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15 the Classical Ring of Left Quotients 15 The Classical Ring of Left Quotients. In Section 12 we introduced the maximal ring Qmax of left quotients of a ring R, and in the last section we considered the collection of all rings of left quotients of R and how they arise from topologies on R. Fortunately, Qmax is the largest of those rings of left quotients in which we are mostly interested. Here we begin the investigation of several important special types of rings of quotients that will nish up the material in these lectures. Specically, in this section we look at the so-called classical ring of left quotients that generalizes the familiar ring of quotients for commutative rings that we studied in garduate algebra. The one downside of our study is that these classical rings of quotients need not exist for many rings, and so part of our assignment is to learn when they do exist. Let R be a ring, and let be the set of non-zero divisors of R. A ring Q and an injective ring homomorphism ϕ : R → Q is a classical ring of left quotients of R if (Cl.1) ϕ(d) is invertible in Q for every d ∈ ; (Cl.2) Each q ∈ Q can be factored q = ϕ(d)1ϕ(a) for some d ∈ and a ∈ R. It is easy, but slightly tedious, to see that if R has a classical ring of left quotients, then it is unique is a fairly stong sense. In fact it satises she following universal property. We will leave the details as an exercise. 15.1. Proposition. Suppose that R has a classical ring of left quotients, ϕ : R → Q.IfS is a ring and : R → S is a ring homomorphism such that (d) is invertible in S for each d ∈ , then there is a unique ring homomophism : R → S with = ϕ. As we shall see in the Exercises not every ring has a classical ring of left quotients. So let’s check for some necessary conditions for R to have a classical ring of left quotients, ϕ : R → Q. Let a ∈ R and d ∈ , so that both ϕ(a) and ϕ(d)1 are in Q. Thus, by (Cl.2) there must exist b ∈ R and d0 ∈ with ϕ(a)ϕ(d)1 = ϕ(d0)1ϕ(b), or ϕ(d0a)=ϕ(d0)ϕ(a)=ϕ(b)ϕ(d)=ϕ(bd), so that d0a = bd. We say that R satises the left Ore condition or simply is left Ore in case for each a ∈ R and d ∈ , Rd ∩ a =6 ∅. Thus, we have that the left Ore condition is necessary for R to have a ring of left quotients. Shortly, we shall be able to prove that it is also sucient, and our rst step will be to nd a topology. 98 Section 15 15.2. Lemma. If R left Ore, then for every d ∈ , the left ideal Rd is dense. Proof. Recall (Lemma 12.6) that Rd is dense i for every a ∈ R, rR(Rd : a) = 0. But since R is 0 0 0 0 left Ore, there exists some d ∈ with d a ∈ Rd ∩ a,sod ∈ (Rd : a) and rR(Rd : a) rR(d )=0. This suggests that we check whether the set of left ideals of the form Rd with d ∈ can produce a topology. So we consider the set O = {I RR : I ∩ =6 ∅}. 15.3. Proposition. For a ring R the following are equivalent: (a) O is a topology; (b) O satises (T.1); (c) R is left Ore. Proof. (a) =⇒ (b) is clear. For (b) =⇒ (a) we need only show that O satises (T.2). So suppose that D ∈Oand that I is a left ideal such that (I : a) ∈Ofor all a ∈ D. But there is some d ∈ D ∩ , so (I : d) ∩ =6 ∅,sayd0 ∈ (I : d) ∩ . Then d0d ∈ I ∩ and I ∈O. (b) ⇐⇒ (c) Let a ∈ R and d ∈ . Then (Rd : a) ∈O ⇐⇒ (Rd : a) ∩ =6 ∅⇐⇒∃d0a ∈ Rd ⇐⇒ Rd ∩ a =6 ∅. 15.4. Theorem. If R is left Ore, then O is a topology and QO is a classical ring of left quotients for R. Proof. By Proposition 15.3, O is a topology, and by Lemma 15.2, OD.SoQO is a subring of Qmax, and we can view R as a subring of QO. Let d ∈ , so there exists an R-homomorphism Rd → R such that ad 7→ a for all a ∈ R. But Rd ∈O, so that there is a unique q ∈ Qmax with q(d) = 1. (See Corollary 12.5.) But then Rd (R : q), so by Corollary 14.5, q ∈ QO. But then dq = dq(1) = q(d)=1. On the other hand d(q(1)d 1)=(dq)d d = d d =0, so that Rd(q(1)d 1) = Rd(qd 1) = 0. However, Rd ∈D,soqd = 1, and d is invertible in QO. Finally, let q ∈ QO. Then by Corollary 14.5 again (R : q) ∈O, so that d ∈ (R : q) for some d ∈ . Thus, dq = a ∈ R and q = d1a. 15.5. Corollary. A ring R has a classical ring of left quotients i R is left Ore. Let R be a left Ore ring, so by Corollary 15.5, it has a classical ring of left quotients. By Proposition 15.1 this is unique to within isomorphism through R. Thus, we will take QO to be the classical ring of Non-Commutative Rings Section 15 99 left quotients of R and we will denote it simply by Qcl. A (not necessarily commutative) integral domain is a left Ore domain in case it satises the left Ore condition. This bring us to where much of this all started with the following result due originally to Ore. 15.6. Corollary. A ring R has a classical division ring of left quotients if and only if R is a left Ore domain. You should be able to note that the left Ore condition, also sometimes referred to as the common left multiple property, is really a substitute for commutativity. We shall see how this works in the Exercises. But for one particularly valuable illustration of this we prove that in Qcl there exist “common denominators”. 15.7. Lemma. Let E be a left Ore ring. If 1 1 ∈ d1 a1,...,qn an Qcl, then there exist d ∈ and b1,...,bn ∈ R with 1 1 di ai = d bi for all i =1,...,n. Proof. Since O is a topology Rd1 ∩∩Rdn ∈O, so there is some d ∈ ∩ Rd1 ∩∩Rdn.So ∈ 1 1 there are c1,...,cn with d = cidi for all i =1,...,n. But then di ai = d ciai for i =1,...,n. Exercises 15. 15.1. Let : R → S be an injective ring homomorphism that satises (Cl.1). 1 1 1 (a) Prove that for each pair d1,d2 ∈ , (d1d2) = (d2) (d1) . (b) Let ϕ : R → Q be a classical ring of left quotients for R. Prove that there exists a unique injective ring homomorphism : Q → S with ϕ = . 15.2. Let A be a commutative integral domain and let R be the ring of 2 2 matrices over A. (a) Prove that R is left Ore. (b) Find the classical ring Qcl of left quotients of R. 100 Section 15 15.3. Let M2(Z) be the ring of all 2 2 matrices over Z. Let 1 1 R = {A ∈ M (Z):A = r for some r ∈ Z}. 2 1 1 (a) Prove that A ∈ R i [ 1 1]A = s [1 1 ] for some s ∈ Z. (b) Prove that R is a subring of M2(Z). (c) Prove that R is left Ore. (d) Find Qcl for R. 15.4. Let R be the ring of all 2 2 upper triangular matrices over Z. (a) Show that R is left Ore. (b) Find the classical ring Qcl of left quotients of R. 15.5. Let R be an integral domain, not necesarily commutative. (a) Prove that R is left Ore i every non-zero principal left ideal Rd is essential in R. (b) Using the above fact show that the ring khx, yi of polynomials over a eld k in the non- commuting indeterimants x and y is not left Ore. 15.6. Let R be a left noetherian integral domain. Prove that R is left Ore. [Hint: Let x, y ∈ R be non-zero. For each n ∈ N let 2 n In = Rx + Rxy + Rxy + Rxy . Then for some n ∈ N we have In = In+1. Now use the previous exercise.] 15.7. Let R be an integral domain. Prove that if every nitely generated left ideal is principal (such a ring is left Bezout), then R is left Ore. [Hint: Let x, y ∈ R be non-zero and suppose that Rx ∩ Ry = 0. Then Rx + Ry = Rz for some z ∈ R. In particular, y = rz for some r ∈ R.Now show that R = Rx = R/Rr, a contradiction.] 15.8. Let R satisfy the DCC for principal left ideals. Prove that R = Qcl. [Hint: Suppose d ∈ . Then there is some n ∈ N with Rdn = Rdn+1,sodn = adn+1 for some a ∈ R. Show that a = d1.].
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