}\in\min(R)}R_{\mathfrak{p}}$ is a direct
product of local rings $(R_{\mathfrak{p}},\mathfrak{p}_{\mathfrak{p}})$ and
the maximal ideal $\mathfrak{p}_{\mathfrak{p}}$ of $R_{\mathfrak{p}}$ is a
nilpotent ideal.

    2. (b)

The rings ${\cal C}^{-1}(R/\mathfrak{p})\simeq(R/\mathfrak{p})_{\mathfrak{p}}$
are fields for all $\mathfrak{p}\in\min(R)$.

    3. (c)

$Q_{c}(R)/\mathfrak{n}_{Q_{c}(R)}\simeq\prod_{\mathfrak{p}\min(R)}(R/\mathfrak{p})_{\mathfrak{p}}$
is a direct product of fields.

Proof. 1. By Theorem 4.2.(5), ${\cal
C}=R\backslash\Big{(}\bigcup_{\mathfrak{p}\in\min(R)}\mathfrak{p}\Big{)}$ is a
multiplicative subset of $R$ and

$Q_{c}(R)={\cal
C}^{-1}R=\bigg{(}R\backslash\Big{(}\bigcup_{\mathfrak{p}\in\min(R)}\mathfrak{p}\Big{)}\bigg{)}^{-1}R.$

Since $|\min(R)|<\infty$, $K{\rm dim}(Q_{c}(R))=0$ and ${\rm
Spec}(Q_{c}(R))=\\{{\cal
C}^{-1}\mathfrak{p}\,|\,\mathfrak{p}\in\min(R)\\}=\min(Q_{c}(R))$.

2\. By statement 1, the set ${\rm Spec}(Q_{c}(R))=\\{{\cal
C}^{-1}\mathfrak{p}\,|\,\mathfrak{p}\in\min(R)\\}$ consists precisely of
maximal ideals of the ring $Q_{c}(R)$. Hence, they are co-prime ideals, that
is ${\cal C}^{-1}\mathfrak{p}+{\cal C}^{-1}\mathfrak{p}^{\prime}=Q_{c}(R)$ for
all $\mathfrak{p}\neq\mathfrak{p}^{\prime}$. Hence,

$Q_{c}(R)/\mathfrak{n}_{Q_{c}(R)}=Q_{c}(R)/\bigcap_{\mathfrak{p}\min(R)}{\cal
C}^{-1}\mathfrak{p}\simeq\prod_{\mathfrak{p}\min(R)}Q_{c}(R)/{\cal
C}^{-1}\mathfrak{p}=\prod_{\mathfrak{p}\min(R)}{\cal C}^{-1}R/{\cal
C}^{-1}\mathfrak{p}=\prod_{\mathfrak{p}\min(R)}{\cal C}^{-1}(R/\mathfrak{p})$

is a direct product of fields. Therefore, ${\cal
C}^{-1}(R/\mathfrak{p})\simeq(R/\mathfrak{p})_{\mathfrak{p}}$ for all
$\mathfrak{p}\in\min(R)$. Since $\mathfrak{n}_{R}$ is a nilpotent ideal, the
orthogonal primitive idempotents that correspond to the direct product
decomposition of the factor ring $Q_{c}(R)/\mathfrak{n}_{Q_{c}(R)}$ above can
be lifted, i.e. the ring $Q_{c}(R)$ is a product of local rings. Hence,

$Q_{c}(R)\simeq\prod_{\mathfrak{p}\in\min(R)}R_{\mathfrak{p}}.\;\;\Box$

For a commutative ring $R$ such that $|\min(R)|<\infty$ and
$\mathfrak{n}_{R}=0$ (eg, $R$ is a commutative semiprime Noetherian ring),
Proposition 4.3 shows that ${\cal C}\mathrm{L}(R)={\cal C}_{R}$ and
$Q_{c}(R)=Q_{cl}(R)$.

###### Theorem 4.4

Let $R$ be a commutative ring with $|\min(R)|<\infty$ and
$\mathfrak{n}_{R}=0$, eg, $R$ is a commutative semiprime Noetherian ring, and
${\cal C}:={\cal C}\mathrm{L}(R)$. Then

  1. 1.

The ring $R$ is a semiprime Goldie ring.

  2. 2.

$Q_{cl}(R)=\prod_{\mathfrak{p}\in\min(R)}R_{\mathfrak{p}}$ is a product of
fields.

  3. 3.

For all $\mathfrak{p}\in\min(R)$, ${\rm
ass}_{R}(S_{\mathfrak{p}})=\mathfrak{p}$.

  4. 4.

${\cal C}={\cal C}_{R}$ and $Q_{c}(R)=Q_{cl}(R)$.

  5. 5.

$\mathfrak{n}_{Q_{c}(R)}=0$.

  6. 6.

$\mathrm{L}{\rm rad}(R)=0$.

  7. 7.

$\mathfrak{c}_{R}=0$.

Proof. 1. By the assumption, the ring $R$ is a commutative ring with
$|\min(R)|<\infty$ and $\mathfrak{n}_{R}=0$. By [2, Theorem 5.1], the ring $R$
is a semiprime Goldie ring since for each minimal prime ideal
$\mathfrak{p}\in\min(R)$, the field $R/\mathfrak{p}$ is a Goldie ring (any
field is a Goldie ring).

2 and 3. Statements 2 and follow from statement 1 and [2, Theorem 4.1].

4\. By statements 2 and 3, the union
$\bigcup_{\mathfrak{p}\min(R)}\mathfrak{p}$ is the set of all zero divisors of
the ring $R$, and so

${\cal C}_{R}=R\backslash\bigcup_{\mathfrak{p}\min(R)}\mathfrak{p}={\cal C},$

by Theorem 4.2.(5). Hence, $Q_{c}(R)=Q_{cl}(R)$.

5\. By Theorem 4.2.(7) (or by statements 2 and 4),
$\mathfrak{n}_{Q_{c}(R)}={\cal C}^{-1}\mathfrak{n}_{R}=0$.

6\. By Theorem 4.2.(4), $\mathrm{L}{\rm rad}(R)=0$.

7\. Since ${\cal C}={\cal C}_{R}$ (statement 4), $\mathfrak{c}_{R}=0$. $\Box$

Corollary 4.5 gives a characterization of commutative semiprime Goldie rings.

###### Corollary 4.5

Let $R$ be a commutative semiprime ring. Then the ring $R$ is a Goldie ring
iff $|\min(R)|<\infty$.

Proof. Suppose that the ring $R$ is a Goldie ring. Then the ring $Q_{cl}(R)$
is a commutative semisimple Artinian ring, i.e. a finite product of fields.
Hence, $|\min(R)|<\infty$.

Suppose that the ring $R$ has only finitely many minimal primes. Then, by
Theorem 4.4.(1), the ring $R$ is a Goldie ring. $\Box$

##  5 Localizations of semiprime left Goldie rings and direct products of
division rings

Proposition 5.1.(4) describes the maximal localizable sets of finite direct
product of rings and their localizations. Proposition 5.1.(1) describes
localizations of the direct product of rings via localizations of its
components. For a direct product of simple rings $A=\prod_{i=1}^{s}A_{i}$ such
that ${\cal L}\mathrm{L}(A_{i})=A_{i}^{\times}$ for $i=1,\ldots,s$, Theorem
5.3 describes the sets $\max\mathrm{L}(A)$, ${\rm ass}_{A}(S)$ for all
$S\in\max\mathrm{L}(A)$, the sets of localizable, non-localizable, and
completely localizable elements of $A$. It also shows that $\mathrm{L}{\rm
rad}(A)=0$. Every semisimple Artinian ring satisfies the assumptions of
Theorem 5.3, see Corollary 5.4 for detail. For a semiprime left Goldie ring,
Lemma 5.8 describes all the maximal localizable sets that contain the set of
regular elements of the ring.

Let $D=\prod_{i\in I}D_{i}$ be a direct product of division rings $D_{i}$
where $I$ is an arbitrary set. Proposition 5.9 describes all the localizable
sets in the ring $D$ and all the localizations of $D$. Theorem 5.11 describes
the following sets: $\max\mathrm{L}(D)$, $\max\mathrm{L}{\rm oc}(D)$,
$\mathrm{L}{\rm rad}(D)$, ${\cal C}\mathrm{L}(D)$, $Q_{c}(D)$, ${\cal
L}\mathrm{L}(D)$, and ${\cal N}{\cal L}\mathrm{L}(D)$. Theorem 5.11.(2) shows
that for every ${\cal S}\in\max\mathrm{L}(D)$, the ring $D\langle{\cal
S}^{-1}\rangle$ is a division ring.

The maximal localizable sets of finite direct product of rings. Let
$A=\prod_{i=1}^{s}A_{i}$ be a finite direct product of rings $A_{i}$. For each
$i=1,\ldots,n$, let $p_{i}:A\rightarrow A_{i}$, $(a_{1},\ldots,a_{n})\mapsto
a_{i}$ be the canonical projection and $\nu_{i}:A_{i}\rightarrow A$,
$a_{i}\mapsto(0,\ldots,0,a_{i},0,\ldots,0)$ is the canonical injection.

Proposition 5.1.(4) describes the maximal localizable sets of finite direct
product of rings and their localizations.

###### Proposition 5.1

Let $A=\prod_{i=1}^{s}A_{i}$ be a direct product of rings, $S\subseteq A$, and
$S_{i}:=p_{i}(S)$ for $i=1,\ldots,n$. Then

  1. 1.

$A\langle S^{-1}\rangle\simeq\prod_{i=1}^{s}A_{i}\langle S_{i}^{-1}\rangle$,
an $A$-isomorphism.

  2. 2.

${\rm ass}_{A}(S)=\prod_{i=1}^{s}{\rm ass}_{A_{i}}(S_{i})$.

  3. 3.

$S\in\mathrm{L}(A)$ iff $S_{i}\in\mathrm{L}(A_{i})$ for some $i$.

  4. 4.

$\max\mathrm{L}(A)=\coprod_{i=1}^{s}\max\mathrm{L}(A_{i})$, where the set
$\max\mathrm{L}(A_{i})$ is identified with its image in $\max\mathrm{L}(A)$
via the injection
$\phi_{i}:\max\mathrm{L}(A_{i})\rightarrow\max\mathrm{L}(A)$, $S_{i}\mapsto
A_{1}\times\cdots\times A_{i-1}\times S_{i}\times A_{i+1}\times\cdots\times
A_{s}$, and $A\langle S_{i}^{-1}\rangle\simeq A_{i}\langle S_{i}^{-1}\rangle$.

Proof. 1. The ring $\prod_{i=1}^{s}A_{i}\langle S_{i}^{-1}\rangle$ satisfies
the universal property of localization for the set $S$, and statement 1
follows.

2 and 3. Statements 2 and 3 follow from statement 1.

4\. By statement 1, $A\langle S_{i}^{-1}\rangle\simeq A_{i}\langle
S_{i}^{-1}\rangle$ for all $S_{i}\in\max\mathrm{L}(A_{i})$. Therefore, the map
$\phi_{i}$ is an injection for every $i=1,\ldots,s$. By the definition of the
maps $\phi_{i}$,

$\max\mathrm{L}(A)\supseteq\coprod_{i=1}^{s}\max\mathrm{L}(A_{i}).$

Let $S\in\max\mathrm{L}(A)$. By statement 1, $A\langle
S^{-1}\rangle\simeq\prod_{i=1}^{s}A_{i}\langle S_{i}^{-1}\rangle$, where
$S_{i}:=p_{i}(S)$ for $i=1,\ldots,s$. There is an index $i$ such that
$A_{i}\langle S_{i}^{-1}\rangle\neq\\{0\\}$. Therefore,

$S\subseteq T_{i}:=A_{1}\times\cdots\times A_{i-1}\times S_{i}\times
A_{i+1}\times\cdots\times A_{s},$

and $A\langle T_{i}^{-1}\rangle\simeq A_{i}\langle S_{i}^{-1}\rangle$. Hence,

$\max\mathrm{L}(A)\subseteq\coprod_{i=1}^{s}\max\mathrm{L}(A_{i}),$

and statement 4 follows. $\Box$

###### Lemma 5.2

Let $A$ be a ring such that ${\cal L}\mathrm{L}(A)=A^{\times}$. Then

  1. 1.

$\max\mathrm{L}(A)=\\{A^{\times}\\}$ and $A\langle(A^{\times})^{-1}\rangle=A$.

  2. 2.

$\mathrm{L}{\rm rad}(A)={\rm ass}_{A}(A^{\times})=0$.

  3. 3.

${\cal C}\mathrm{L}(A)={\cal L}\mathrm{L}(A)=A^{\times}$, and $Q_{c}(A)=A$,
$\mathfrak{c}_{A}=0$.

Proof. 1. By Proposition 2.8.(1), $A^{\times}\in\mathrm{L}(A)$ and
$A\langle{A^{\times}}^{-1}\rangle=A$, and so ${\rm ass}_{A}(A^{\times})=0$.
Hence, $A^{\times}\in\mathrm{L}(A)$. If $S\in\mathrm{L}(A)$ then
$S\subseteq{\cal L}\mathrm{L}(A)=A^{\times}$, by Corollary 3.13.(1).
Therefore, $\max\mathrm{L}(A)=\\{A^{\times}\\}$ and statement 1 follows.

2\. By statement 1, $\max\mathrm{L}(A)=\\{A^{\times}\\}$, and so
$\mathrm{L}{\rm rad}(A)={\rm ass}_{A}(A^{\times})=0$.

3\. Statement 3 follows from statement 1. $\Box$

Example. Let $(A,\mathfrak{m})$ be a local ring where $\mathfrak{m}$ is a
maximal ideal of $A$ which is assumed to be a nil ideal (that is every element
of $\mathfrak{m}$ is a nilpotent element). Then

${\cal L}\mathrm{L}(A)=A^{\times}\;\;{\rm and}\;\;{\cal N}{\cal
L}\mathrm{L}(A)=\mathfrak{m}.$

In particular, the ring $A$ satisfies the conditions of Lemma 5.2, and so
$\max\mathrm{L}(A)=\\{A^{\times}\\}$, $A\langle(A^{\times})^{-1}\rangle=A$,
and $\mathrm{L}{\rm rad}(A)={\rm ass}_{A}(A^{\times})=0$.

Let $A=\prod_{i=1}^{s}A_{i}$ be a direct product of simple rings $A_{i}$. Then
$\min(A)=\\{\mathfrak{p}_{i}\,|\,i=1,\ldots,s\\}$ where
$\mathfrak{p}_{i}:=A_{1}\times\cdots\times A_{i-1}\times\\{0\\}\times
A_{i+1}\times\cdots\times A_{s}$, and $A/\mathfrak{p}_{i}\simeq A_{i}$.

For the ring $A=\prod_{i=1}^{s}A_{i}$ such that ${\cal
L}\mathrm{L}(A_{i})=A_{i}^{\times}$ for $i=1,\ldots,n$, Theorem 5.3 describes
in an explicit way the following sets: $\max\mathrm{L}(A)$, $\mathrm{L}{\rm
rad}(A)$, ${\cal C}\mathrm{L}(A)$, $Q_{c}(A)$, ${\cal L}\mathrm{L}(A)$, and
${\cal N}{\cal L}\mathrm{L}(A)$.

###### Theorem 5.3

Let $A=\prod_{i=1}^{s}A_{i}$ be a direct product of rings $A_{i}$ such that
${\cal L}\mathrm{L}(A_{i})=A_{i}^{\times}$ for $i=1,\ldots,n$. Then

  1. 1.

$\max\mathrm{L}(A)=\\{S_{i}\,|\,i=1,\ldots,s\\}$ where
$S_{i}:=p_{i}^{-1}(A_{i}^{\times})=\\{(a_{1},\ldots,a_{s})\in A\,|\,a_{i}\in
A_{i}^{\times},a_{j}\in A_{j}$ for all $j\neq i\\}$.

  2. 2.

${\rm ass}_{A}(S_{i})=\mathfrak{p}_{i}$ and $A\langle S_{i}^{-1}\rangle\simeq
A/\mathfrak{p}_{i}=A_{i}$.

  3. 3.

$\mathrm{L}{\rm rad}(A)=0$.

  4. 4.

${\cal C}\mathrm{L}(A)=A^{\times}$ and $Q_{c}(A)=A$.

  5. 5.

${\cal L}\mathrm{L}(A)=\\{(a_{1},\ldots,a_{s}))\in A\,|\,a_{i}\in
A_{i}^{\times}$ for some $i\\}$ and ${\cal N}{\cal
L}\mathrm{L}(A)=\prod_{i=1}^{s}A_{i}\backslash A_{i}^{\times}$.

Proof. 1. Statement 1 follows from Proposition 5.1.(4) and Lemma 5.2.(1).

2\. Statement 2 follows from statement 1.

3\. By statement 2, $\mathrm{L}{\rm
rad}(A)=\bigcap_{i=1}^{s}\mathfrak{p}_{i}=0$.

4\. By statement 1, ${\cal C}\mathrm{L}(A)=\bigcap_{i=1}^{s}S_{i}=A^{\times}$,
and so $Q_{c}(A)=A\langle{\cal C}\mathrm{L}(A)^{-1}\rangle=A$.

5\. Statement 5 follows from statement 1. $\Box$

Example. $A=\prod_{i=1}^{s}A_{i}$ be a direct product of local rings
$(A_{i},\mathfrak{m}_{i})$ where the maximal ideals $\mathfrak{m}_{i}$ are nil