ANNIHILATOR-SMALL RIGHT IDEALS
W.K. Nicholson Yiqiang Zhou Department of Mathematics Department of Mathematics University of Calgary Memorial University of Newfoundland Calgary T2N 1N4, Canada St. John’sA1C 5S7, Canada [email protected] [email protected]
April 22, 2008
Abstract A right ideal A of a ring R is called annihilator-small if A + T = R;T a right ideal, implies that l(T ) = 0; where l() indicates the left annihilator. The sum Ar of all such right ideals turns out to be a two-sided ideal that contains the Jacobson radical and the left singular ideal, and is contained in the ideal generated by the total of the ring. The ideal Ar is studied, conditions when it is annihilator small are given, its relationship to the total of the ring is examined, and its connection with related rings is investigated.
Keywords: Small right ideals, annihilators, total of a ring Mathematics Subject Classi…cation (2000): 16A34, 16A21, 16D25 1. Introduction
It is well known that if a; b are elements of a ring R; then 1 ab is a unit if and only if 1 ba is a unit. The starting point in this paper is that the analogue for left annihilators is also true: l(1 ab) = 0 if and only if l(1 ba) = 0: This leads to the concept of an “annihilator-small” right ideal. In fact, the set Kr of all elements k such that kR is annihilator small turns out to be a semi-ideal that contains both the Jacobson radical and the left singular ideal and is contained in the total of the ring. The ideal Ar generated by Kr is an “annihilator-small” analogue of the Jacobson radical that contains every annihilator small right ideal, contains both the Jacobson radical and the left singular ideal, and is contained in the ideal generated by the total. Moreover, Ar is quite well behaved when passing to rings related to R (corners, matrix rings and power series rings). On the other hand, an example is given showing that Ar may not be annihilator small, and conditions when this happens are investigated. Along the way, a number of examples are given to rule out possible extensions of theorems. Throughout the paper every ring R is associative with unity 1 = 0; and all modules are 6 unitary. We abbreviate the Jacobson radical as J(R) = J; and we write Sr;Sl;Zr and Zl for the right and left socles, and the right and left singular ideals of R; respectively. We write Z ess for the ring of integers and Zn for the ring of integers modulo n: The notations N M and N max M denote respectively that a submodule N M is essential and maximal in the moduleM; and we write N M to mean that N is a direct summand of M: Left and right annihilators of a subset X R are denoted by l(X) and r(X) respectively. Annihilator-Small Right Ideals — 2
2. Annihilator-Small Right Ideals
Recall that a right ideal K of a ring R is called small in R if the only right ideal X of R such that K + X = R is X = R: By analogy we say that a right ideal K of R is annihilator-small (a-small) if K + X = R;X a right ideal of R; implies that l(X) = 0: as We write K RR in this case, and we de…ne a-small left ideals analogously. The following observation is clear from the de…nition, and will be used repeatedly.
as as Lemma 1. If T K RR; where T is a right ideal of R; then T RR: The ring R is never a-small as a right ideal; if R is a domain, a right ideal K is a-small if and only if K = R: Small right ideals are clearly a-small (but the converse is false as we shall see); in 6 particular the Jacobson radical J is a-small as a right ideal. The left singular ideal Zl; de…ned ess by Zl = z R l(z) RR ; is also a-small as a right ideal. In fact we have: f 2 j g
Proposition 2. If K is an a-small right ideal of RR; so also is K + J + Zl:
Proof. Let (K + J + Zl) + X = R where X is a right ideal. Then K + Zl + X = R because J is small in RR; say k + z + x = 1 where k K; z Zl and x X: Hence K + zR + X = R so 2as 2 2 ess 0 = l(zR + X) = l(z) l(X) because K RR: Thus l(X) = 0 because l(z) RR: \ ess as Lemma 3. Let T be a right ideal of R and assume l(T ) RR: Then rl(T ) RR; in as particular, we obtain T RR: Proof. Let rl(T ) + X = R: Then 0 = l(R) = lrl(T ) l(X) = l(T ) l(X); so l(X) = 0 by \ \ hypothesis. The last observation is by Lemma 1 because T rl(T ) always holds. Note that the converse to Lemma 3 is true if r[(l(T ) Rb] = rl(T ) + r(b) holds for all right \ ideals T of R and all b R: In fact, if l(T ) Rb = 0 this gives rl(T ) + r(b) = R so lr(b) = 0 2 \ ess by hypothesis. Hence b = 0 because Rb lr(b); proving that l(T ) RR: In particular, the converse holds if R is a right Ikeda-Nakayama ring [2], that is r[L K] = r(L) + r(K) for any left ideals L and K: \
3. The ideal Ar:
In characterizing the Jacobson radical of a ring R one needs the observation that 1 ab is a unit in R if and only if 1 ba is a unit. The analogue for left annihilators is true: l(1 ab) = 0 if and only if l(1 ba) = 0: In fact we have: Lemma 4. If R is a ring, the following are equivalent for k R : (1) kR is a-small in R: 2 (2) bR bkR for all 0 = b R: 6 2 (3) l(1 kr) = 0 for all r R: 2 (4) l(1 rk) = 0 for all r R: 2 (5) l(k krk) = l(k) for all r R: 2 Annihilator-Small Right Ideals — 3
Proof. (1) (2). If b bkR write b = bkr; r R; so that b l(1 kr): But kR+(1 kr)R = R; ) 2 2 2 so l(1 kr) = 0 by (1). Thus b = 0: (2) (3). If b l(1 kr); r R; then b = bkr bkR: Hence b = 0 by (2). (3))(4). If b(12 rk ) = 0 then2 br(1 kr) = b(2r rkr) = b(1 rk)r = 0: Hence br = 0 by (3), and) so b = brk = 0: (4) (5). If b(k krk) = 0 then bk(1 rk) = 0 so bk = 0 by (4). This shows that ) l(k krk) l(k); the other inclusion always holds. (5) (1). If kR + X = R;X a right ideal, write 1 = kr + x; r R; x X: If b l(X) then b = bkr;)so b(k krk) = b(1 kr)k = bxk = 0: Hence bk = 0 by (5),2 and2 so b = bkr2 = 0: We observe in passing that in fact l(k) l(1 kr) = l(k krk) and l(1 kr) = l(1 rk) hold for all k; r R; proving again that (3) (4) (5) in Lemma 4. Note that condition (2) in Lemma 4 implies2 that if k R is right a-small, and, not nilpotent then kR k2R k3R is strictly decreasing. 2 as An element k R is called right a-small if kR RR; that is k satis…es the conditions in Lemma 4; left2 a-small elements are de…ned analogously. A unit is never right a-small; if R is a domain an element k is right a-small if and only if kR = R; that is if and only if k is a non-unit. However Example 15 below shows that an element6 in a ring can be a-small on one side but not the other. For convenience, de…ne as Kr = Kr(R) = k R k is right a-small in R = k R kR RR : f 2 j g f 2 j g Note that Zl Kr and J Kr by Proposition 2, but Kr may not be closed under addition (consider 2 and 3 in R = Z): In general Kr is contained in the set of nonunits (with equality in a domain or a local ring). Conditions (3) and (4) in Lemma 4 show that:
Corollary 5. If k Kr then kR Kr and Rk Kr: 2
By virtue of the property in Corollary 5, J = Kr is called a semi-ideal of the ring R: Note that we do not insist that a semi-ideal be closed under addition, but when this holds the semi-ideal is an ideal.
2 Corollary 6. If e = e Kr then e = 0: 2 Proposition 7. The following are equivalent for a right ideal K of a ring R : (1) K is a-small in R: (2) K Kr; that is every element of K is right a-small. (3) l(1 k) = 0 for every k K: 2 Proof. (1) (2) by Lemma 1, and (2) (3) by Lemma 4. Given (3), let K + X = R;X a right ) ) ideal of R: If 1 = k + x; k K; x X; then l(X) l(1 k) = 0 by (3). Hence (3) (1). 2 2 ) As we have seen, the sum of a-small right ideals need not be a-small (consider 3Z + ( 2)Z in Z): With this in mind we de…ne the right AS-ideal Ar = Ar(R) of R to be the sum of all the a-small right ideals of R : as Ar = Ar(R) = K K RR : f j g We de…ne the left AS-ideal Al = Al(R) analogously. Clearly Kr Ar in every ring, but this may not be equality (consider Z): Annihilator-Small Right Ideals — 4
Theorem 8. If R is any ring, then: (1) Ar is an ideal (two sided) of R that contains every a-small right ideal of R: (2) Ar = k1 + k2 + + kn ki Kr for each i; n 1 : f j 2 g (3) Ar = KrR = RKr: (4) J Ar and Zl Ar:
Proof. (1). Clearly Ar is a right ideal; it is a left ideal by (2) below and Corollary 5. (2). Write X = k1+k2+ +kn ki Kr for each i; n 1 : If x Ar then x K1+ +Kn as f j 2 g 2 2 as where Ki RR for each i: If x = k1 + + kn; ki Ki; then kiR Ki so kiR RR by 2 Lemma 1. Hence ki Kr for each i; proving that Ar X: Conversely, if k Kr then as 2 2 kR RR; so k Ar: It follows that X Ar: 2 (3). This follows from (1) and (2) and the fact that Kr Ar: (4). This follows from Proposition 2.
Example 15 below shows that:
Example 9. Ar Al and Al Ar can happen. 6 6
Note that Ar = R is possible, indeed Ar(Z) = Z; so the ideal Ar need not be a-small as a right ideal. Also, the sum of a-small right ideals need not be a-small (consider 2Z + 3Z Z): These properties are equivalent.
Theorem 10. The following are equivalent for a ring R : as as as (1) If K RR and L RR then K + L RR: (2) Kr is closed under addition, that is a sum of right a-small elements is right a-small. (3) Ar = Kr: as (4) Ar RR: If this is the case, we have: (a) Ar is the unique largest a-small right ideal of R: (b) Ar = k l(1 kr) = 0 for all r R = k l(1 rk) = 0 for all r R : f j max 2 g f j 2 g (c) Ar = M RR Ar M : \f j g Proof. (1) (2). This follows from Lemma 1 because (k + l)R kR + lR: ) (2) (3). Always Kr Ar; and Ar Kr by (2) and Theorem 8(2). ) (3) (4). Let Ar + X = R; so Kr + X = R by (3). If 1 = k + x with k Kr and x X; ) as 2 2 then R = kR + X and kR RR: Hence l(X) = 0; proving (4). as as (4) (1). Let K RR and L RR: Then K Ar and L Ar; so K + L Ar: Hence )as K + L RR by (4) and Lemma 1. Finally, (a) is clear by (4), and (b) follows from (3) and Lemma 4. As to (c): If a = Ar then 2 aR is not right a-small by (3) so aR + X = R for some right ideal X with l(X) = 0: Since as max 6 Ar RR by (4), it follows that Ar + X = R: If Ar + X M RR; then a = M; proving 6 2 (c).
We say that Ar = Ar(R) is closed if the conditions in Theorem 10 are satis…ed.
Corollary 11. Ar is closed if Ar J + Zl: The converse is false. Annihilator-Small Right Ideals — 5
Proof. The …rst assertion is because J +Zl is a-small as a right ideal by Proposition 2. Example 15 below provides a counterexample to the converse.
It follows from Lemma 1 and the de…nition of Ar that Ar = J if and only if every a-small right ideal is small. Hence Corollary 11 gives
Corollary 12. If every a-small right ideal is small then Ar is closed (and Ar = J):
4. Examples
We have referred to Example 15 several times above. This example comes from the following RV result about the split-null extension 0 S of two rings R and S by a bimodule V = RVS using matrix operations. h i
RV Proposition 13. Consider the split-null extension T = 0 S . Then Ar(R) V Ar(R) V (1) Ar(T ) : h i 0 0 0 Ar(S) Al(R) lR(V ) V Al(R) V (2) h \ i hAl(T ) i : 0 Al(S) 0 Al(S) h i h i Proof. (1). As Ar(R) is a sum of a-small right ideals of R; it su¢ ces for the …rst inclusion to KV as as k v KV show that TT whenever K RR: By Proposition 7, if we 0 0 0 0 2 0 0 1 k v r w 1 k v must showh that lTi = 0: But if = 0h we havei rh(1 k)i = 0; 0 1 0 s 0 1 r w w = rv and s = 0; soh =i 0 because lRh(1 k)i =h 0: i 0 s a v Turning to the secondh inclusion,i let Ar(T ); we must show that a Ar(R) and 0 b 2 2 a v a v b Ar(S): Since is a sum of righth a-smalli matrices, we may assume that is 2 0 b 0 b right a-small in T;h and hencei that a 0 is right a-small in T (because 0 v h J(T )i 0 b 0 0 2 c 0 Ar(T )): Given r R and s S; leth c ilR(1 ar) and d lS(1 bs)h: Theni 2 2 2 2 0 d 2 1 0 a 0 r 0 lM ; so c = 0 = d: Hence lR(1 ar) = 0 and lS(1 hbs) = 0i; so 0 1 0 b 0 s a nhAr(R) andi bh Ar(Si) h as required.io 2 2 a v Al(R) lR(V ) V (2). For the …rst inclusion, let \ ; we must show that 0 b 2 0 Al(S) a v a 0 r w Al(T ); equivalently thath i h Al(T ): Given i in T; we have 0 b 2 0 b 2 0 s h 1 0 i a 0 r w 1 ar h 0 i p u h 1 i ar 0 0 1 0 b 0 s = 0 1 bs : Hence 0 q rM 0 1 bs gives 2 ph rR(1i har) = 0i; h q rR(1i bsh ) = 0 and u =ia(ru) = 0h; as required.i h i 2 2 a v For the other inclusion, let Al(T ); we must show that a Al(R) and b Al(S): 0 b 2 2 2 a v h i a 0 We may assume that 0 b is right a-small in T , whence 0 b is right a-small in T: p 0 Suppose that p rR(1h ar)i = 0 and q rR(1 bs): Thenh we seei that lies in 2 2 0 q 1 0 a 0 r 0 rT = 0 so p = 0 and q = 0; as required. h i 0 1 0 b 0 s nh i h i h io Annihilator-Small Right Ideals — 6
Example 14. Let R = ZZ2 : Then: 0 Z2 ZZ2 h i 2ZZ2 (1) Ar(R) = 0 0 and Al(R) = 0 0 : (2) Ar(R) ish not closedi but Al(R) ish closed. i
Proof. We have Ar(Z) = Z = Al(Z);Ar(Z2) = 0 = Al(Z2) and lZ(Z2) = 2Z: Hence, by 2ZZ2 ZZ2 n a Proposition 13, Ak(R) : But if Al(R); it is easy to 0 0 0 0 0 0 2 h i h 1i 0 h i show that n is even. Next, the idempotent 0 0 is in Ar(R); so Ar(R) is not closed by 2n x Corollary 6. Finally, to see that Al(R) is closed,h leti Al(R): Then we see that 0 0 2 1 0 2n x k p 1 2nk xq = and thish has zeroi right annihilator in R: Hence 0 1 0 0 0 q 0 1 2n x sm h i ish left a-smalli h in Ri byh Proposition 7i so Al(R) RR: 0 0 h i Example 15. Let R be a ring. Then: (1) Ar Al and Al Ar can happen. (2) An6 element can6 be a-small on one side but not on the other. (3) If Ar is closed it need not happen that Ar J + Zl: Proof. We work in the ring R in Example 14 0 Z2 (1). We have J = so J Al Ar: Similarly we can have J Ar Al: 0 0 2 0 (2).The matrix k =h i R is left a-small by Proposition 7 because k Al(R) which is 0 0 2 2 closed by Exampleh 14.i On the other hand, k is not right a-small because 1 0 2 0 1 0 1 0 = has nonzero left annihilator. 0 1 0 0 0 0 0 0 2ZZ2 0 Z2 h (3).iAl h= i h isi closedh but Ail = J + Zr because J = and Zr = 0— in fact 0 0 6 0 0 ess h i 2hmZZ2 i T RR if and only if T Tm for some m = 0 where Tm = : 6 0 Z2 h i If R is a ring and RXR is a bimodule, the trivial extension of R by X is de…ned to be the additive abelian group T = T (R;X) = R X with multiplication (r; x)(r ; x ) = (rr ; rx +xr ): 0 0 0 0 0 We identify R T and X T so that T = R X;X is an ideal of T , X2 = 0; and T=X = R:
Proposition 16. Let T = R X be the trivial extension of the ring R by the bimodule RXR: (1) Ar(T ) Ar(R) X: (2) If Ar(R) = J(R) then Ar(T ) = J(T ): (3) If rR(x) = 0; x X; implies that x = 0; then Ar(T ) = Ar(R) X: 2 Proof. Observe …rst that if a R then (r + x)a = 0 if and only if ra = 0 and xa = 0; that is 2 lT (a) = lR(a) lX (a): (*) (1). It su¢ ces to show that if a + x is right a-small in T then a is right a-small in R: Since x J(T ) then a is right a-small in T: Hence, if r R; we have lT (1 ar) = 0 so lR(1 ar) = 0 2 2 by (*). It follows that a Ar(R): 2 (2). If Ar(R) = J(R) then Ar(T ) J(R) X = J(T ) using (1), proving (2). (3). We have X Ar(T ) because X J(T ): If a Ar(R) we must show that a Ar(T ); and we may assume that a is right a-small in R: If t =2r + x in T then (*) gives 2
lT (1 at) = lT [(1 ar) ax] = lR(1 ar) + lX (1 ar) = 0 + lX (1 ar): Annihilator-Small Right Ideals — 7
But lX (1 ar) = 0 by hypothesis because 1 ar = 0 (in fact, lR(1 ar) = 0): 6 5. Relation to the Total
If a is an element in a ring R; it is easy to see that Ra contains a nonzero idempotent if and only if aR contains a nonzero idempotent. Kasch and Mader [3] say that a has a partial inverse in this case, and they de…ne the total of the ring R as follows: tot(R) = a R a has no partial inverse} f 2 j The total is a semi-ideal but may not be closed under addition. In fact, if 0 and 1 are the only idempotents in R; then tot(R) is the set of nonunits. In particular, tot(R) = Kr(R) if R is a domain. Call a subset I of a ring R idempotent-free if it contains no nonzero idempotent. If I is idempotent free, and either aR I or Ra I for each a I; we claim that I tot(R): Indeed, if a I but a = tot(R); then a has a partial inverse, so2 there exists 0 = e2 = e Ra and 22 2 6 2 0 = f = f aR; contrary to assumption. In particular, tot(R) contains J(R);Zr(R) and 6 2 Zl(R) for any ring R: Since (by Corollary 6) Kr and Kl are idempotent free subsets, Corollary 5 gives
Proposition 17. If R is any ring then Kr tot(R) and Kl tot(R): If I is a subset of a ring R; we say that R is I-semipotent (see [6]) if every right (equivalently left) ideal not contained in I contains a nonzero idempotent, equivalently if every element a = I has a partial inverse. 2
Lemma 18. Let I be a subset of a ring R: The following are equivalent: (1) R is I-semipotent. (2) tot(R) I: Proof. If a tot(R) then a has no partial inverse by de…nition, so a I by (1). This proves that (1) (2).2 Conversely, if a = I then a = tot(R) by (2), so a has a2 partial inverse, proving that (2))(1). 2 2 ) A J(R)-semipotent ring R is called semipotent. The following answers a question of Kasch and Schneider [4, Page 173].
Corollary 19. If R is any ring then R is semipotent if and only if J(R) = tot(R):
Proof. If R is semipotent, we have tot(R) J by Lemma 18; the other inclusion always holds. Conversely, if J = tot(R) then R is J-semipotent by the de…nition of tot(R):
Theorem 20. The following are equivalent for a ring R : (1) R is Kr-semipotent. (2) If l(b) = 0 where b R; then l(b) contains a nonzero idempotent. 6 2 (3) If l(b) = 0 where b R; then (1 b)R contains a nonzero idempotent. 6 2 (4) Kr = tot(R): Annihilator-Small Right Ideals — 8
Proof. (4) (1) is clear from the de…nitions, and (1) (4) by Proposition 17 and Lemma 18. ) ) (1) (2). Let ab = 0 where a = 0: Then a = Kr; since otherwise R = l(1) = l(1 ab) = 0 ) 6 2 by Lemma 4. So there exists 0 = e2 = e Ra by (1), whence eb (Ra)b = 0; that is e l(b): 6 2 2 2 (2) (3). If 0 = e2 = e l(b); then e = e(1 b) R(1 b): Now (3) follows. ) 6 2 2 (3) (1). Let a = Ar: By (4) of Lemma 4 we have l(1 ra) = 0 for some r R: By (3) let ) 2 2 0 = e2 = e l(1 ra): Hence e(1 ra) = 0; and so e = era Ra: 6 2 2
Note that Kr(Z) = Z 1; 1 ; and so Kr(Z) = tot(Z) because tot(Z) contains no units. f g However, Ar(Z) = Z = Kr(Z) so Ar(Z) is not closed. 6
Recall that Ar is closed if it is contained in J + Zl because this ideal is always right a-small. Many of our examples arise in this way. If Ar is closed it contains no nonzero idempotent (Corollary 6), and this points to a large class of rings in which Ar = Al is closed. The ring R is called semipotent if it is J-semipotent. The class of semipotent rings is large, including all exchange rings (for every a R; there exists e2 = e aR such that 1 e (1 a)R); and all semiregular rings (R=J is (von2 Neumann) regular and2 idempotents lift modulo2 J):
Proposition 21. If R is semipotent then Ar = J = Al = tot(R): In particular, both Ar and Al are closed.
Proof. We always have J Kr; with equality by Corollary 6 because R is J-semipotent. Thus Kr = J is closed under addition, so Ar is closed and Ar = Kr = J by Theorem 10. Moreover, this shows that R is Kr-semipotent, so J = tot(R) by Theorem 20. Finally, a similar argument shows that Al = J:
More generally, the proof of Proposition 21 goes through to show that if R is I-semipotent where I = Zr or I = Zr + J; then Ar = I = tot(R) is closed. 6. Other Results
A ring R is called reversible if ab = 0 in R implies ba = 0; equivalently if, for any a R; we have 2 l(a) = 0 if and only if r(a) = 0: Every commutative is clearly reversible, as is every reduced ring (no nonzero nilpotent elements). If R is reversible then Ar = Al by Lemma 4. There is a natural class of rings that contains the reversible ones: A ring R is called semicommutative if ab = 0 in R implies that aRb = 0; equivalently if l(a) (respectively r(a)) is an ideal for each a R: This raises the question: 2
Question. If R is semicommutative is Ar = Al?
Note that the converse is false. Call a ring R directly …nite if ab = 1 in R implies ba = 1: If R is regular but not directly …nite then Ar = Al by Proposition 21 but R is not even reversible. Indeed, if ab = 1 and ba = 1; then l(a) = 0 but r(a) = (1 ba)R = 0: 6 6 Recall that the right socle of a ring is denoted Sr; and that J r(Sr) always holds. The ideal Ar lies between J and r(Sr):
Proposition 22. For any ring R we have SrAr = 0; so that J Ar r(Sr): Annihilator-Small Right Ideals — 9
Proof. We have already observed that J Ar; so it remains to show that SrAr = 0: By Theorem 8, it su¢ ces to show that bk = 0 whenever bR R is simple and k is right a-small. But if bk = 0 then bkR = bR because bR is simple, say b = bkr: Hence b l(1 kr) = 0 by 6 2 Lemma 4, a contradiction.
Observe that, while SrAr = 0; we need not have r(Sr) = Ar or l(Ar) = Sr: Indeed: FF FF 0 F (1) If R = then r(Sr) = = = Ar: 0 F 0 0 6 0 0 h ZZ2 i h 0 0 i h 0 Z2i (2) If R = then r(Al) = = = Sl: 0 Z2 0 Z2 6 0 0 h i h i h i 0 F Note that (1) also shows that Sr Ar = 0 can happen (because Sr = in that case). \ 6 0 F h i It is a routine matter to verify that
r(Sr) = a R Ba = 0 for every simple module BR that embeds in RR f 2 jmax g = M RR l(M) = 0 : \f j 6 g In view of condition (c) in Theorem 10, it is tempting to ask whether Ar = r(Sr) implies that Ar is closed, or conversely. However both implications are false: FF FF 0 F If R = 0 F where F is a …eld, then M = 0 0 and N = 0 F are the only maximal righth idealsi of R; l(M) = 0; and l(N) =h 0: Hence,i in this case,h r(iSr) = M but 6 0 F ZZ2 Ar = J = = r(Sr): On the other hand, if R = then the maximal right ideals 0 0 6 0 Z2 ZZ2 pZZ2 are M = h i and Mp = where p is a prime.h Sincei l(Mp) = 0 and l(M) = 0 we 0 0 0 Z2 6 have r(Srh) = M =i Ar in this caseh (usingi Example 14 below), but Ar is not closed by Corollary 6. max If a R is right a-small and X RR satis…es l(X) = 0; then a X (since otherwise 2 max6 2 aR + X = R and so l(X) = 0): It follows that Ar X RR l(X) = 0 ; but the next \f j 6 g example shows that this need not be equality, even in a right artinian ring for which Ar = J is closed.
FF Example 23. Let R = 0 F where F is a …eld. Then 0 F max FF Ar = J = h iwhile M RR l(M) = 0 = : 0 0 \f j 6 g 0 0 h i h i 0 F Proof. Note …rst that Ar = J = 0 0 by Proposition 21. The only maximal right ideals FF 0 F of R are X1 = = e11R; andh X2 =i = Sr: Since l(X1) = R(1 e11) = 0 and 0 0 0 F 6 l(X2) = 0; theh result follows.i h i
as as Remark. If K RR; Zorn’s lemma provides maximal members of T K T RR ; f j g call them max-as right ideals. Then Ar(R) = M M is max-as}, and Ar(R) is closed if f j and only if Ar(R) is the unique max-as right ideal. Moreover, Zl + J M M is max-as} as \f j because M + Zl + J RR by Proposition 2. But Zl + J = M M is max-as} is possible FF 6 \f j because J * Zl can happen: For example if R = 0 F where F is a …eld then Zl = 0 = Zr 0 F h i but J = 0 0 : h i Annihilator-Small Right Ideals — 10
max Lemma 24. The following are equivalent for a maximal left ideal M RR : as (1) r(M) RR: ess (2) M RR: Proof. (1) (2). If (2) fails, let M K = 0 where K = 0: Since M is maximal, it follows that ) 2 \ 6 M = Re where e = e: Hence r(M) = (1 e)R; so (1) shows that 1 e Ar: Hence e = 1 by Corollary 6, a contradiction. 2 (2) (1). If r(M) + X = RR then 0 = l(R) = lr(M) l(X): Since M lr(M); (2) gives ) \ l(X) = 0; as required.
It follows from Theorem 8 that Ar = J if and only if every a-small right ideal of a ring R is small. This holds if R is semipotent (Proposition 21) and we are going to give several other cases where it is true. The unifying result is:
Proposition 25. If R is a ring in which l(a) = 0; a R; implies aR = R; then Ar = J: 2
Proof. Always J Kr; we show that this is equality (then Kr is closed under addition so Ar is closed and Ar = Kr = J): If k Kr then k is right a-small so l(1 kr) = 0 for all r R by Lemma 4. Hence (1 kr)R = R 2by hypothesis, and it follows that k J: 2 2
The converse to Proposition 25 is false. If R = Z2[[x]] is the power series ring then R is local (hence semipotent) and so Ar = J: However l(x) = 0 but x is not a unit in R: A ring R is called right Kasch if each simple right module embeds in R; equivalently if l(M) = 0 for every maximal right ideal M of R: Call R left principally injective if every R- linear6 map Ra R; a R; extends to R R; equivalently if aR is a right annihilator in R for each a R:!Finally,2 call R a left C2 ring! if every left ideal that is isomorphic to a direct summand of2 R is itself a direct summand.
Example 26. In each of the following cases we have J = Ar (so Ar is closed). (1) R is semipotent. (2) R is right Kasch. (3) R is left principally injective. (4) R is a left C2 ring.
Proof. Proposition 21 gives (1); for the rest we show that l(a) = 0; a R; implies aR = R; and invoke Proposition 25. 2 max (2). If l(a) = 0; a R; and aR = R then aR M RR so l(M) = 0 contrary to the Kasch hypothesis. Hence2 aR = R and6 Proposition 25 applies. (3). If l(a) = 0; a R; write aR = r(X) by (3). Then X l(a) = 0 so aR = r(X) = R; as required. 2 (4). If l(a) = 0; a R; then aR = R so a is regular by (4), say aba = a: But then 2 0 = l(a) = l(ab) = R(1 ab); so aR = R and again Proposition 25 applies.
Note that Example 15 shows that the fact that Ar is closed does not imply that Ar = J: Note further that in case (3) of Example 26 we have J = Zl = Ar by [5, Theorem 2.1]. A ring R with Ar(R) = J(R) need not be in any of the classes in Example 26. We have Ar(Z) = J(Z) = Z 1; 1 but Z is not semipotent; the localization Z(2) of the integers at 2 f g Annihilator-Small Right Ideals — 11
is a domain that satis…es Ar = J (it is local and so semipotent) but, as is readily veri…ed, it is not right Kasch, not left principally injective, and not left C2. We conclude this section by identifying some modules for which the endomorphism ring has Ar = J: Recall that a module RM is said to generate a module RK if K = M : M K ; and that M cogenerates K if K given 0 = k K then k = 0 for some : fK jM: ! g 6 2 6 !
Proposition 27. Consider a module RM and write E = end(RM): (1) If M generates ker( ) for each E; and monomorphisms in E are epic, 2 then Ar(E) = J(E): (2) If M cogenerates M=M for each E; and epimorphisms in E are monic, 2 then Al(E) = J(E):
Proof. (1). If E is right a-small, then lE(1 ) = 0 for all E; so it su¢ ces by 2 2 hypothesis to show that lE() = 0; E; implies that is monic. Write K = ker() and, by hypothesis, let K = M : M K2 : But given : M K we have (M) K = 0; so f j ! g ! = 0: Hence lE() = 0; and it follows that K = 0; as required. 2 (2). If E is left a-small, then rE(1 ) = 0 for all E; so it su¢ ces by hypothesis 2 2 to show that rE() = 0; E implies that is epic. But if m0 = M choose : M=M M 2 2 ! such that (m0 + M) = 0: If E is de…ned by m = (m + M) for all m M; then 6 2 2 (M) = 0 so rE() = 0; a contradiction because m0 = 0: 2 7. Related Rings
The ideal Ar(R) is relatively well behaved when the ring R changes. We begin with direct products.
Proposition 28. If R = i I Ri is a direct product of rings, then Ar(R) = iAr(Ri): 2
Proof. Observe …rst that if a = ai R then a is right a-small in R if and only if each ai is h i 2 right a-small in Ri: Indeed, given r = ri R we have lR(1 ar) = ilR (1 airi): The result h i 2 i follows because a Ar(R) if and only if it is a …nite sum of right a-small elements. 2 Proposition 29. Let R denote any ring and let e2 = e R: 2 (1) eAr(R)e Ar(eRe): (2) If ReR = R then eAr(R)e = Ar(eRe):
Proof. For convenience denote S = eRe: (1). Let a eAr(R)e Ar(R); say a = iai where each ai is right a-small in R: Then 2 a = ieaie so it su¢ ces to prove that, if a is right a-small in R; then eae is right a-small in S: So let s S be arbitrary and consider x lS[e (eae)s]: Then x = xe lS[1 (ae)s] = 0 2 2 2 because ae Ar(R): 2 (2). Now assume that ReR = R; and let a Ar(eRe); it su¢ ces to show that a Ar(R); We may assume that a is a-small in S; and we2 show that a is right a-small in R;2 that is lR(1 ra) = 0 for all r R: If b(1 ra) = 0 then b = be and we have 2 0 = exb(1 ra) = exbe(1 ra) = exb(e era) = exb(e (ere)a): Hence exb = 0 (because a Ar(S)); and it follows that ReRb = 0: Thus b = 0 by hypothesis. 2 Annihilator-Small Right Ideals — 12
ZZ2 1 1 Example 30. If R = and e = then eAl(R)e Al(eRe): 0 Z2 0 0 h i h i 2ZZ2 2Z 0 Proof. By Example 14, eAl(R)e = e 0 0 e = 0 0 ; so it su¢ ces to show that n nh i h i3 3 2 0 Al(eRe) = eRe: We have eRe = n Z : Since e = it su¢ ces to 0 0 j 2 0 0 0 0 3 3 nh 2i 0 o h i h i show that both a = 0 0 and b = 0 0 are left a-small in eRe: But a routine calculation shows that reRe(e hax) = 0i = reRe(eh ax) ifor all x eRe: 2
Let Mn(R) denote the ring of n n matrices over the ring R:
Proposition 31. If n 1 then Ar[Mn(R)] = Mn[Ar(R)] for any ring R:
Proof. Because of Proposition 29, it su¢ ces to prove it for n = 2: Write S = Mn(R): If a = [aij] Ar(S) we must show that Mn[Ar(R)]; we may assume that is right 2 2 a-small in S: Let "ij denote the matrix units in S: Hence aij"11 = "1i "j1 is right a-small in S by Corollary 5, so it su¢ ces to show that "11 right a-small in S implies that is right a-small in R: But if r R is arbitrary and q l(1 ar); then q"11 l(1S "11r"11) = 0 so q = 0: 2 2 2 Conversely, let = [aij] Mn[Ar(R)]; we must show that Ar(S): Since each aij 2 2 2 Ar(R); we may assume that aij is right a-small in R for all i and j: Moreover, since = i;jaij"ij and aij"ij = "i1(aij"11)"1j; it su¢ ces to show that a right a-small in R implies that a"11 = a 0 1 0 a 0 r s is right a-small in S: Hence we must show that lS = 0 0 0 0 1 0 0 t q forh all r;i s; t; q R: But if nh i h i h io 2 u v 1 0 a 0 r s u v 1 ar as u(1 ar) v as w z 0 1 0 0 t q = w z 0 1 = w(1 ar) z ws h i h i h i h i h i h i h i vanishes we obtain u = v = w = z = 0 because lR(1 ar) = 0: If we combine Propositions 29 and 31, we obtain
Theorem 32. The following conditions on a ring R are Morita invariants: (1) Ar(R) = 0: (2) Ar(R) = R: (3) Ar(R) = J: (4) Ar(R) = Zl: ess (5) Ar(R) RR: 2 Proof. If we write M = Mn(R) and e = e M satis…es MeM = M; we must show that S = eMe inherits each of these conditions. Then2 Propositions 29 and 31 give (1) and (2), while (3) and (4) are because J[Mn(R)] = Mn(J);J(eRe) = eJe; Zl[Mn(R)] = Mn[Zl(R)] and (if ess ReR = R) Zl(eRe) = eZl(R)e: Finally, (5) follows because if T RR;T a right ideal of the ess ess ring R; then Mn(T ) Mn(R) and (if ReR = R) eT e eRe: If R is a ring let R[[x]] denote formal power series ring over R in an indeterminant x:
Proposition 33. If x is an indeterminant in R then Ar(R[[x]]) = Ar(R) + xR[[x]]: Annihilator-Small Right Ideals — 13
Proof. Write S = R[[x]]: We begin with: Claim. x is right a-small in S:
Proof. We show that l(1 xf) = 0 for every f S: If g = b0 + b1x + l(1 xf) where 2 2 each bi R; we have g = gxf so b0 = 0 and hence g = xg1 where g1 = b1 + b2x + : But then 2 xg1(1 xf) = 0; so g1 l(1 xf) and we obtain b1 = 0: Continuing in this way, we see that g = 0; proving the Claim.2
Write I = Ar(R) + xS: To show that I Ar(S); let f = a + xg I where a Ar(R); 2 2 g S; we must show that f Ar(S): We may assume that a is a-small in R; and we show 2 2 that f = a + xg Ar(S): By the claim, it su¢ ces to show that a is a-small in S: To this end, let aS + =2S where is a right ideal of S: Looking at constant coe¢ cients, we see T T that aR + T = R where T = t R t + xg for some g S : But then lR(T ) = 0 by f 2 j 2 T 2m g m+1 hypothesis, and we claim that l ( ) = 0: If not, let f = a0x + a1x + l ( ) R[x] T 2 R[x] T where a0 = 0: Then for each t T; f(t + xg) = 0 for some g S; which implies that a0t = 0: 6 2 2 Hence a0 lR(T ); a contradiction. This proves that I Ar(S): 2 Conversely, let f = a + xg Ar(S) where a R and g S; we must show that a Ar(R): Since f is a sum of a-small elements2 in R[x]; comparing2 constant2 coe¢ cients shows that2 we may assume that f = a + xg is a-small in S: We show that a is a-small in R: So let aR + T = R;T a right ideal of R: Then aS+T [x] = S; so (a+gx)S+T [[x]]+xS = S: Hence (a+gx)S+T [[x]] = S because xS J(S): But (a + gx) is a-small in S; so lS(T ) = 0: This gives lR(T ) = 0; and so proves that a Ar(R): 2 The situation for the polynomial ring R[x] is quite di¤erent from R[[x]]: A ring R is called i j an Armendariz ring if whenever (aix )(bjx ) = 0 in R[x] we have aibj = 0 for all i and n j: Examples of Armendariz rings include reduced rings, R[x]=(x ) where R is reduced, Zn for n 2; and polynomial rings over Armendariz rings.
Proposition 34. If R is an Armendariz ring then Ar(R[x]) = R[x]:
Proof. Write S = R[x]: The proof of the Claim in Proposition 33 goes through to show that x is a-small in S: Hence x is a-small in S so it su¢ ces to show that a = 1 + x is a-small in S (since then 1 = (1 + x) + ( x) Ar(S)): Hence we prove the 2 Claim. lS(1 + ab) = 0 for all b S: 2 Write b = b0 + b1x + + bn where n 0: Then 2 n n+1 1 + ab = (1 + b0) + (b0 + b1)x + (b1 + b2)x + + (bn 1 + bn)x + bnx : m Suppose that lS(1 + ab) = 0: Then there exists 0 = c0 + c1x + + cmx in S such that 6 m 6 0 = (c0 + c1x + + cmx )(1 + ab) m n n+1 = (c0 + c1x + + cmx )((1 + b0) + (b0 + b1)x + + (bn 1 + bn)x + bnx ): We may assume that c0 = 0 with no loss in generality. Since R is Armendariz, it follows that 6 c0(1 + b0) = 0 c0(b0 + b1) = 0 . . c0(bn 1 + bn) = 0 c0(bn) = 0 Annihilator-Small Right Ideals — 14
This gives c0bi = 0 for i = n; n 1; ; 2; 1; 0: So c0 = 0; a contradiction. This proves the Claim, and hence the proposition.
A result of Anderson and Camillo [1] shows that if R is Armendariz then R[x1; x2; ; xn] is Armendariz for each n: Hence:
Corollary 35. If R is Armendariz then Ar(R[x1; x2; ; xn]) = R[x1; x2; ; xn]:
RR R 0 R R . . . If R is a ring, let Tn(R) = . . . . . denote the ring of n n upper triangular 2 . . . . 3 0 0 R 6 7 matrices over R: 4 5
Proposition 36. If R is any ring then
Ar(R) R R Al(R) R R 0 Ar(R) R 0 Al(R) R . . . . . . Ar[Tn(R)] = . . . . . and Al[Tn(R)] = . . . . . : 2 . . . . 3 2 . . . . 3 0 0 Ar(R) 0 0 Al(R) 6 7 6 7 4 5 4 5 Tn 1(R) V Proof. Induct on n; the result being clear if n = 1: If n 2 write Tn(R) = 0 R n 1 where V = R ; written as column matrices, is a Tn 1(R)-R bimodule in a naturalh way. Byi Ar[Tn 1(R)] V Ar[Tn 1(R)] V Proposition 13 we have Ar[Tn(R)] : By induction, 0 0 0 Ar(R) this gives h i h i Ar(R) R RR Ar(R) R RR 0 Ar(R) RR 0 Ar(R) RR . . ...... 2 0 0 . . . 3 Ar[Tn(R)] 2 0 0 . . . 3 . . ...... 6 . . Ar(R) R 7 6 . . Ar(R) R 7 6 7 6 7 6 0 0 0 0 7 6 0 0 0 Ar(R) 7 4 5 4 5 If k Kr and let A = [aij] Tn(R); write kEnn for the matrix in Tn(R) with k in the 2 2 (n; n)-position and 0s elsewhere. Then, for each B = [bij] Tn(R) we have 2 b11 b12 b1 n 1 b1n(1 annk) 0 b22 b2 n 1 b2n(1 annk) . . . . . B l[1 A(kEnn)] if and only if 2 . . . . . 3 = 0: 2 . . . . . 0 0 bn 1 n 1 bn n 1(1 annk) 6 7 6 0 0 0 bnn(1 annk) 7 4 5 Since lR(1 annk) = 0; this holds if and only if B = 0: Hence kEnn Ar[Tn(R)]; that is 2 Ar(R) Enn Ar(R): This proves the …rst of the assertions in the Proposition; the proof of the other assertion is similar.
Corollary 37. Let R be an Armendariz ring, and write T = Tn(R): Then Ar(T [x]) = T [x]:
Proof. We haved T [x] = Tn(R[x]) via a natural isomorphism. Since R is Armendariz, Propo- sition 36 gives REFERENCES Annihilator-Small Right Ideals — 15
Ar(R[x]) R[x] R[x] R[x] R[x] R[x] 0 Ar(R[x]) R[x] 0 R[x] R[x] . . . . . . Ar(Tn(R[x])) = . . . . . = . . . . . = Tn(R[x]) 2 . . . . 3 2 . . . . 3 0 0 Ar(R[x]) 0 0 R[x] 6 7 6 7 4 5 4 5 as required.
Remark. Note that it is known that Tn(R) is not Armendariz unless n = 1: Hence being Armendariz is su¢ cient, but not necessary, in showing that Ar(R[x]) = R[x]:
Acknowledgement: The work of the …rst author was supported by NSERC Grant A8075, and that of the second author by Grant OGP0194196. The …rst author is grateful to the Department of Mathematics and Statistics of Memorial University for their hospitality and …nancial support during a visit there, and the second author is grateful to the Department of Mathematics and Statistics of the University of Calgary for the hospitality and …nancial support received when this work was done during his sabbatical year there.
References
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