Chapter II. Rings of Fractions

In § 10 (Example 3) of the preceding chapter it was briefly indicated how the well• known construction of the of fractions of a commutative integral could be generalized to arbitrary commutative rings. When trying to extend this construction to non-commutative rings, one finds that this is not always possible, but that one can give necessary and sufficient conditions for the existence of a of fractions. Such a condition was first found by Qj. Ore [1] around 1930 for the case of a skew- of a domain. The existence of a total ring of fractions of an arbitrary ring was first considered by K. Asano [1]. General rings offractions were studied by Elizarov [1], and a systematic theory of rings and modules of fractions was developed by P. Gabriel [1, 2] in connection with his theory of general rings of quotients.

§ 1. The Ring of Fractions

Let A be a ring and let S be a multiplicatively closed subset of A, i.e. s, tES implies s tES and 1ES. We define a right ring of fractions of A with respect to S as a ring A[S-1] together with a ring homomorphism qJ: A-+A[S-1] satisfying: Fl.

Proposition 1.1. When A [S-1] exists, it has the following universal property: for every ring homomorphism t/I: A -+ B such that t/I (s) is invertible in B for every SES, there exists a unique homomorphism (J: A[S-1] -+ B such that (J qJ =t/I.

Proof (J is defined as (J(qJ(a) qJ(s)-1)=t/I(a) t/I(S)-1. We have to verify that this is well-defined. So suppose qJ(a) qJ(S)-1 =qJ(b) qJ(t)-1. Then qJ(a)=qJ(b) qJ(t)-1 qJ(s)

B. Stenström, Rings of Quotients © Springer-Verlag Berlin Heidelberg 1975 § 1. The Ring of Fractions 51

=q>(b) q>(c) q>(U)-l for some CEA and UES, by F2. This gives

q> (a) q> (u) = q> (b) q> (c) and q>(s) q>(u)=q>(t) q>(c), from which we obtain with the use of F 3 that a u v = b c v and s u Vi = t C Vi for some v and Vi in S. Since tjJ (v) and tjJ (Vi) are invertible, this implies

tjJ (a) tjJ (u) = tjJ (b) tjJ (c) and tjJ(s) tjJ(u)=tjJ(t) tjJ(c), and we can go backwards to find that tjJ(a) tjJ(S)-l = tjJ (b) tjJ(t)-l. We leave to the reader to verify that (J is a homomorphism. It is clear that (J q> = tjJ and that (J is unique. 0 Because of the unicity of the solution of a universal problem we may draw the following two conclusions:

Corollary 1.2. A [S -1 J is unique up to isomorphism. Corollary 1.3. If both A[S-lJ and [S-lJ A exist, then they are naturally iso• morphic. The existence of a solution of the universal problem stated in Prop.l.1 does not imply that A[S-lJ exists. This is clear since it may happen that [S-lJ A exists but A[S-lJ does not exist (see Example 5 below). Note that it follows from F 3 that the canonical homomorphism q>: A ~ A [S-l J is a monomorphism if and only if S is contained in the set of non-zero-divisors. When so is the case, A is usually identified with q>(A), and elements in A[S-lJ may be written as as-i. We now turn to the question of the existence of A[S-l]. It is rather easy to show that the universal problem in Prop. 1.1 always has a solution (in a presenta• tion of A by generators and relations, one adjoins a new generator s for each SE S, and adds the new relations s s = s s = 1). However, this solution is practically useless because in general it is difficult to tell what its elements look like, e.g. it is hard to determine the kernel of the canonical homomorphism from A into it. Proposition 1.4. Let S be a multiplicatively closed subset of A. A[S-lJ exists if and only if S satisfies: S1. If SES and aEA, there exist tES and bEA such that s b=a t. S2. If sa=O with SES, then at=O for some tES. Men A[S-lJ exists, it has the form A[S-lJ =A x Sf'" where'" is the equivalence relation defined as (a, s)",(b, t) if there exist c, dEA such that ac=bd and sc=tdES.

Proof Suppose A[S-lJ exists and let q>: A~A[S-lJ be the canonical map. We first verify S 1. If aEA and SE S, then F2 gives q> (S)-l q> (a) = q> (b) q>(t)-l for some bEA and tES. Hence q>(a t)=q>(s b), and by F3 this means that at u=s b u 52 Chapter II. Rings of Fractions

for some UES. Since tUES, this gives S1. If sa=O with SES, then qJ(a)=O by F1, and F3 implies that a t=O for some tES. Thus we also have S2. Now suppose instead that Sl and S2 are satisfied. It is routine matter to verify that the stated relation ~ is an equivalence relation. We form A x SI ~ and define addition and multiplication in the obvious way: (a,s)+(b,t)=(ac+bd,u) where u=sc=tdES, (a,s)· (b,t)=(ac,tu) where sc=bu and UES. One then has to verify that these definitions are independent of the choice of representing couples. This we leave to the industrious reader, as well as the verification that A x SI ~ is a ring and that aH(a, 1) is a ring homomorphism qJ: A _ A x SI ~. It is easy to see that we have arrived at a right ring of fractions: the inverse of qJ(s) is (1, s) when SES, and (a, s)=qJ(a) qJ(S)-l; finally, the kernel of qJ is {aEAla s=O for some SES}. D S is a right denominator set when it is multiplicatively closed and satisfies S 1 and S2. Similarly one defines left and two-sided denominator sets. Sometimes S is called right permutable if it satisfies S 1 and right reversible if it satisfies S 2. We will now show that Sl implies S2 under certain circumstances. A right of the form r(S)= {aEA IS a=O}, for some subset S of A, is called a right annihilator. Proposition 1.5. Assume A satisfies ACC on right annihilators. If S is multi• plicatively closed and satisfies S 1, then S is a right denominator set. Proof Suppose s a=O for some SES. There exists an n such that r(i)= r(sn) for all k ~n. By Slone can find tES and bEA such that ~ b=a t. Then sn+l b= sa t=O, so bEr(~+l)=r(sn). Hence a t=sn b=O, which establishes S2. D

Examples 1. Commutative rings. When A is a , both Sl and S2 are automatically satisfied, so every multiplicatively closed set S is a denominator set. 2. Ore rings. The most important example of a multiplicatively closed set is the set Sreg of all regular elements (i.e. non-zero-divisors) of A. The ring of fractions A [S;;;~] is usually called the classical right ring of quotients (or sometimes the total right ring of fractions) of A. We will denote it by Q~I(A), or more often simply by Qcl' The corresponding left-hand notation is Q~I(A). Recall that when both Q~l (A) and Q~l (A) exist, they coincide. Since condition S2 is automatically satisfied, we have: Proposition 1.6 (Ore [1 ]). Q~l (A) exists if and only if A satisfies the right Ore condition, i.e. for a and s in A with s regular, there exist band t in A with t regular, such that a t=s b. 3. Ore domains. If A has no zero-divisors, the right Ore condition states that a An sA :f: 0 for all non-zero elements a, s of A, and this is the same as saying that an b:f: 0 for all non-zero right ideals a and b. A ring without zero-divisors and satisfying the right Ore condition is called a right Ore domain, and its classical right ring of quotients is a skew-field. Numerous examples of Ore domains are obtained from: § 1. The Ring of Fractions 53

Proposition 1.7. Every right without zero-divisors is a right Ore domain.

Proof. Let a and b be non-zero elements of A. Consider the right ideals an = b A + a b A + ... + an b A. Since A is right noetherian, there exists a smallest n for which 0n=on+l' and then an+lb=bco+abcl+···+anbcn. This gives

b Co =a(an b -b C1 - •.• _an- 1 b cn}=t::O where the minimality of n implies that the bracket is =1= O. Thus bAn a A =1= 0, which is the right Ore condition. 0 This result will be considerably improved in the next section. 4. Bezout domains. A right Bezout domain is a ring without zero-divisors in which every finitely generated right ideal is principal. Proposition 1.8. Every right Bezout domain is a right Ore domain. Proof. If A does not satisfy the Ore condition, then there exist non-zero elements a and b of A such that aAnbA=O. Put aA+bA=cA and write b=cd. Then aA=aA/(aAnbA)~(aA+bA)/bA=cA/bA~A/dA. Since d=l=O, this is a con• tradiction. 0 5. Skew polynomial rings. Our aim is to construct examples of rings which are Ore domains to the right but not to the left. Let K be a field and qJ: K ---+ K an endomorphism of K. The elements of the skew polynomial ring Kq> [X] are formal polynomials ao + X a1 + ... + xn an with addition defined as usual, but with multiplication given by aX =X qJ(a), i.e. xm a. xn b= xm+n qJn(a) b. This gives rise to a ring which contains K as a sub ring. Each element f of Kq>[X] has a well-defined degree deg(f), and one has deg(fg) = deg(f) + deg(g). It follows in particular that Kq>[X] has no zero-divisors. It is easily ve.rified that Kq>[X] is a right euclidean domain, i.e. for any f and g there exist q and r with deg r < deg q such that f = g q + r. This implies that Kq> [X] is a right principal domain and therefore a right Ore domain by Prop. 1.7 (or 1.8). If qJ is an automorphism of K, one can perform the euclidean division algorithm also to the left, and Kq> [X] is a left Ore domain. But if there exists aEK which is not in the image of qJ, then the left Ore condition fails for the couple (X, X a), as one easily verifies. Thus Kq> [X] is a left Ore domain if and only if qJ is an auto• morphism of K. 6. Rings of quotients. A ring A is called a ring of quotients if every non-zero• divisor of A is invertible, i.e. A is its own classical right and left ring of quotients. E.g. every regular ring is a ring of quotients, as one easily sees. Also rings with minimum condition on right ideals are rings of quotients, in fact one has more generally: Proposition 1.9. If A satisfies DCC on principal right ideals, then A is a ring of quotients. 54 Chapter II. Rings of Fractions

Here DCC stands for the" descending chain condition ", i.e. every strictly de• scending chain (of principal right ideals) is finite. The assertion is a consequence of: Lemma 1.10. Suppose A satisfies DCC on principal right ideals. The following properties of an element a are equivalent: (a) a is invertible. (b) a is regular. (c) r(a)=O.

Proof Obviously (a) ~ (b) ~ (c). Assume the element a satisfies condition (c). Consider the descending chain aA=>a 2 A=> .. ·. By hypothesis, an A=a"+1 A for some n. Thus an=an+1 b for some bEA, and then a"(l-a b)=O, and (c) implies 1 =a b. We then also have a b a=a, and (c) implies b a= 1. Hence (c) ~ (a). 0 Given a ring Q of quotients, one may be interested in investigating those subrings of Q for which Q is the classical right ring of quotients. Such subrings are called right orders in Q.

§ 2. Orders in a Semi-Simple Ring

In this section we prove a theorem by Goldie [2] which characterizes those rings which have a semi-simple classical ring of quotients. Before stating the theorem we have to make some definitions. The ring A has finite right rank if there are no infinite direct sums of non-zero right ideals within A, i.e. A does not contain any right ideal of the form a = a1 Ef>a2 Ef> .. · with ai=FO, iEN. Every right noetherian ring has of course finite right rank. A right ideal a is nilpotent if an = °for some n, and is nil if every element of a is nilpotent. The ring A is prime if there are no non-zero two-sided ideals a and b such that a b=O, and A is semi-prime if it has no non-zero nilpotent ideals. Finally, the right ideal a is essential if an b =F 0 holds for every right ideal b =F O. We note:

Lemma 2.1. The ring A is semi-simple if and only if it has no proper essential right ideals.

Proof If A is semi-simple, then every proper right ideal of A is a direct summand and can therefore not be essentia1. Conversely, suppose there are no essential right ideals =F A. If a is any right ideal of A, one can use Zorn's lemma to find a right ideal b such that b is maximal with the property that an b = 0. But then a + b is an essen• tial right ideal of A, for if c is a right ideal such that (a + b) n c = 0, then one has an(b+c)=O, which implies that c=O by the maximality of b. It follows that a Ef> b = A, so every right ideal of A is a direct summand. Hence A is semi-simple. 0

Theorem 2.2 (Goldie). The following properties of the ring A are equivalent: (a) A is a right order in a semi-simple ring. (b) A has finite right rank, satisfies ACC on right annihilators and is a semi• prime ring. (c) A right ideal of A is essential if and only if it contains a regular element. § 2. Orders in a Semi-Simple Ring 55

Proof (a) => (b): Let Q be the semi-simple right ring of quotients of A.1f 0 and b are right ideals of A such that 0 n b = 0, then it is easy to see that also 0 Q n b Q = 0. Since Q has finite right rank, this observation implies that also A has finite right

rank. A satisfies ACC on right annihilators because rA (S) = rQ(S) n A for any S c A, where the subscripts indicate in which ring the annihilator is taken. Finally, let 0 be a nilpotent two-sided ideal in A. 1(0) is essential as a right ideal of A, because if b is any non-zero right ideal of A, then there is a largest integer n2':O such that bon=l=O, and clearly bnl(o):=>bon=l=O. But if 1(0) is essential in A, then l(o)Q is an essential right ideal in Q, as one easily verifies. Since Q is semi• simple, this means that 1(0) Q=Q, and we may write 1 = L ai qi with aiEl(o). There exists a regular element SE A such that all qi SE A. Then s 0 = L ai qi soc L ai 0 = 0, and it follows that 0 = 0. (b) => (c): Every non-zero-divisor generates an essential right ideal: Lemma 2.3. If A has finite right rank and r(s)=O, then sA is an essential right ideal. Proof If 0 is a non-zero right ideal such that sAn 0 = 0, then it is easy to see ~hat one gets a direct sum 0 Et> s 0 EEl ... Et> sn 0 Et> •.. in A. 0 It requires more work to show that every essential right ideal contains a regular element. Here we follow the proof given by Goldie in [7]. We need: Lemma 2.4. If A is semi-prime and satisfies ACC on right annihilator ideals, then A has no right or left nil ideals =1=0. Proof Since Aa is nil if and only if aA is nil, it is enough to consider a left nil ideal Aa. Assume Aa=l=O. Among the non-zero elements of Aa, choose one bEAa with maximal right annihilator. For each cEA, let k be the smallest integer such that (c b)k=O. Since r(b)cr((c b)k-1), the choice of b implies that we actually have r(b)=r((c bt-1). It follows that c bEr(b) and hence bAb=O by the arbitrary choice of c. From this we obtain (AbA)2=0, but every nilpotent ideal is zero, so AbA=O. Hence b=O, which is a contradiction, and we must have Aa=O. 0

Now let 0 be an essential right ideal of A. We have to find a regular element in o. Since A has ACC on right annihilators and 0 is not a nil ideal (by the preceding lemma), there exists al =1=0 in 0 such that r(al)=r(af).1f onr(al)=O, we continue and choosea2 EO n r(al)such thata2 =1= Oandr(a2)= r(a~).lfthen 0 n r(a1) n r(a2) =1=0, we go on and get a3 EO n r(al) n r(a2), and so on. At each step we obtain a direct sum al A Et> ... Et> ak A. This is proved by induction: suppose a1 A Et> ... Et> ak- l A is direct and ak b = al bl + ... + ak- l bk- l ; since for each i < k we have ai ak = 0, we k-l get biEr(af)=r(aJ and hence L ai bi=O. But A has finite right rank, so the process I must stop at some stage, where we have onr(a1)n ... nr(ak)=O. Then r(at)n··· nr(ak)=O, so if c=at + ... +akEo, then r(c)=O. c is thus the kind of element we are looking for, but it remains to show that also l(c)=O. Note that cA is an essential right ideal by Lemma 2.3. Define the right singular ideal of A as Z={aEAlr(a) is an essential right ideal of A}. We leave as an exercise to show that Z is a two-sided ideal. Since l(c)cZ and A contains no nilpotent ideals =1=0, we may conclude that l(c)=O by using: 56 Chapter II. Rings of Fractions

Lemma 2.5. If A satisfies ACC on right annihilators, then the right singular ideal is nilpotent.

Proof We will show that the ascending chain r(Z)cr(Z2)c ... would be strictly ascending if Z were not nilpotent. If Zn=l=O, choose an element aEZ with zn -1 a =1= 0 and largest possible right annihilator. F or each bE Z we have r(b) n a A =1= 0 since r(b) is essential in A. So there exists c E A such that a c =1= 0, but b a c = O. This means that reb a) is strictly larger than rea), and by the choice of a we must therefore have zn-l b a=O. Since bEZ is arbitrary, we get zn a=O and hence r(Zn-l) is strictly contained in r(Zn). 0

(c) => (a): We first verify the right Ore condition. Let s be a regular element and a arbitrary in A. Since sA is essential, also (sA:a)= {cia cEsA} is an essential right ideal, as is easily checked, and it therefore contains a regular element t. Thus a t = s b for some bE A. To prove that Q = Qcl (A) is semi-simple, it suffices by Lemma 2.1 to show that Q has no proper essential right ideals. If a is any right ideal of Q, then (a n A) Q = a since every element qEa can be written as q=as- 1 with a=qsEanA. If fur• thermore a is an essential right ideal of Q, then a n A is an essential right ideal of A and thus contains a regular element t. Since t is invertible in Q, a = (a n A) Q = Q. This concludes the proof of the theorem. 0

As a special case of Goldie's theorem we obtain a characterization of those rings A for which Qcl(A) is a simple ring (Goldie [1], Lesieur and Croisot [2]):

Proposition 2.6. The following properties of A are equivalent: (a) A is a right order in a simple ring. (b) A has finite right rank, satisfies ACC on right annihilators and is a prime ring.

Proof (a) => (b): It only remains to show that A is a prime ring. Let a and b be two-sided ideals of A with a b = O. Then b Qa n A is a two-sided ideal of A and (b Q a" A)2 = O. Since A has no non-zero nilpotent ideals by the theorem, we have b Q a n A = O. But this clearly implies that b Q a = O. Since Q is a simple ring, it is prime, and therefore b or a is zero. (b) => (a): Since every semi-simple prime ring is simple, it suffices to show that Q is a prime ring. Suppose a and b are two-sided ideals of Q such that a b = O. Then (anA)(bnA)=O, and A prime implies that anA or bnA is zero, which means that a or b is zero. 0

Examples 1. Noetherian rings. Every semi-prime right noetherian ring has a semi-simple classical right ring of quotients.

2. Group rings. Let A be a commutative with field offractions K, and let G be a finite group. Then K[G] is an artinian ring, and is therefore a ring of quotients by Prop. 1.9. Clearly A [G] is a two-sided order in K[G]. By Prop. 1.7.9, K [G] is a semi -simple ring if the characteristic of K does not divide the order of G. §3. Modules of Fractions 57

§ 3. Modules of Fractions

Let S be a right denominator set in the ring A. For each right A- M we define its module of fractions with respect to S as M[S-I] = M ®A A[S-I] with its canonical structure as a right A[S-I]-module. This is a reasonable definition in view of the following two results:

Proposition 3.1. The canonical A-linear map JiM: M ~ M [S-I] has the following universal property: for each right A[S-I]-module N and A-linear map a: M ~N there exists a unique A[S-I]-linear map a: M[S-I]~N such that a JiM=a. Proof HomA[s-l](M@A A[S-I], N)~HomA(M, HomA[s-l](A[S-I], N))~ HomA (M, N). 0

Proposition 3.2. M [S -1] ~ M x S / "', where '" is the equivalence relation defined as (x, s)",(y, t) if there exist c, dEA such that x c= y d and s c=tdES. Proof This is proved similarly to Prop. lA, by verifying that M x S/ '" solves the universal problem stated in Prop. 3.1. The operations in M x S/ '" are defined as:

(x,s)+(y,t)=(xc+yd,u) where u=sc=tdES (usingSl), (x,s)· (b,t)=(xc,tu) for (b,t)EA[S-I], wheresc=buanduES. 0

Corollary 3.3. The kernel of JiM: M ~M[S-I] consists of those XEM for which there exist s E S with x s = O. From this corollary it follows that t(M)={XEMlxs=O for some SES} is a submodule of M, called the S- submodule. M is an S-torsion module if M =t(M), i.e. if M[S-I] =0, and is S-torsion-free if t(M)=O. Lemma 3.4. M/t(M) is S-torsion-free. Proof If XEM and xEt(M/t(M)), then x sEt(M) for some SES, and this means that x s t=O for some tES. Hence xEt(M) and X=O. 0

The assignment Mt-->M[S-I] is a functor Mod-A~Mod-A[S-I]. A basic fact is that this functor is exact, or in other words: Proposition 3.5. A[S-I] is flat as a left A-module.

Proof Let a: L~M be a monomorphisIl1 iu Mod-A. We must show that the induced homomorphism L[S-I]~M[S-I] is a monomorphism. Suppose x S-1 (with xEL, SES) has the property that a(x) S-1 =0 in M[S-I]. By Prop. 3.2 this means that a(x) c=o in M for some cEA such that s CES. Since a is a mono• morphism, this implies xc = 0 in L, and then x S-1 = 0 in L [S-I]. 0 Obviously we have: Proposition 3.6. If M is a right A [S-1 ]-module, then the canonical map M ~M[S-I] is an isomorphism. This result means that the category of modules of fractions, considered as a full subcategory of Mod-A, is "equivalent" (in a sense to be made precise in Chap. IV) to the category Mod-A [S-I]. 58 Chapter II. Rings of Fractions

We want to have an intrinsic characterization of those A-modules which are modules of fractions, i.e. may be considered as A [S-1 J-modules. Definitions. (i) An A-module M is S-injective if for each right ideal a of A such that Il n S =!= 11 and each homomorphism IX: Il - M, there exists x EM such that IX(a)=x a for all aEIl. (ii) M is S-divisible if M =M s for each SES. Proposition 3.7. Assume S is a right denominator set in A. The following properties of a right A -module M are equivalent: (a) M~M[S-1]. (b) M is S-torsion-free and S-injective. (c) M is S-torsion-free and S-divisible.

Proof. (a) =:> (c): The condition (a) means that M is an A [S-1 J-module. If now XEM and x s=OforsomesES, then X=X s· S-1 =0. HenceMis S-torsion-free. For each xEM and SES we can write X=X S-l. SEM S, so M is S-divisible. (c) =:> (a): M S-torsion-free implies that M is a submodule of M[S-1]. For each X®S-lEM[S-1] we can write x=ys for some YEM by S-divisibility. Then x®s-1=y®ss-1=y®1, and so the map M_M[S-1] is an epimorphism also. (c) =:> (b): Suppose we are given IX: a-M, where a is a right ideal with SEllnS. Write IX(S)=X s for some xEM. We want to show that IX(a)=x a holds for all aEIl. By the condition Sl we may write a t=s b for some tES and bEA. Then IX(a) t=IX(S) b=x s b=x a t, and M S-torsion-free implies that IX(a)=x a. (b) =:> (c): Given XEM and SES, we define a homomorphism IX: sA-M by IX(S a)=x a. IX is well-defined because if s a=s b, then a t=b t for some tES by S2, and x a t = x b t implies x a = x b since M is S-torsion-free. By hypothesis there exists YEM such that IX(S) = y s, and thus X= y s. This shows that M =M s. 0

Examples 1. Sreg-torsion. The definition of Sreg-torsion elements was first proposed by Levy [1]. Assume Sreg is a right denominator set in A. Instead of" Sreg-torsion" and "Sreg-divisible ", one uses the shorter terms "torsion" and" divisible". Note that every Sreg-injective module is divisible, because the homomorphism IX: sA - M, given by IX(S a)=x a, is always well-defined when s is regular, and can be extended to A. Also, every is torsion-free (by the argument used in Example 1.10.1). 2. Orders in a semi-simple ring. Suppose A has a classical right ring of quotients Q. Then Q is semi-simple if and only if Sreg-injective modules are injective:

Proposition 3.8. The following properties of a ring A with a classical right ring of quotients Q are equivalent: (a) Every torsion-free and divisible right A-module is injective. (b) Every right Q-module is injective over A. (c) Every Sreg-injective right A-module is injective. (d) Q is a semi-simple ring. §4. Invertible Ideals and Hereditary Orders 59

Proof (a) <=> (b) by Prop. 3.7, and also (c) => (b) follows from this proposition. (b) => (d): Let q be any right ideal in Q. Since q is injective over A, Q splits as an A-module as Q = q EEl K. Now K is torsion-free and divisible, and is therefore a Q-module. Q thus splits as a right Q-module, so Q must be a semi-simple ring. (d) => (c): Let M be an Sreg-injective module. Consider any homomorphism IX: a ~ M from a right ideal a of A. Let b be a right ideal which is maximal with respect to on b = O. IX can trivially be extended to ex': a + b ~ M. But a + b is an essential right ideal of A (see the proof of Lemma 2.1) and therefore contains a regular element s (Theorem 2.2). Since M is Sreg -injective, rl can be extended to the whole of A. 0

§ 4. Invertible Ideals and Hereditary Orders

Let A be a right order in the ring Q. As usual we let Sreg denote the set of regular elements of A. For each right A-submodule I of Q we put 1* = {qEQlqI cAl.

Proposition 4.1. If I is a right A-submodule of Q with I nSreg=l=~, then I*~ HomA(I, A).

Proof If q E1*, then x H q x defines a homomorphism I ~ A. Conversely, suppose cP: I ~A is a homomorphism. Let SEI nSreg . For each XEI there exists tESreg with xtEA, and by the Ore condition there exists bEA and UESreg such that sb=xtu. Then

cp(x) tu = cp(x tu) = cp(s b) = cp(s) b = cp(s) S-1 S b = cp(s) S-1 X tu,

which implies cp(x) = cp(s) S-1 x with cp(S)S-IEI*. 0 Definition. A right A-submodule I of Q is right invertible if there exist aI' ... ,anEI and ql' ... ,qnEI* such that l=Laiqi' Note that under these circumstances we have for each XEI that X= L aiqix with qixEA, so aI' ... ,an generate the module I. Thus: Proposition 4.2. Every right invertible A-submodule of Q is finitely generated. Proposition 4.3. A right A-submodule I of Q is right invertible if and only if InS reg =l= ~ and I is a projective module.

Proof Suppose I is right invertible, i.e. 1 = L aiqi with aiEl and qJ c A. Define homomorphisms CPi: I ~A as CPi(X)=qiX, The family (CPi' ai) clearly satis• fies the requirements of Prop. 1.6.3, and therefore makes I into a finitely generated projective module. Choose SESreg such that all qisEA. Then s= L aiqisEI nSreg . Assume conversely that I is a projective A-module with I nSreg=l=~. Let (cpj' a)J be a family of projective coordinates for I. Each CP/ I ~ A is of the form cpix)=qjx for some qjEI*. If SEI nSreg , then s= L ajcpis) = L ajqjs, and s regular implies 1 = L ajqj' Hence I is right invertible. 0 A divisible module has the following injectivity property with respect to in• vertible right ideals: 60 Chapter II. Rings of Fractions

Proposition 4.4. Suppose A is a two-sided order in Q. Let 0 be a right invertible right ideal of A and let M be a divisible module. For each homomorphism a: 0 -+ M there exists xeM such that a(a)=xafor all aeo. Proof Let al> ... ,anEO and ql' .. ·,qnEI* with Laiqi=l. Since A is assumed to be also a left order in Q, there exists SESreg such that all sqiEA. The divisibility of M implies that we can write a(ai)=xis with xl> ... ,XnEM. For every aEO we obtain a(a)=a(L aiqi a)= L a(ai)qi a= L xisqia=xa with X= LxisqiEM. 0 This result is used to prove the following characterization of hereditary orders (due to Levy [1]): Proposition 4.5. Suppose A is a two-sided order in a semi-simple ring. The following properties of A are equivalent: (a) A is right hereditary. (b) Every right ideal containing a regular element is right invertible. (c) Every divisible module is injective.

Proof (a) => (b) is a consequence of Prop. 4.3. (b) => (c): Every divisible module is Sreg-injective by Prop. 4.4, and is therefore injective by Prop. 3.8. (c) => (a): Every quotient module of an injective module is divisible, and is therefore injective by (c). Hence A is right hereditary by Prop. 1.9.5. 0 The following is another important property of orders in semi-simple rings: Proposition 4.6. If A is a two-sided order in a semi-simple ring, then every finitely generated torsion-free module is a submodule of a free module.

Proof Let M be a torsion-free module with generators Xl' ... , xm • By Cor. 3.3 there is a monomorphism M -+ M ®A Q, where Q is the classical right ring of quotients of A. Now M ®A Q is a finitely generated Q-module, and since Q is semi-simple, this implies that M ®A Q is a submodule of a finitely generated free Q-module F. Let vl' ... , Vn be a basis for F over Q. Then xj= L Vi% for certain i qijEQ. Choose seSreg such that all sqijEA. Then xj= L ViS- l . sqij' so M is im• i bedded in the free A-submodule of F generated by V1S- l, ... , vns- l. 0 Corollary 4.7. Let A be a two-sided order in a semi-simple ring. The following properties of A equivalent: (a) A is right semi-hereditary. (b) Every finitely generated torsion-free module is projective. (c) Every torsion-free module is flat. (d) Every right ideal is flat.

Proof (a) => (b): Every finitely generated torsion-free module is a submodule of a free module (Prop. 4.6), and therefore projective (Prop. I.6.9). (b) => (c): Every module is the direct limit of its finitely generated submodules, and the direct limit of projective modules is flat (Prop. LlO.3). Exercises 61

(c) ~ (d) is trivial. (d) ~ (a) follows from Prop. 1.11.6. 0

Examples 1. Dedekind domains. When A is a commutative integral domain, Prop. 4.5 gives: Proposition 4.8. The following properties of a commutative integral domain A are equivalent: (a) A is hereditary. (b) Every non-zero ideal is invertible. (c) Every divisible module is injective. A ring satisfying these conditions is called a Dedekind domain. Every Dedekind domain is a noetherian ring (by Prop. 4.2), and its invertible modules form an abelian group under multiplication. 2. Priifer domains. A semi-hereditary commutative integral domain is called a Prufer domain.

Exercises

§1 Let S be a right denominator set in A, and cp: A -> A [S-l] the canonical homomorphism. 1. Show that cp: A->A[S-l] is an epimorphism in the category of rings (as defined in Exer- cise 1.17). 2. Show that if q1' ... , q.eA[S-l], then there exists seS such that q1 s, ... , q.seq>(A). 3. Let A be a right Ore domain. Show that also the polynomial ring A[X] is a right Ore domain. 4. Consider S as a subset of M.(A) in the obvious way (diagonal matrices). Show that: (i) S is a right denominator set of M.(A). (ii) M.(A)[S-l] ~ M.(A[S-l]). (iii) UT.(A)[S-l]~UT.(A[S-l]). 5. Show that if A is injective both as a right and as a left A-module, then A is a ring of quotients.

§2 6. Let Sand T be subsets of A. Show that:

r(/(r(S»)) = r(S), r(Su T)=r(S)nr(T),

SeT implies r(S)=>r(T).

7. Show that A satisfies ACC on right annihilators if and only if A satisfies DCC on left annihilators. 8. Show that if A is a prime ring, then every two-sided ideal is essential in A. 9. Show that (i) Q•• (M.(A»)~M.(Q •.(A»). (ii) Q•• (UT.(A») ~ UT.(Q •.(A»). (Cf. Exercise 4.) 62 Chapter II. Rings of Fractions

§3 10. Let A be a right Ore domain with skew-field of fractions Q. Show that A is a left Ore domain if and only if Q is flat as a right A-module. (Hint: the Ore condition for a, b in A is obtained by noting that the relation 1 . a=ab-'· b in Q comes from a relation in A by flatness.) 11. Let S be a right denominator set in A. Let M be a right A-module and Jl: M ..... M[S-'] the canonical homomorphism. A submodule L of M is called S-saturated in M if it satisfies: if xEM and xSEL for some SES, then xEL. Show that: (i) There is a bijective correspondence between S-saturated submodules of M and A [S-']- submodules of M[S-l], given by LHL[S-'] and NHJl-l(N). (ii) Mis S-torsion-free if and only if Ann (x) is S-saturated in A for every xEM. (iii) If A is a right noetherian ring, then so is also A [S-']. 12. Let S be a right denominator set in A. Show that: (i) If M is a right A-module and N is a right A [S-']-module, then

HomA[s-,](M[S-'], N)~HomA(M, N).

(ii) A right A [S-l ]-module is injective if and only if it is injective as a right A-module. 13. Suppose S,.. is a right denominator set in A. Show that the following conditions are equivalent: (a) A is a ring of quotients. (b) There are no non-zero torsion modules. (c) Every direct product of torsion modules is a torsion module. 14. Let A be a right order in Q. Show that: (i) Every finitely generated right ideal of Q is of the form aQ, where a is a finitely generated right ideal of A. (ii) If A is right semi-hereditary, then so is also Q.

§4 15. Let A be a right order in Q. Show that if every divisible right A-module is injective, then A is right hereditary and Q is semi-simple. 16. Let R be a commutative integral domain with field of fractions K. Let Q be a semi-simple K - (Exercise l.l), which is finite-dimensional over K. (i) Let I be an R-submodule of Q. Show that the following properties of I are equivalent: (a) I contains a basis for Q as a vector space over K. (b)IK=Q. (c) For each qEQ there exists O~rER such that rqE/. (ii) Let A be an R-subalgebra of Q. Show that A is an order in Q if and only if AK = Q. (iii) Suppose R is a noetherian ring, and let A be an R-subalgebra of Q such that A is finitely generated as an R-module and AK = Q (A is then called an R-order in Q). Show that the following properties of a right A-submodule I of Q are equivalent: (a) I is finitely generated as an R-modu1e, and IK = Q. (b) There exist non-zero r, SER such that rEI and sl cA. 17. Show that if A is a right hereditary ring and is a right order in a semi-simple ring, then A is right noetherian.