Download 12 The Maximal Ring of Quotients.

12
The Maximal Ring of Quotients.
Let R be a commutative ring with set ∆ of non-zero divisiors. In 600 Algebra we learn that there is a
commutative overring Q of R in which each d ∈ ∆ is invertible and for which each q ∈ Q can be written
in the form q = ad−1 for some a ∈ R and d ∈ ∆; moreover, this ring Q is unique to within isomorphism
over R. For non-commutative rings, however, such rings of fractions need not exist. Sometimes they
do, but in any case there are various alternatives. In the next few Sections we shall look at some of
these. In this one we shall look at what is the largest of the viable alternatives.
We begin with a not necessarily commutative ring R. Let E = E(R R) be the injective envelope of
the regular left module; we will adopt the point of view that R R is a submodule of R E. Next, let
S = End(R E)
be the endomorphism ring of
R E.
So
R ES
is a bimodule. Now for each x ∈ E right multiplication
ρx : a 7−→ ax is an R-homomorphism from R to E. Since E is an injective extension of R, there is an
extension ρx : E −→ E of ρx over R. Then since (1)ρx = x, this shows that each element x ∈ E has
the form (1)s for at least one s ∈ S. Thus,
12.1. Lemma. The map s 7−→ (1)s is an S-epimorphism SS −→ ES .
Now the ring that we are really looking for is
Q = End(ES ) = BiEnd(R E),
the biendomorphism ring of R E. We call the ring Q the maximal ring of left quotients of R. Since
RE
is faithful, the map λ : R −→ Q defined by λa : x 7−→ ax for all x ∈ E and a ∈ R is an injective
ring homomorphism. We usually identify R with its image in Q, and so treat R as a subring of Q. All
of this identification could be dangerous except fortunately, we have
12.2. Lemma. The map η : R Q −→ R E defined by η : q 7−→ q(1) ∈ E is an R-monomorphism.
Proof. The identification of R with a subring of Q is via the map λ, so for each r ∈ R and q ∈ Q,
as endomorphisms of ES , rq = λr ◦ q. Thus,
η(rq) = (λr ◦ q)(1) = λr (q(1)) = r(q(1)) = rη(q),
and η is an R-homomorphism. So suppose η(q) = q(1) = 0. Then for each s ∈ S
(q(1))s = q((1)s) = 0
Thus, by Lemma 12.1, q(E) = 0 and q = 0.
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Thus, we shall often think of Q as being a Q-submodule of E. Of course, if at any time we want to
be careful to distinguish between q ∈ Q as an element of E and as a bi-endomorphism of E, we shall
treat the former as η(q) = q(1). With that in mind we want to characterize Q as a submodule of E.
12.3. Proposition. For an element x ∈ E, we have x ∈ Q iff rS (R) ⊆ rS (x). In particular,
Q = lE rS (R).
Proof. (=⇒) Consider η(q) = q(1). Then Rs = 0 gives that
qs = η(q)s = (q(1))s = q((1)s) = q(0) = 0.
(⇐=) Let x ∈ E and suppose rS (R) ⊆ rS (x). Since E = (1)S (see Lemma 12.1) this means there is
q ∈ Q with q : (1)s 7−→ xs. Thus xs = q(1s) = q(1)s, so if s = 1E , then x = q(1).
For the final statement observe that we have proved that rS (R) ⊆ rS (Q). On the other hand since
R ⊆ Q, we have rS (Q) ⊆ rS (R).
Now we come to one of the key notions in the study of Q. A left ideal D ≤ R is said to be dense
in case
rS (D) = rS (R).
Note that if D is dense and D ≤ I ≤ R are left ideals, then (see Lemma 11.1)
rS (R) ⊆ rS (I) ⊆ rS (D)
and since D is dense, rS (D) = rS (R), so rS (I) = rS (R) and I is dense.
12.4. Lemma. Let D be a dense left ideal of R. Then
(1) (D : q) = {r ∈ R : rq ∈ D} is dense for every q ∈ Q;
(2) If x ∈ E, then Dx = 0 =⇒ x = 0.
Proof. For (1) suppose that (D : q)s = 0 for some s ∈ S. Then for each d ∈ D and r ∈ R
if d + rq = 0, then rq ∈ D, so that rs = 0. So there is an R-homomorphism from the submodule
D + Rq of E to E given by d + rq 7−→ rs. Then since
RE
is injective, there is some s0 ∈ S such that
(d + rq)s0 = rs for all d + rq ∈ D + Rq. But D is dense and Ds0 = 0, so Rs0 = 0. Therefore,
1s = qs0 = (q(1))s0 = q(1s0 ) = q(0) = 0,
and so Rs = (R1)s = R((1)s) = 0.
For (2), suppose that Dx = 0. Now by Lemma 12.1, there is an s ∈ S with x = (1)s. So
Dx = D((1)s)) = Dx = 0, and thus, 0 = Rs = (R1)s = R((1)s)) = Rx, so that x = 0.
Non-Commutative Rings
Section 12
81
It follows at once from part (1) of Lemma 12.4 that if q ∈ Q, then (R : q) is dense. Thus, right
multiplication by the element q in Q is an R-homomorphism from the dense left ideal (R : q) into R.
Moreover, by part (2) of that Lemma, q is uniquely determined by its behavior on any dense left ideal.
In fact, we have
12.5. Corollary. Let D be a dense left ideal of R and let f ∈ HomR (D, R). Then there is a unique
q ∈ Q with q(d) = f (d) for all d ∈ D.
Proof. Since E is an injective extension of both D and R, the homomorphism extends to an
endomorphism of E. That is, there is some s ∈ S with f (d) = (d)s for all d ∈ D. We claim that
(1)s ∈ Q, and to show that we use Lemma 12.3 and show that rS (R) ⊆ rS ((1)s). So suppose that
s0 ∈ S with Rs0 = 0; but then D(1s)s0 = ((D)s)s0 = (f (D))s0 ⊆ Rs0 = 0. Thus, Dss0 = 0; so since D is
dense, Rss0 = 0. But then by Lemma 12.3, we have Q((1)s)s0 = Qss0 = 0. Thus, rS (R) ⊆ rS (Qs), so
by Lemma 12.3, again, Q((1))s ⊆ Q. That means (1)s = q ∈ Q. Then q(d) = d(1)s = (d)s = f (d) for
all d ∈ D. Uniqueness of q follows immediately from Lemma 12.4.
Thus, thanks to the last two results alone, the dense left ideals generate some real interest. As we
shall see in the next Section they form one of a class of very special left ideals. So next, we seek an
“internal” characterization of the dense ideals. First, though, let’s give them a label. So we let
D = {D ≤R R : D is dense in R}
be the set of all dense left ideals of R.
12.6. Lemma. For a left ideal I of R, the following statements are equivalent:
(a) I ∈ D;
(b) There is a D ∈ D such that rR (I : d) = 0 for all d ∈ D;
(c) For every d ∈ R, rR (I : d) = 0.
Proof. (a) =⇒ (c) Let d ∈ R and a ∈ rR (I : d). By Lemma 12.1 there is an s ∈ S with (1)s = a,
so s ∈ rS (I : d). Since I ∈ D, (I : d) ∈ D for all d ∈ R by Lemma 12.4, and we have Rs = 0. But then
Ra = R(1)s = Rs = 0, so a = 0.
(c) =⇒ (b) Let D = R.
(b) =⇒ (a) Suppose that there is some s ∈ S with Is = 0 and Rs 6= 0. Since D is dense, Ds 6= 0.
But R /
– E, so there is some 0 6= r ∈ Ds ∩ R. Say r = ds with d ∈ D. Then
(I : d)r = (I : d)ds ⊆ Is = 0,
contradicting (b). So Rs = 0 and I ∈ D.
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Here is a useful little Corollary of this characterization.
12.7. Corollary. Every left ideal D ≤ R that is dense in R is essential in R.
Proof. Suppose that D is a left ideal and x 6= 0 in R with D ∩ Rx = 0. Then (D : x) = lR (x), so
that rR (D : x) = rR lR (x). But x ∈ rR lR (x), so by Lemma 12.6, D is not dense.
We conclude with a lemma that will be of significance in the next Section.
12.8. Lemma. Let I ≤ R be a left ideal and let D ∈ D be dense. If (I : d) ∈ D for each d ∈ D,
then I ∈ D.
Proof. Let a ∈ R; we claim that rR (I : a) = 0, so that by Lemma 12.6 we would have that I is
dense. Suppose that x ∈ R is not zero. Since D is dense, we have by Lemma 12.6, (D : a)x 6= 0, say
that ra ∈ D and rx 6= 0. But ra ∈ D, so (I : ra) ∈ D, and (I : ra)rx 6= 0 by Lemma 12.4. However,
(I : ra)r ⊆ (I : a), so (I : a)x 6= 0. Since this holds for all a ∈ R, we have I ∈ D.
Exercises 12.
12.1. The goal of this exercise is to show that if Q is the maximal ring of left quotients of R, then Q is
also the maximal ring of left quotients of Q. So let E = E(R R) and S = End(R E).
(a) Show that
QE
is an injective envelope of its submodule
Q Q.
[Hint: Let I ≤
QQ
be a
left ideal and let ϕ : I −→ E be a Q-homomorphism. So there is an R-endomorphism
ϕ : E −→ E extending ϕ. For each q ∈ Q define the R-endomorphism sq : E −→ E by
(x)sq = (xq)ϕ−x(q)ϕ. Note that (R)sq = 0, so by 12.3 (Q)sq = 0 and ϕ is a Q-endomorphism,
and
QE
is injective. Then easily Q /–
Q E.
]
(b) Conclude that Q is it own maximal ring of left quotients. [Hint: The main thing is to show
that S = End(Q E).]
12.2. Prove that the poset D of dense left ideals of R is closed under intersection. Thus, deduce that D
is a lattice. [Hint: Let D1 , D2 ∈ D. Show that (D1 ∩ D2 : d) ∈ D for all d ∈ D2 . Use Lemma 12.8.]
12.3. Suppose that the lattice D of dense left ideals of R has a least element, D0 . Prove that the maximal
ring Q of left quotients of R is isomorphic to End(R D0 ). [Hint: For each q ∈ Q, the dense left
ideal (D0 : q) contains D0 .]
12.4. Prove that if R is commutative, then its maximal ring Q of leftt quotients is isomorphic to the
center of S = End(R E). In particular, Q is commutative.
Non-Commutative Rings
Section 12
83
12.5. Let K be a field and let R be the 3-dimenional commutative K algebra R = K1 + Kx + Ky with
x2 = xy = y 2 = 0.
(a) Show that if S = End(R E), then S is not commutative. Thus, the maximal ring Q of left
quotients is not self-injective.
(b) Since R is Artinian it has a least dense left ideal D0 . Determine D0 and find the maximal
ring Q of left quotients of R. (See Exercises 12.2 and 12.3.)