The largest left quotient ring of a ring

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by  V. V. Bavula

The largest left quotient ring of a ring
V. V. Bavula
1
1
0
2 Abstract

n The left quotient ring (i.e. the left classical ring of fractions) Qcl(R) of a ring R does
a
not always exist and still, in general, there is no good understanding of the reason why this
J
happens. In this paper, it is proved existence of the largest left quotient ring Ql(R), i.e.
6 Ql(R)=S0(R)−1R where S0(R) is thelargest left regular denominator set of R. It is proved
2
that Ql(Ql(R)) = Ql(R); the ring Ql(R) is semi-simple iff Qcl(R) exists and is semi-simple;
] moreover,iftheringQl(R)isleftartinianthenQcl(R)existsandQl(R)=Qcl(R). Thegroup
A of units Ql(R)∗ of Ql(R) is equal to the set {s−1t|s,t ∈ S0(R)} and S0(R) = R∩Ql(R)∗.
R If there exists a finitely generated flat left R-module which is not projective then Ql(R) is
notasemi-simplering. WeextendslightlyOre’smethodoflocalization tolocalizable left Ore
.
h sets, give a criterion of when a left Ore set is localizable, and prove that all left and right
t Ore sets of an arbitrary ring are localizable (not just denominator sets as in Ore’s method
a
m of localization). Applications are given for certain classes of rings (semi-prime Goldie rings,
Noetherian commutativerings, thealgebra of polynomial integro-differential operators).
[
Key Words: the largest left quotient ring of a ring, the largest left (regular) denominator

1 set of a ring, the classical left quotient ring of a ring, denominator set, the maximal left
v quotient rings of a ring.
7 Mathematics subject classification 2000: 16U20, 16P40, 16S32, 13N10.
0
1 Contents
5
. 1. Introduction.
1
0 2. The largest left quotient ring of a ring.
1 3. The maximal left quotient rings of a ring.
1
: 4. The largest quotient ring of a ring.
v
5. Examples.
i
X
r
a 1 Introduction
The aim of the paper is to introduce the following new concepts and to prove their existence for
an arbitrary ring: the largest left quotient ring of a ring, the largest left (regular) denominator set
of a ring, the maximal left quotient rings of a ring, the largest (two-sided) quotient ring of a ring,
the maximal (two-sided) quotient rings of a ring, a (left) localization maximal ring.
Throughout, module means a left module; R is a ring with 1; C is the set of (left and
R
right) regular elements of the ring R (i.e. C is the set of non-zero-divisors of R); Q (R) is the
R cl
left quotient ring of R (if exists); Den (R,0) is the set of regular left denominator sets S in R
l
(S ⊆C ).
R
The largestleftregular denominatorsetand the largestleftquotient ring ofaring.
• (Theorem2.1)There exists the largest (w.r.t. inclusion) left regular denominators set S (R)
0
in R and so Q (R):=S (R)−1R is the largest left quotient ring of R.
l 0
• (Corollary2.5) The set Den(R,0) of left regular denominator sets of R is a complete lattice
l
and an abelian monoid.
1
The next theorem describes the group of units Q (R)∗ of the ring Q (R) and its connection
l l
with S (R).
0
• (Theorem 2.8)
1. S (Q (R))=Q (R)∗ and S (Q (R))∩R=S (R).
0 l l 0 l 0
2. Q (R)∗ = hS (R),S (R)−1i, i.e. the group of units of the ring Q (R) is generated by
l 0 0 l
the sets S (R) and S (R)−1 :={s−1|s∈S (R)}.
0 0 0
3. Q (R)∗ ={s−1t|s,t∈S (R)}.
l 0
4. Q (Q (R))=Q (R).
l l l
The next theorem gives an answer to the question of when the ring Q (R) is semi-simple
l
(Theorem 2.8 is a key step in proving Theorem 2.9, statements 2-5 is Goldie’s Theorem).
• (Theorem 2.9) The following statements are equivalent.
1. Q (R) is a semi-simple ring.
l
2. Q (R) exists and is a semi-simple ring.
cl
3. R is a left order in a semi-simple ring.
4. R has finite left rank, satisfies the ascending chain condition on left annihilators, and
is a semi-prime ring.
5. A left ideal of the ring R is essential iff it contains a regular element.
If one of these conditions hold then S (R)=C and Q (R)=Q (R).
0 R l cl
The next result is a sufficient condition for the ring Q (R) not being a semi-simple ring.
l
• (Corollary2.12)If there exists a finitely generated flat R-module which is not projective then
the ring Q (R) is not semi-simple.
l
The next corollary gives a sufficient condition for existing of Q (R).
cl
• (Corollary 2.10) If Q (R) is a left artinian ring then S (R)=C and Q (R)=Q (R).
l 0 R l cl
Let Sr(R) and Q (R) := RSr(R)−1 be the largest regular right denominator set in R and
0 r 0
the largest right quotient ring of R, respectively. In general, the following natural questions have
negativeanswersasthe algebraI1 :=Khx,ddx, iofpolynomialintegro-differentialoperatorsover
a field K of characteristic zero demonstrates [2R]:
Question 1. Is S (R)=Sr(R)?
0 0
Question 2. Is S (R)⊆Sr(R) or S (R)⊇Sr(R)?
0 0 0 0
Question 3. Are the rings Q (R) and Q (R) isomorphic?
l r
Though, for the algebra I1 the next question has positive answer [2].
Question 4. Are the rings Q (R) and Q (R) anti-isomorphic?
l r
Remark. The algebra I1 is neither left nor right Noetherian, it contains infinite direct sums of
nonzero left (and right) ideals, neither left nor right quotient ring of I1 exists [2].
Notation:
• Ore(R):={S|S is a left Ore set in R};
l
• Den(R):={S|S is a left denominator set in R};
l
• Loc (R):={S−1R|S ∈Den(R)};
l l
• Ass (R):={ass(S)|S ∈Den (R)} where ass(S):={r∈R|sr =0 for some s=s(r)∈S};
l l
2
• Den(R,a):={S ∈Den (R)|ass(S)=a} where a∈Ass (R);
l l l
• S = S (R) = S (R) is the largest element of the poset (Den (R,a),⊆) and Q (R) :=
a a l,a l a
Q (R):=S−1Risthelargest left quotient ring associated toa,S exists(Theorem2.1.(2));
l,a a a
• In particular, S = S (R) = S (R) is the largest element of the poset (Den (R,0),⊆) and
0 0 l,0 l
Q (R):=S−1R is the largest left quotient ring of R;
l 0
• Loc (R,a):={S−1R|S ∈Den (R,a)}.
l l
For each denominator set S ∈Den (R,a) where a:=ass(S), there are natural ring homomor-
l
phisms R →π R/a → S−1R. In Section 3, connections between the denominator sets Den (R,a),
l
Den (R/a,0) and Den (S−1R,0) are established (Lemma 3.2, Lemma 3.3 and Proposition 3.4;
l l
these results are too technical to explain in the Introduction). They are used to prove the follow-
ing results.
• (Lemma3.3.(2))Leta∈Ass (R)andπ :R→R/a,a7→a+a. Thenπ−1(S (R/a))=S (R),
l 0 a
π(S (R))=S (R/a) and Q (R)=Q (R/a).
a 0 a l
• (Lemma3.7.(2))The set max.Den (R) of maximal elements of the poset (Den (R),⊆) of left
l l
denominator sets of an arbitrary ring R is a non-empty set.
This means that the set max.Loc(R) of maximal left quotient rings of an arbitrary ring R is a
l
non-empty set, and max.Loc (R)={S−1R|S ∈max.Den(R)}.
l l
Example. max.Denl(I1) = {I1\F} (Proposition 5.8.(3)) where F is the only proper ideal
of the algebra of polynomial integro-differential operators I1 and max.Locl(I1) = {Frac(A1)}
(Proposition 5.8.(2)) where Frac(A ) is the Weyl skew field, i.e. the classical (left and right) ring
1
of fractions of the first Weyl algebra A =Khx, x i.
1 dx
A new class of rings is introduced, the class of left localization maximal rings, see Section 3
(intuitively, one can invert nothing more in these rings). For an arbitrary ring R, a criterion
is given in terms of this class of rings of when a left localization S−1R of R is a maximal left
localization of R.
• (Theorem 3.11) Let a ring A be a left localization of a ring R, i.e. A∈Loc (R,a) for some
l
a ∈ Ass (R). Then A ∈ max.Loc (R) iff Q (A) = A and Ass (A) = {0}, i.e. A is a left
l l l l
localization maximal ring.
Example. Let V be an infinite dimensional vector space with countable basis over a field K,
let C := {ϕ ∈ End (V)|dim (im(ϕ)) < ∞} be the ideal of compact operators/linear maps of
K K
the algebra End (V). Then the factor algebra End (V)/C is a left localization maximal ring
K K
(Theorem 5.2.(5)). Moreover, the set F of Fredholm linear maps in V is the maximal left (resp.
right; left and right) denominator set of the ring End (V) and
K
F−1End (V)≃End (V)F−1 ≃End (V)/C
K K K
is the maximal left (resp. right; left and right) localization ring of End (V) (Theorem 5.2).
K
A slight extension/generalization of Ore’s method of localization. Ore’s method of
left localization says that we can localize a ring R precisely at the left denominator sets Den (R)
l
ofthe ringR. Notice that eachleft denominatorsetis a left Oresetbut notthe other wayround,
in general. We introduce the concept of a localizable left Ore set and give a criterion of when a
left Ore set is localizable (Theorem 3.15). Each left denominator set is a localizable left Ore set
but not vice versa, in general. We extend Ore’s method of localization to localizable left Ore sets
and prove an analogue of Ore’s Theorem (by using Ore’s Theorem) for localizable left Ore sets.
3
• (Corollary3.16) Let S be a localizable left Ore set in a ring R. Then there exists an ordered
pair (Q,f) where Q is a ring and f :R→Q is a ring homomorphism such that
(i) for all s∈S, f(s) is a unit in Q;
and if (Q′,f′) is another pair satisfying the condition (i) then there is a unique ring ho-
momorphism h : Q → Q′ such that f′ = hf. The ring Q is unique up to isomorphism.
The ring Q is isomorphic to the left localization of the ring R/p(S) at the left denomina-
tor set π(S) ∈ Den (R/p(S),0) where the ideal p(S) of the ring R is defined in (11) and
l
π :R→R/p(S), a7→a+p(S).
In Section 4, we prove two-sided analogues of some of the results of Sections 2 and 3. In most
casesthe proofs areeasy corollariesofthe correspondingone-sided(i.e. left andright) resultsbut
there are some surprises. In particular,
• (Theorem 4.15) Every (left and right) Ore set is localizable.
Therefore we can localize at all the (left and right) Ore sets not just at the (left and right)
denominator sets as in Ore’s method of localization.
• (Corollary 4.16) Let S be an Ore set in a ring R. Then there exists an ordered pair (Q,f)
where Q is a ring and f :R→Q is a ring homomorphism such that
(i) for all s∈S, f(s) is a unit in Q;
and if (Q′,f′) is another pair satisfying the condition (i) then there is a unique ring ho-
momorphism h : Q → Q′ such that f′ = hf. The ring Q is unique up to isomor-
phism. The ring Q is isomorphic to the localization of the ring R/p(S) at the denomi-
nator set π(S) ∈ Den(R/p(S),0) where the ideal p(S) of the ring R is defined in (19) and
π :R→R/p(S), a7→a+p(S).
In Section 5, the largest (left; right; left and right) and maximal (left; right; left and right)
quotient rings are found for following rings: the endomorphism algebra End (V) of an infinite
K
dimensional vector space V with countable basis, semi-prime Goldie rings,the algebra I1 of poly-
nomial integro-differential operators and Noetherian commutative rings.
2 The largest left quotient ring of a ring
Inthis section,existenceofthe largestleftquotientringofaringis proved(Theorem2.1). Proofs
of Theorems 2.8 and 2.9 are given.
The largest left quotient ring of a ring. Let R be a ring. A multiplicatively closed subset
S of R (i.e. a multiplicative sub-semigroupof (R,·) suchthat 1∈S and 06∈S)is saidto be a left
Ore set if it satisfies the left Ore condition: for each r ∈ R and s ∈ S, Sr Rs 6= ∅. Let S be a
(non-empty)multiplicatively closedsubset ofR,and letass(S):={r ∈R|sTr=0 for somes∈S}
(if, in addition, S is a left Ore set then ass(S) is an ideal of the ring R). Then a left quotient ring
of R with respect to S (a left localization of R at S) is a ring Q together with a homomorphism
ϕ:R→Q such that
(i) for all s∈S, ϕ(s) is a unit of Q,
(ii) for all q ∈Q, q =ϕ(s)−1ϕ(r) for some r ∈R, s∈S, and
(iii) ker(ϕ)=ass(S).
If exists the ring Q is unique up to isomorphism, usually it is denoted by S−1R. Recall that
S−1R exists iff S is a left Ore set and the set S = {s + ass(S) ∈ R/ass(S)|s ∈ S} consists
of regular elements ([5], 2.1.12). If the last two conditions are satisfied then S is called a left
denominator set. Similarly,aright Ore set,the right Ore condition, theright denominator setand
the right quotient ring RS−1 are defined. If both S−1R and RS−1 exist then they are isomorphic
([5], 2.1.4.(ii)). The left quotient ring of R with respect to the set C of all regular elements is
R
4
calledthe left quotient ringofR,ifexists,itisdenotedbyFrac (R)orQ (R). Similarly,theright
l cl
quotient ring, Frac (R)=Qr(R), is defined. If both left and right quotient rings of R exist then
r cl
they are isomorphic and we write simply Frac(R) or Q(R) in this case.
Theorem 2.1 1. For each a ∈ Ass (R), the set Den (R,a) is an ordered abelian semigroup
l l
(S S =S S , and S ⊆S implies S S ⊆S S ) where the product S S =hS ,S i is the
1 2 2 1 1 2 1 3 2 3 1 2 1 2
multiplicative subsemigroup of (R,·) generated by S and S .
1 2
2. S := S (R) := S is the largest element (w.r.t. ⊆) in Den (R,a). The set S
a a S∈Denl(R,a) l a
is called the largSest left denominator set associated to a.
3. Let S ∈Den (R,a), i∈I, where I is an arbitrary non-empty set. Then
i l
hS |i∈Ii:= S ∈Den (R,a) (1)
i j l
[ Y
∅6=J⊆I,|J|<∞j∈J
theleftdenominator setgeneratedbytheleft denominators setsS , itistheleast upperbound
i
of the set {S } in Den (R,a), i.e. hS |i∈Ii= S .
i i∈I l i i∈I i
W
Remark. Clearly, S S =S S , the join of S and S .
1 2 1 2 1 2
Proof. 1. It remains to showWthat S S ∈Den (R,a), that is, S S is a left Ore set in R with
1 2 l 1 2
ass(S S ) = a and rs = 0 where r ∈ R and s ∈ S implies tr = 0 for some t ∈ S. To prove that
1 2
the left Ore condition holds for the multiplicatively closed set S = S S we have to show that,
1 2
for each s ∈ S and a ∈ R, there exist elements t ∈ S and b ∈ R such that ta = bs. We use
induction on the length l(s) = min{n|s = s ···s where all s ∈ S ∪S } of the element s. If
1 n i 1 2
l(s)=1,i.e. s∈S ∪S ,then the statementis obvioussince S andS areleft Oresets. Suppose
1 2 1 2
that n := l(s) > 1, and the statement holds for n′ < n. Fix a presentation s = s ···s s
1 n−1 n
where all s ∈ S ∪S . Then t a = b s for some elements t ∈ S ∪S and b ∈ R, and, by
i 1 2 1 1 n 1 1 2 1
induction, t b = bs ···s for some elements t ∈ S and b ∈ R. Clearly, t = t t ∈ S and
2 1 1 n−1 2 2 1
ta=t (t a)=(t b )s =bs, as required.
2 1 2 1 n
Let us show that ass(S) = a. Clearly, ass(S S ) ⊇ a. Let r ∈ ass(S S ). Then sr = 0
1 2 1 2
for some element s = s ···s ∈ S where all s ∈ S ∪S . Without loss of generality we may
1 m i 1 2
assume that s ∈S and s ∈S . Then s ···s r ∈a, and so there exists s′ ∈S such that
odd 1 even 2 2 m 2 2
(s′s )s ···s r = s′′s ···s r = 0 where s′′ = s′s ∈ S . By induction on m, we conclude that
2 2 3 m 2 3 m 2 2 2 2
r ∈a. Therefore, ass(S)=a.
Finally, suppose that rs = 0 where r ∈ R and s ∈ S. We have to show that tr = 0 for some
t ∈ S. Fix a presentation s = s ...s where s ∈ S ∪S . Then 0 = rs = rs ···s implies that
1 n i 1 2 1 n
t rs ···s = 0 for some t ∈ S ∪S since S ,S ∈ Den (R,a). Similarly, t t rs ···s = 0
1 1 n−1 1 1 2 1 2 l 2 1 1 n−2
for some t ∈ S ∪S . Repeating the same argument several times we see that t t ···t r =0
2 1 2 n n−1 1
for some t ∈S ∪S . Notice that t:=t t ···t ∈S and tr =0, as required.
i 1 2 n n−1 1
2. Let s ,s ∈ S . Then s ∈ S and s ∈ S for some S ,S ∈ Den (R,a), and so s ,s ∈
1 2 a 1 1 2 2 1 2 l 1 2
S S ∈ Den (R,a), by statement 1. Therefore, S ∈ Den (R,a), and so S is the largest element
1 2 l a l a
in Den (R,a).
l
3. Statement 3 follows from statement 1. (cid:3)
Definition. For each ideal a ∈ Ass (R), the ring Q (R) := S−1R is called the largest left
l a a
quotient ring associated to a. When a=0, the ring Q (R):=Q (R):=S−1R is called the largest
l a 0
left quotient ring of R and S =S (R) is called the largest left regular denominator set of R.
0 0
The next obvious corollary shows that Q (R) is a generalization of the classical left quotient
l
ring Q (R).
cl
Corollary 2.2 1. IftheclassicalleftquotientringQ (R):=C−1Rexiststhenthesetofregular
cl R
elements C of the ring R is the largest left regular denominator set and Q (R)=Q (R).
R l cl
2. Let R ,...,R be rings. Then Q ( n R )≃ n Q (R ).
1 n l i=1 i i=1 l i
Q Q
5
Proof. It is obvious. (cid:3)
Question 2.3 Can a ring monomorphism f : A → B be lifted (necessarily uniquely) to a ring
monomorphism f′ :Q (A)→Q (B)?
l l
In general, the answer is no.
Proposition 2.4 (Corollary9.9,[2]) Let K be a field of characteristic zero and A =Khx, d i be
1 dx
the ring of polynomial differential operators (the first Weyl algebra). Then the inclusion A1 →I1
cannot be lifted neither to a ring homomorphism Frac(A1)=Ql(A1)→Ql(I1) nor to Frac(A1)=
Qr(A1)→Qr(I1).
Let (P,≤) be a partially ordered set, a poset, for short (a ≤ a for all a ∈ P; a ≤ b and b ≤ a
implies a=b; a≤b and b≤c implies a≤c). For a subset S of P, an element x∈P is called an
upper bound (resp. a lower bound) if s ≤x (resp. s≥ x) for all s ∈S. The least upper bound for
S is the least element in the the set of all upper bounds for S. Similarly, the greatest lower bound
for S is defined. A lattice is a poset such that eachpair x, y of elements of the set S has both the
least upper bound x y (called the join for x and y) and the greatest lower bound x y (called
the meet of x and y).WIt follows then by induction that everynon-empty finite set of eleVments has
the join and the meet. A lattice L is complete if every subset S of L has the least upper bound
(the join of S) denoted sup(S) or s and the greatest lower bound (the meet of S) denoted
s∈S
inf(S)or s. Inacomplete lattWice thereexists the greatestelementsup(L)denotedby1,and
s∈S
the smalleVst element inf(L) denoted by 0. By definition, sup(∅) = 0. Let (L,≤) be a lattice and
a,b ∈ L with a ≤ b. The set [a,b] := {x ∈ L|a ≤ x ≤ b} is the interval between a and b. The
interval [a,b] is a sublattice of L with inf([a,b])=a and sup([a,b])=b.
Corollary 2.5 The abelian monoid Den (R,0)is a complete lattice suchthat S S =S S and
l 1 2 1 1
S = S where all S ∈Den (R,0). W
i∈I i S∈Denl(R,0),S⊆∩i∈ISi i i l
V W
Proof. The largest and least elements of the poset Den (R,0) are S (R) and {1} respec-
l 0
tively. Clearly, {1}is the identity element ofthe semigroupDen (R,0). By Theorem2.1.(3), each
l
nonempty subset of Den(R,0) has the least upper bound, hence Den (R,0) is a complete lattice
l l
by Proposition 1.2, Sect. 3, [6] and S = S . (cid:3)
i∈I i S∈Denl(R,0),S⊆∩i∈ISi i
V W
Clearly, S is the largest element of the set {S|S ∈Den (R,0),S ⊆ S }.
i∈I i l i∈I i
V T
Corollary 2.6 1. Let R be a ring. Each ring automorphism σ ∈Aut(R) of the ring R has the
unique extension σ ∈ Aut(Q (R)) to an automorphism of the ring Q (R) given by the rule
l l
σ(s−1r)=σ(s)−1σ(r) where s∈S (R) and r ∈R.
0
2. The group Aut(R) is a subgroup of the group Aut(Q (R)). Moreover, Aut(R) = {τ ∈
l
Aut(Q (R))|τ(R)=R}.
l
Proof. 1. By the uniqueness of the set S (R), σ(S (R))=S (R) for all elements σ ∈Aut(R).
0 0 0
Now, statement 1 follows from the universal property of the localization at S (R).
0
2. Statement 2 follows from statement 1. (cid:3)
For a ring R, let R∗ be its group of units and Inn(R) := {ω |u ∈ R∗} be the group of inner
u
automorphisms of the ring R where ω (r)=uru−1 is the inner automorphism determined by the
u
element u. The next proposition is used in the proof of Theorem 2.8.
Proposition 2.7 Let R be a ring, S ∈Den(R,0), and T ∈Den (S−1R,0); and so R⊆S−1R⊆
l l
T−1(S−1R) are natural inclusions of rings. Then
1. T−1(S−1R)= T−1(S−1R) for some T ∈ Den (S−1R,0) such that S,S−1 ⊆T ; in particu-
1 1 l 1
lar, sT s−1 =T for all s∈S.
1 1
6
2. If, in addition, S ⊆T then T ∩R∈Den (R,0) and S ⊆T ∩R.
l
Proof. 1. For each s ∈ S, T−1(S−1R) = ω (T−1(S−1R)) = ω (T)−1(S−1R) and ω (T) ∈
s s s
Den (S−1R,0). It suffices to take T := hS,S−1,ω (T)|s ∈ Si in Den (S−1R,0) since clearly
l 1 s l
S,S−1 ∈ Den (S−1R,0) and, for each non-empty finite subset J of S, ( ω (T))−1(S−1R) =
l s∈J s
T−1(S−1R). In more detail, Q
T−1(S−1R)= ( ω (T))−1(S−1R)= T−1(S−1R)=T−1(S−1R).
1 s
[ Y [
∅6=J⊆I,|J|<∞ s∈J ∅6=J⊆I,|J|<∞
2. The set T′ := T ∩R is a multiplicatively closed subset of C that contains the set S. It
R
remainstoshowthattheleftOreconditionholdsforT′ inR: foreachelementst′ ∈T′ andr ∈R,
T′r ∩Rt′ 6= ∅. Since T ∈ Den (S−1R,0), Tr∩S−1Rt′ 6= ∅. Take an element, say u, from the
l
intersection. The element u can be written in two different ways as follows
s−1t ·r =s−1r t′
1 1 2 2
for some s−1t ∈T and s−1r ∈S−1R where s ,s ∈S and t ,r ∈R. Clearly, t =s ·s−1t ∈
1 1 2 2 1 2 1 2 1 1 1 1
R∩T =T′ (since S ⊆T), and s s−1 =s−1r for some s ∈S and r ∈R. Then the element
1 2 3 3 3 3
s t ·r =s s ·s−1t r =s s s−1r t′ =r r t′
3 1 3 1 1 1 3 1 2 2 3 2
belongs to the intersection Tr∩Rt′ since s t ∈T (since S ⊆T) and r r ∈R. (cid:3)
3 1 3 2
The group of units Q (R)∗ of Q (R). For a ring R and its largest left quotient ring Q (R),
l l l
Theorem 2.8 is used in the proof of Theorem 2.9 and gives an answers to the following natural
questions:
• What is S (Q (R))?
0 l
• What is S (Q (R))∩R?
0 l
• What is the group Q (R)∗ of units of the ring Q (R)?
l l
• Is the natural inclusion Q (R)⊆Q (Q (R)) an equality?
l l l
Theorem 2.8 1. S (Q (R))=Q (R)∗ and S (Q (R))∩R=S (R).
0 l l 0 l 0
2. Q (R)∗ = hS (R),S (R)−1i, i.e. the group of units of the ring Q (R) is generated by the
l 0 0 l
sets S (R) and S (R)−1 :={s−1|s∈S (R)}.
0 0 0
3. Q (R)∗ ={s−1t|s,t∈S (R)}.
l 0
4. Q (Q (R))=Q (R).
l l l
Proof. 1–3. It is obvious that G := hS (R),S (R)−1i ⊆ Q (R)∗ ⊆ S (Q (R)). Applying
0 0 l 0 l
Proposition 2.7.(2) in the situation where S =S (R) and T =S (Q (R)) we see that
0 0 l
S (R)⊆T′ :=S (Q (R))∩R∈Den(R,0),
0 0 l l
and so S (R) = T′, by the maximality of S (R). Let q ∈ S (Q (R)). Then q = s−1t for some
0 0 0 l
elements s ∈ S (R) and t = sq ∈ S (Q (R))∩R = S (R). Therefore, S (Q (R)) ⊆ {s−1t|s,t ∈
0 0 l 0 0 l
S (R)} ⊆ G, and so G = Q (R)∗ = S (Q (R)) = {s−1t|s,t ∈ S (R)}. This proves statements
0 l 0 l 0
1–3.
4. Statement 4 follows from statement 1. (cid:3)
Necessary and sufficient conditions for Q (R) to be a semi-simple ring. A ring Q is
l
called a ring of quotients if every element c∈C is invertible. A subring R of a ring of quotients
Q
Q is called a left order in Q if C is a left Ore set and C−1R = Q. A ring R has finite left rank
R R
(i.e. finite left uniform dimension) if there are no infinite direct sums of nonzero left ideals in R.
The next theorem gives an answer to the question of when Q (R) is a semi-simple ring.
l
7
Theorem 2.9 The following properties of a ring R are equivalent.
1. Q (R) is a semi-simple ring.
l
2. Q (R) exists and is a semi-simple ring.
cl
3. R is a left order in a semi-simple ring.
4. R has finite left rank, satisfies the ascending chain condition on left annihilators and is a
semi-prime ring.
5. A left ideal of R is essential iff it contains a regular element.
If one of the equivalent conditions hold then S (R)=C and Q (R)=Q (R).
0 R l cl
Proof. Goldie’s Theorem states that 2⇔3⇔4⇔5.
(3⇒1) If statement 3 holds then Q (R)=C−1R is a semi-simple ring.
l R
(1⇒3)WehavetoshowthatC isaleftOresetsincethenC =S (R)andQ (R)=Q (R)is
R R 0 l cl
asemi-simplering. Notice thatS (R)⊆C . Letq ∈C . ThentheQ (R)-modulemonomorphism
0 R R l
·q : Q (R) → Q (R), x 7→ xq, is an isomorphism, and its inverse is necessarily of the form ·p for
l l
some element p ∈ Q (R). Clearly, pq = 1 and qp = 1, i.e. q ∈ Q (R)∗. By Theorem 2.8.(3),
l l
q =s−1t for some elements s,t∈S (R). We have to show that, for each q ∈C and r ∈R, there
0 R
exists an element c∈C such that crq−1 =crt−1s∈R. Since t∈S (R), there exists an element
R 0
s ∈S (R) such that s rt−1 ∈R. It suffices to take c=s since S (R)⊆C . (cid:3)
1 0 1 1 0 R
The next corollary gives an interesting criterion of when the classical quotient ring Q (R) =
cl
C−1R exists.
R
Corollary 2.10 If the ring Q (R) is a left artinian ring then S (R)=C and Q (R)=Q (R).
l 0 R cl l
Proof. Each left artinian ring is left Noetherian, hence the left Q (R)-module Q (R) has finite
l l
length. Now, we can repeat the proof of the implication 1⇒3 of Theorem 2.9, where, in fact, we
only used the fact that the left Q (R)-module Q (R) has finite length. (cid:3)
l l
Proposition 2.11 (Proposition 11.6, [6]; [4]) Let A be a subring of a ring B. If M is a finitely
generated flat A-module such that B ⊗ M is a projective B-module then M is a projective A-
A
module.
Corollary 2.12 If there exists a finitely generated flat R-module M which is not projective the
the ring Q (R) is not a semi-simple ring.
l
Proof. IfQ (R)wereasemi-simpleringthenQ (R)⊗ M wouldbeaprojectiveQ (R)-module,
l l R l
and so M would be a projective R-module, by Proposition 2.11, a contradiction. (cid:3)
3 The maximal left quotient rings of a ring
In this section, a new class of rings, the class of left localization maximal rings, is introduced. It
is proved that, for an arbitrary ring R, the set of maximal elements of the poset (Den (R),⊆)
l
is a non-empty set (Lemma 3.7.(2)), and therefore the set of maximal left quotient rings of the
ring R is a non-empty set. A criterion is given (Theorem 3.11) for a left quotient ring of a ring
to be a maximal left quotient ring of the ring. Many results on denominator sets are proved. In
particular, for each denominator set S ∈Den (R,a), connections are established between the left
l
denominator sets Den (R,a), Den (R/a,0) and Den (S−1R,0).
l l l
Proposition 3.1 1. For each ring A∈Loc (R,a) where a∈Ass (R), the set Den(R,a,A):=
l l l
{S ∈Den (R,a)|S−1R=A} is an ordered submonoid of Den (R,a), and
l l
8
2. S(R,a,A):= S is its largest element. In particular, S (R)=S(R,0,Q(R)).
S∈Denl(R,a,A) 0 l
S
3. Let S ∈ Den (R,a,A), i ∈ I, where I is an arbitrary non-empty set. Then hS |i ∈ Ii ∈
i l i
Den(R,a,A) (see (13)). Moreover, hS |i∈Ii is the least upper bound of the set {S } in
l i i i∈I
Den(R,a,A) and in Den (R,a).
l l
Proof. 1. In view of Theorem 2.1, it suffices to show that if S ,S ∈ Den (R,a,A) then
1 2 l
S S ∈ Den(R,a,A), i.e. (S S )−1R = A. By Theorem 2.1, S S ∈ Den (R,a). Notice that
1 2 l 1 2 1 2 l
A=S−1R=S−1R. For each s=s ···s ∈S S where all s ∈S ∪S , and, for each r ∈R,
1 2 1 n 1 2 i 1 2
s−1r =s−1···s−1r ∈s−1···(s−1A)⊆s−1···(s−1A)⊆···⊆A.
n 1 n 2 n 3
Therefore, (S S )−1R=A.
1 2
2. By Theorem 2.1, S(R,a,A) = hS|S ∈ Den (R,a,A)i ∈ Den (R,a); and S(R,a,A)−1R =
l l
injlimS−1R=injlimA=A where the injective limit is over S ∈Den (R,a,A).
l
3. Repeat the proof of statement 2 replacing Den(R,a,A) by I. (cid:3)
l
Lemma 3.2 1. Let S ∈Den (R,a), b be an ideal of the ring R such b⊆a, and π :R→R/b,
l
a7→a=a+b. Then π(S)∈Den (R/b,a/b) and S−1R≃π(S)−1(R/b).
l
2. Let S ,S ∈Den(R) and S ⊆S . Then
1 2 l 1 2
(a) a := ass(S ) ⊆ a := ass(S ); there is the R-ring homomorphism ϕ : S−1R → S−1R,
1 1 2 2 1 2
s−1a7→s−1a; and ker(ϕ)=S−1(a /a ).
1 2 1
(b) Let π : R → R/a , a 7→ a = a + a , and S be the multiplicative submonoid of
1 1 1 2
(S−1(R/a ),·) generated by π (S ) and π (S )−1 = {s−1|s ∈ S }. Then π (S ),S ∈
1 1 1 2 1 1 e 1 1 2 2
Den (S−1R,S−1(a /a )) and S−1(S−1R)≃π (S )−1(S−1R)≃S−1R.
l 1 1 2 1 2 1 1 2 1 2 e
e
Proof. 1. Since π is an epimorphism and b ⊆ a, π(S) is a left Ore set in R/a. Clearly,
a/b ⊆ ass(π(S)). To prove that the inverse inclusion holds, let a ∈ ass(π(S)). Then 0 = sa
for some element s ∈ π(S). Then b := sa ∈ b. Since b ⊆ a, we can find an element t ∈ S
such that tb = 0. Then (ts)a = 0, and so a ∈ a. Therefore, ass(π(S)) = a/b. To prove that
π(S) ∈ Den (R/b), it remains us to show that as = 0, for some a ∈ R/b and s ∈ π(S), implies
l
that ta = 0 for some t ∈ S. Clearly, b := as ∈ b ⊆ a, and so 0 = s b = (s a)s for some s ∈ S.
1 1 1
It follows that s a∈a since S ∈Den (R,a). We can find an element s ∈S such that s s a=0.
1 l 2 2 1
Therefore, π(S)∈Den (R/b,a/b). It suffices to take t=s s ∈S.
l 2 1
By the universal property of the ring S−1R, there is the ring epimorphism
S−1R→π(S)−1(R/b), s−1r 7→π(s)−1π(r).
Bytheuniversalpropertyoftheringπ(S)−1(R/b),thereistheringepimorphismπ(S)−1(R/b)→
S−1R, π(s)−1π(r)7→s−1r. Therefore, S−1R≃π(S)−1(R/b).
2a. The inclusion a ⊆ a is obvious. The image S of the Ore set S under the ring epi-
1 2 1 1
morphism π : R → R/a is an Ore set in R/a since S ⊆ S and S ∈ Den (R,a ). Therefore,
2 2 1 2 2 l 2
eachelementof the subring ofS−1R generatedby π(R) and S has the formπ(s)−1π(r) for some
2 1
s ∈ S and r ∈ R. By the universal property of the ring S−1R, the map ϕ exists. An element
1 1
s−1r ∈S−1R, where s∈S and r ∈R, belongs to the kernel of ϕ iff s−1r =0 in S−1R iff tr =0
1 1 2
for some t∈S iff r∈a . Therefore, ker(ϕ)=S−1(a /a ).
2 2 1 2 1
2b. First, we prove that the set π (S ) (resp. S ) is a left Ore set in S−1R, i.e. we have to
1 2 2 1
prove that for each element s ∈ π (S ) (resp. s ∈ S ) and s−1r ∈ S−1R where s ∈ S and
2 1 2 2e 2 1 1 1 1
r ∈R, there exist elements t∈π (S ) (resp. t∈S ) and a∈S−1R such that ts−1r =as . In the
1 2 2 e 1 1 2
ring S−1R, the element s−1rs−1 can be written as s−1r for some s ∈S and r ∈R. In S−1R,
2 1 2 e 3 1 3 2 1 1
b:=s s−1r−r s ∈ker(ϕ)=S−1(a /a ), by statement 2(a). Therefore, there exists an element
3 1 1 2 1 2 1
s ∈S such that s b=0, and so s s ·s−1r=s r·s . It suffices to take t=s s ∈π (S )⊆S
4 2 4 4 3 1 4 2 4 3 1 2 2
and a=s r.
4 e
9
It is obvious that S−1(a /a ) ⊆ ass(π (S )) ⊆ ass(S ). If u ∈ ass(S ) then su = 0 for some
1 2 1 1 2 2 2
s ∈ S . Then 0 = ϕ(su) = ϕ(s)ϕ(u), and so u ∈ ker(ϕ) = S−1(a /a ) (by statement 2(a)) since
2 e 1 2 1 e
ϕ(s) is a unit. Therefore, S−1(a /a )=ass(π (S ))=ass(S ).
e 1 2 1 1 2 2
Ifvs=0forsomeelementsv ∈S−1Rands∈π (S )(resp. s∈S ). Then0=ϕ(v)ϕ(s),andso
1 1 2 e 2
ϕ(v)∈ker(ϕ)=S−1(a /a )sinceϕ(s)isaunit. Thisprovesthatπ (S )∈Den (S−1R,S−1(a /a ))
1 2 1 1e 2 l 1 2 1
(resp. S ∈ Den (S−1R,S−1(a /a ))). Now, it is obvious that π (S )−1(S−1R) ≃ S−1R ≃
2 l 1 2 1 1 2 1 2
S−1(S−1R). (cid:3)
2 1 e
e
The set (Loc (R,a),→) is a poset where A → A if A = S−1R and A = S−1R for some
l 1 2 1 1 2 2
denominator sets S ,S ∈Den (R,a) with S ⊆S , and A →A is the inclusion map in Lemma
1 2 l 1 2 1 2
3.2.(2a). If (S′,S′) is another such a pair then, by Proposition 3.1.(1), A = S−1R = S′−1R =
1 2 1 1 1
(S S′)−1R→A =S−1R=S′−1R=(S S′)−1R; S S′,S S′ ∈Den (R,a) with S S′ ⊆S S′.
1 1 2 2 2 2 2 1 1 2 2 l 1 1 2 2
In the same way, the poset (Loc (R),→) is defined, i.e. A → A if there exists S ,S ∈
l 1 2 1 2
Den (R) such that S ⊆ S , A = S−1R and A = S−1R, A → A stands for the map ϕ :
l 1 2 1 1 2 2 1 2
S−1R→S−1R (Lemma 3.2.(2a)). The map
1 2
(·)−1R:Den (R)→Loc (R), S 7→S−1R, (2)
l l
is an epimorphism of the posets (Den (R),⊆) and (Loc (R),→). For each ideal a ∈ Ass (R), it
l l l
induces the epimorphism of posets (Den (R,a),⊆) and (Loc (R,a),→),
l l
(·)−1R:Den (R,a)→Loc (R,a), S 7→S−1R. (3)
l l
The sets Den(R) and Loc (R) are the disjoint unions
l l
Den (R)= Den (R,a), Loc (R)= Loc (R,a). (4)
l l l l
G G
a∈Assl(R) a∈Assl(R)
For each ideal a∈Ass (R), the set Den (R,a) is the disjoint union
l l
Den (R,a))= Den (R,a,A). (5)
l l
G
A∈Locl(R,a))
Let LDen (R,a):={S(R,a,A)|A∈Loc (R,a)}, see Proposition 3.1.(2). The map
l l
(·)−1R:LDen (R,a)→Loc (R,a), S 7→S−1R, (6)
l l
is an isomorphism of posets.
For a left denominator set S ∈Den (R,a), there are natural ring homomorphisms
l
R→R/a→S−1R.
Lemma3.3andProposition3.4establishconnectionsbetweentheleftdenominatorsetsDen (R,a),
l
Den (R/a,0) and Den (S−1R,0).
l l
Let S,T ∈ Den (R). The denominator set T is called S-saturated if sr ∈ T, for some s ∈ S
l
and r ∈R, then r ∈T, and if r′s′ ∈T, for some s′ ∈S and r′ ∈R, then r′ ∈T.
Lemma 3.3 Let S ∈Den(R,a), π :R→R/a, a7→a+a, and σ :R→S−1R, r 7→r/1.
l
1. Let T ∈Den (S−1R,0) be such that π(S),π(S−1)⊆T. Then T′ :=σ−1(T)∈Den (R,a), T′
l l
is S-saturated, T ={s−1t′|s∈S,t′ ∈T′}, and S−1R⊆T′−1R=T−1R.
2. π−1(S (R/a))=S (R), π(S (R))=S (R/a)) and Q (R)=S (R)−1R=Q (R/a).
0 a a 0 a a l
10