An Example of Osofsky and Essential Overrings

Unknown Book Proceedings Series Volume 00, XXXX

An Example of Osofsky and Essential Overrings

Gary F. Birkenmeier, Jae Keol Park and S. Tariq Rizvi

Dedicated to Professor Carl Faith and Professor Barbara L. Osofsky

Abstract. Osofsky has shown that the right injective hull E(RR) of a ring R, in general, does not have a ring multiplication which extends its R-module „ « Z4 2Z4 multiplication by using the ring R = , where Z4 is the ring of 0 Z4 integers modulo 4. We explicitly characterize all right essential overrings of this ring R and investigate various properties as well as interrelationships between these right essential overrings. As a byproduct, we determine various ring hulls for the ring R (e.g., right FI-extending, right extending, right quasi- continuous, right continuous, and right self-injective ring hulls). Moreover, we ﬁnd an intermediate R-module SR between RR and ER which has a compatible ring structure that is right self-injective and another compatible ring structure on SR which is even not right FI-extending. Generalizing Osofsky’s example, we provide a class of A-algebras R such that E(RR) has no compatible ring structure, where A is a commutative QF-ring.

1. Introduction

Throughout this paper all rings are associative with unity and R denotes such a ring. All modules are unitary and we use MR to denote a right R-module. If NR is a submodule of MR, then NR is said to be essential (resp., dense, also called rational ) in MR if for any 0 6= x ∈ M, there exists r ∈ R such that 0 6= xr ∈ N (resp., for any x, y ∈ M with 0 6= x, there exists r ∈ R such that xr 6= 0, and yr ∈ N). Recall that a right ring of quotients T of R is an overring of R such that RR is dense in TR. The maximal right ring of quotients of R is denoted by Q(R), and the right injective hull of R is denoted by E(RR). We say that T is a right essential overring of a ring R if T is an overring of R such that RR is essential in TR. Note that, for an overring T of a ring R, if RR is

2000 Mathematics Subject Classiﬁcation. Primary 16D50, 16L60 Key words and phrases. Essential extension, Essential overring, (FI-) extending, Kasch ring, Osofsky compatibility, QF-ring, Ring hull. The authors thank the referee for his/her comments and suggestions for the improvement of this paper. The second author was supported in part by Busan National University, 2008-2010.

c XXXX American Mathematical Society 1 2 GARY F. BIRKENMEIER, JAE KEOL PARK AND S. TARIQ RIZVI

dense in TR then RR is essential in TR while the converse is not true. Thus a right essential overring T of a ring R can be considered as a “generalized version” of a right ring of quotients of R.

Deﬁnition 1.1. Let R be a ring, E(RR) the injective hull of RR, and TR an intermediate module between RR and E(RR). Then we say that a ring structure (T, +, •) on T is compatible if the ring multiplication • extends the R-module multiplication of T over R. We notice that for an intermediate R-module TR between RR and E(RR), (T, +, •) is a compatible ring structure if and only if (T, +, •) is a right essential overring of R.

It is well known that if the injective hull E(RR) is a rational extension of the right R-module RR, then it has a unique compatible ring structure [Lam, Theorem 13.11, p.367]. Also it is well known that a rational extension TR of RR has a unique compatible ring structure whenever such a ring structure on TR exists. In a general setting, Osofsky [O1, O2, and O3] investigated rings R such that E(RR) has a compatible ring structure, and she provided an example of a ring R such that no injective hull of RR has a compatible ring structure. Also Lang [Lang] showed that for a commutative Artinian ring R, E(RR) has a compatible ring structure if and only if R = E(RR). Embedding an arbitrary ring into a right self-injective ring has always been a challenge in view of Osofsky’s work, Faith [F, p.308], has described the solution provided by Menal and Vamos [MV] to the question of “embedding of any ring in an FP-injective ring” as the realization of “a three-decade old dream of Ring Theory”. Menal and Vamos [MV] also presented a right FP-injective ring R such that no right injective hull of RR has a compatible ring structure. In honor of Osofsky’s contributions to the study of injective hulls of rings, in [BPR3] a ring R is called right Osofsky compatible if an injective hull E(RR) of R has a compatible ring structure.

Osofsky [O2] has shown that the right injective hull E(RR) of a ring R, in general, does not have a ring multiplication which extends its R-module multiplication Z4 2Z4 by using the ring R = , where Z4 is the ring of integers modulo 4. 0 Z4 In this paper, we explicitly characterize all right essential overrings of this ring R and investigate various properties as well as interrelationships between these right essential overrings. As a byproduct, we determine various ring hulls for this ring R (e.g., right FI-extending, right extending, right quasi-continuous, right continuous, and right self-injective ring hulls). In particular, although E(RR) has no compatible ring structure, there does exist a right self-injective right essential overring for R. This result also illustrates the general fact that if R is an arbitrary ring with right essential overrings S and V with S a subring of V and S is right self-injective, then S = V . So if R is right Osofsky compatible, then E(RR) hjas no proper subring intermediate between R and E(RR) which is right self-injective. Generalizing Osofsky’s example, we provide a class of A-algebras R such that E(RR) has no compatible ring structure, where A is a commutative QF-ring.

Let MR be a right R-module. Then MR is said to be (FI-) extending if every (fully invariant) submodule of MR is essential in a direct summand of MR (see [BMR], [DHSW], and [BPR1]). A ring R is called right (FI-) extending if RR is AN EXAMPLE OF OSOFSKY AND ESSENTIAL OVERRINGS 3

(FI-) extending. Right extending rings have also been called right CS rings [CH]. A right extending ring R is right (quasi-) continuous (if whenever AR and BR are direct summands of RR with A ∩ B = 0, then AR ⊕ BR is a direct summand of ∼ RR) if whenever XR = YR ≤ RR and XR is a direct summand of RR, then YR is a direct summand of RR. It is well known that injective ⇒ continuous ⇒ quasi-continuous ⇒ extending ⇒ FI-extending Reverse implications are not true in general. A ring R is called right Kasch [Lam, p.280] if every maximal right ideal of R has a nonzero left annihilator. If R is a right Kasch ring, then R = Q(R). ess For R-modules MR and NR, we use NR ≤ MR and NR ≤ MR to denote that NR is a submodule of MR and NR is an essential submodule of MR, respectively. We use J(−), Soc(−), and | | to denote the Jacobson radical of a ring, the right socle of a ring, and the cardinality of a set, respectively.

2. Osofsky’s Example and Nonisomorphic Essential Overrings

Let A = Z4, the ring of integers modulo 4, and let A 2A R = , 0 A which is a subring of the 2 × 2 upper triangular matrix ring over A. It is shown in [O2] that no injective hull of RR has a compatible ring structure (i.e., the ring R is not right Osofsky compatible). Interestingly, Osofsky was able to obtain this result without describing the injective hull of RR explicitly. Note that Q(R) = R since R is a right Kasch ring. In this section, we determine all possible right essential overrings of the ring R and their interrelationships. Moreover, we investigate various properties of these right essential overrings. Also we explicitly provide various ring hulls of the ring R.

For f ∈ Hom(2AA,AA) and x ∈ A, we let (f · x)(a) = f(xa) for all a ∈ A. Let

A ⊕ Hom(2A ,A ) A E = A A , Hom(2AA,AA) A where the addition on E is componentwise and the R-module multiplication of E over R is given by a + f b x y ax + f · x ay + f(y) + bz = g c 0 z g · x g(y) + cz a + f b x y for ∈ E and ∈ R, where a, b, c, x, y, z ∈ A and f, g ∈ g c 0 z Hom (2AA,AA).

Proposition 2.1. ([BPR3, Proposition 2.6]) E is an injective hull of RR.

Put f0 ∈ Hom(2AA,AA) such that f0(2a) = 2a for a ∈ A. By [BPR3, Lemma 2.5], Hom(2AA,AA) = f0 · A. Thus if f ∈ Hom(2AA,AA), then f = f0 · s for some s ∈ A. Note that Hom(2AA,AA) = {0, f0}. Therefore all possible intermediate R-modules between RR and ER are: 4 GARY F. BIRKENMEIER, JAE KEOL PARK AND S. TARIQ RIZVI

A ⊕ Hom (2A ,A ) 2A E,V = A A , Hom (2AA,AA) A

A ⊕ Hom (2A ,A ) A AA Y = A A ,W = , 0 A Hom (2AA,AA) A

A 2A S = , Hom (2AA,AA) A

A ⊕ Hom (2A ,A ) 2A AA U = A A ,T = , and R. 0 A 0 A

For the convenience of the reader, we list the following multiplications on the aforementioned overmodules of RR which will be used in Theorem 2.2 to determine the right essential overrings of R. a1 + f0 · r1 2b1 a2 + f0 · r2 2b2 Multiplications on V . Set v1 = , v2 = in f0 · s1 c1 f0 · s2 c2 V , and deﬁne the multiplications •1, •2, •3, and •4 as follows:

a1a2 + f0 · r1a2 + f0 · a1r2 + f0 · r1r2 2a1b2 + 2b2r1 + 2b1c2 v1 •1 v2 = . f0 · s1a2 + f0 · s1r2 + f0 · c1s2 2s1b2 + c1c2

x y v • v = , where x = a a + 2r r + f · r a + f · a r + f · r r , 1 2 2 z w 1 2 1 2 0 1 2 0 1 2 0 1 2 y = 2a1b2 + 2r1b2 + 2b1c2, z = f0 · s1a2 + f0 · s1r2 + f0 · c1s2, w = 2s1b2 + c1c2. x y v • v = , where x = a a +2s r +2a s +2c s +f ·r a +f ·a r + 1 3 2 z w 1 2 1 2 1 2 1 2 0 1 2 0 1 2 f0·r1r2, y = 2a1b2+2r1b2+2b1c2, z = f0·s1a2+f0·c1s2+f0·s1r2, w = 2s1b2+c1c2. x y v • v = , where x = a a +2r r +2s r +2a s +2c s +f ·r a + 1 4 2 z w 1 2 1 2 1 2 1 2 1 2 0 1 2 f0 · a1r2 + f0 · r1r2, y = 2a1b2 + 2r1b2 + 2b1c2, z = f0 · s1a2 + f0 · s1r2 + f0 · c1s2, w = 2s1b2 + c1c2. a1 2b1 a2 2b2 Multiplications on S. For s1 = , s2 = ∈ S, deﬁne f0 · r1 c1 f0 · r2 c2 multiplications ◦(k,t), where k ∈ A and t ∈ 2A: a1a2 − ta1r2 + 2kb1r2 + tc1r2 2a1b2 + 2b1c2 s1 ◦(k,t) s2 = , f0 · r1a2 + f0 · c1r2 2r1b2 + c1c2

a1 + f0 · r1 2b1 a2 + f0 · r2 2b2 Multiplications on U. For u1 = , u2 = ∈ 0 c1 0 c2 U, deﬁne multiplications }1 and }2:

a1a2 + f0 · a1r2 + f0 · r1a2 + f0 · r1r2 2a1b2 + 2r1b2 + 2b1c2 u1 }1 u2 = . 0 c1c2 AN EXAMPLE OF OSOFSKY AND ESSENTIAL OVERRINGS 5

x y u u = , where x = a a + 2r r + f · a r + f · r a + f · r r , 1 }2 2 z w 1 2 1 2 0 1 2 0 1 2 0 1 2 y = 2a1b2 + 2r1b2 + 2b1c2, z = 0, w = c1c2. a1 b1 a2 b2 Finally multiplications on T . For t1 = , t2 = ∈ T , deﬁne 0 c1 0 c2 multiplications 1 and 2: a1a2 a1b2 + b1c2 t1 1 t2 = . 0 c1c2 a1a2 a1b2 + 2b1b2 t1 2 t2 = . 0 c1c2 + 2a1b2 + 2c1b2

Theorem 2.2. (i) There are exactly thirteen right essential overrings of R which are:

(V, +, •1), (V, +, •2), (V, +, •3), (V, +, •4), (S, +, ◦(0,0)), (S, +, ◦(0,2)), (S, +, ◦(1,0)),

(S, +, ◦(1,2)), (U, +, }1), (U, +, }2), (T, +, 1), (T, +, 2), and R such that ∼ ∼ ∼ (V, +, •1) = (V, +, •2) = (V, +, •3) = (V, +, •4), ∼ ∼ (S, +, ◦(0,0)) = (S, +, ◦(0,2)) and (S, +, ◦(1,0)) = (S, +, ◦(1,2)), ∼ ∼ (U, +, }1) = (U, +, }2) and (T, +, 1) = (T, +, 2).

(ii) (S, +, ◦(0,0)) and (U, +, }1) are subrings of both (V, +, •1) and (V, +, •2), while (S, +, ◦(0,2)) and (U, +, }2) are subrings of both (V, +, •3) and (V, +, •4). Proof. (i) The proof proceeds by the following seven (7) steps. ¯ 1. R is not right Osofsky compatible (i.e., E does not have a compatible ring structure) by Osofsky [O2] (see also [Lam, Osofsky’s Example 3.45, p.79]). 2. Y does not have a compatible ring structure. Proof of 2. Assume that Y has a compatible ring structure. Then 0 2 f 0 0 2 f 0 2 0 0 1 = 0 = 0 = 0 0 0 0 0 0 0 0 0 2 0 0

f 0 2 0 0 1 0 0 0 1 0 = = 0, 0 0 0 0 0 0 0 0 0 0 a contradiction. Thus Y cannot have a compatible ring structure. 3. W also does not have a compatible ring structure. Proof of 3. Assume that W has a compatible ring structure. Then 0 0 0 0 0 2 0 0 2 0 0 1 = = = 0 2 f0 0 0 0 f0 0 0 2 0 0

0 0 2 0 0 1 0 0 0 1 = = 0, f0 0 0 2 0 0 0 0 0 0 a contradiction. Hence W also cannot have a compatible ring structure. 6 GARY F. BIRKENMEIER, JAE KEOL PARK AND S. TARIQ RIZVI

4. Compatible ring structures on V : there are exactly four compatible ring structures and they are mutually isomorphic. Proof of 4. Assume that V has a compatible ring structure. Note that ess RR ≤ VR. If 1V − 1R 6= 0, then there is r ∈ R such that 0 6= (1V − 1R)r = 1 0 1 r − 1 r = r − r = 0, a contradiction. Thus 1 = 1 . Let e = and V R V R 1 0 0 0 0 e = . Then V = e V e + e V e + e V e + e V e . 2 0 1 1 1 1 2 2 1 2 2

0 2A 0 0 Claim 1. e V e = and e V e = . 1 2 0 0 2 2 0 A 0 2A Proof of Claim 1. Note that = e Re ⊆ e V e . Take v = 0 0 1 2 1 2 a + f 2b ∈ e V e , where a, b, c ∈ A and f, g ∈ Hom (2A ,A ). Then v = g c 1 2 A A a + f 2b 0 0 0 2b 0 2b 0 0 ve = = . Since ∈ e V e , ∈ 2 g c 0 1 0 c 0 0 1 2 0 c 0 0 0 0 0 0 1 0 0 0 e V e . If 0 6= , then = e ; but = 0, 1 2 0 c 0 c 1 0 c 0 0 0 c 0 2b a contradiction. Thus c = 0, hence v = ∈ e Re . So e V e = e Re = 0 0 1 2 1 2 1 2 0 2A 0 0 . Also observe that = e Re ⊆ e V e . Then using a similar 0 0 0 A 2 2 2 2 0 0 argument as above, we see that e V e = . 2 2 0 A

A ⊕ Hom(2A ,A ) 0 Claim 2. e V e = A A . 1 1 0 0 A 0 a + f 2b Proof of Claim 2. Note that ⊆ e V e . Let v = ∈ 0 0 1 1 g c a + f 0 a 0 e V e . Then v = ve = . Since ∈ e Re ⊆ e V e , it follows 1 1 1 g 0 0 0 1 1 1 1 f 0 that w = ∈ e V e . Suppose that g 6= 0. Then g = f . Since R ≤ess V , g 0 1 1 0 R R x 2y f 0 x 2y f · x f(2y) there exists ∈ R such that 0 6= = ∈ 0 z g 0 0 z f0 · x 2y f · x f(2y) R, so 0 6= ∈ e1R. Hence f0 · x = 0 and 2y = 0. Now f = f0 · a0 f0 · x 2y for some a0 ∈ A. Hence f · x = (f0 · a0) · x = f0 · a0x = (f0 · x) · a0 = 0 f · x f(2y) and f(2y) = 0. Thus 0 = , a contradiction. Thus g = 0, so f0 · x 2y a + f 0 a + f 0 A ⊕ Hom(2A ,A ) 0 v = = ∈ A A . Hence g 0 0 0 0 0

A 0 A ⊕ Hom(2A ,A ) 0 ⊆ e V e ⊆ A A . 0 0 1 1 0 0 AN EXAMPLE OF OSOFSKY AND ESSENTIAL OVERRINGS 7

A 0 A ⊕ Hom (2A ,A ) 0 Thus e V e = or e V e = A A . Suppose that 1 1 0 0 1 1 0 0 A 0 p + f 2q e V e = . Note that if v = ∈ e V e , then 2q = 0 and r = 0 1 1 0 0 g r 2 1 because v = ve1. By Claim 1, and noting that V = e1V e1 +e1V e2 +e2V e1 +e2V e2, f 0 a 0 0 2b p + f 0 0 0 0 = + + + 0 0 0 0 0 0 g 0 0 c

f 0 a + p + f 2b with a, b, p, c ∈ A and f, g ∈ Hom (2A ,A ). So 0 = , A A 0 0 g c hence g = 0, c = 0, 2b = 0, and f0 = a + p + f. Thus a + p = 0 and f = f0. Hence f 0 a 0 −a + f 0 −a + f 0 0 = + 0 with 0 ∈ e V e . 0 0 0 0 0 0 0 0 2 1

x 2y Since R ≤ess V , there is ∈ R such that R R 0 z

−a + f 0 x 2y −ax + f · x −2ay + 2y 0 6= 0 = 0 ∈ R, 0 0 0 z 0 0

−ax + f · x −2ay + 2y A 0 so 0 ∈ e R, a contradiction, hence e V e 6= . 0 0 2 2 2 0 0 A ⊕ Hom (2A ,A ) 0 Therefore e V e = A A . 1 1 0 0

0 0 2 0 Claim 3. e2V e1 = or e2V e1 = 0, . Hom(2AA,AA) 0 f0 0 a + f 2b Proof of Claim 3. Take v = ∈ e V e . Then v = ve = g c 2 1 1 a + f 0 0 0 a + f 0 0 0 a + f 0 . Also v = e v = = + g 0 2 0 1 g 0 0 1 0 0 0 0 0 0 0 0 a + f 0 . But ∈ e · e V e = 0. So it follows that 0 1 g 0 0 1 0 0 2 1 1

0 0 0 0 v = e v = . 2 0 1 g 0

0 0 0 0 0 0 0 0 0 0 Hence v = = or v = . Thus 0 1 0 0 0 0 0 1 f0 0

0 0 0 0 e2V e1 = 0, 0 1 f0 0

0 0 0 0 a + f 0 a + f 0 since |e2V e1| = 2. If v = , v = = . 0 1 f0 0 g 0 f0 0 Note that v + v = 0, so 2a + 2f = 0 since |e2V e1| = 2. Thus 2a = 0, hence a = 0 f 0 2 + f 0 or a = 2. So v = or v = . f0 0 f0 0 8 GARY F. BIRKENMEIER, JAE KEOL PARK AND S. TARIQ RIZVI f0 0 2 + f0 0 f0 0 If f = f0, then v = or v = . If v = , then f0 0 f0 0 f0 0 x 2y f 0 x 2y f · x 2y there is ∈ R with 0 6= 0 = 0 ∈ R, thus 0 z f0 0 0 z f0 · x 2y f0 · x 2y 0 6= ∈ e2R, a contradiction. f0 · x 2y 2 + f 0 x 2y Also if v = 0 , then there is ∈ R with f0 0 0 z 2 + f 0 x 2y 2x + f · x 2y 0 6= 0 = 0 ∈ R, f0 0 0 z f0 · x 2y 2x + f0 · x 2y so 0 6= ∈ e2R, also a contradiction. Therefore f = 0. Thus f0 · x 2y 0 0 2 0 v = or v = . Consequently, f0 0 f0 0 0 0 0 0 2 0 e2V e1 = or e2V e1 = , . Hom(2AA,AA) 0 0 0 f0 0

By Claims 1, 2, and 3, we have the following two cases: A ⊕ Hom(2A ,A ) 0 0 2A Case 1. e V e = A A , e V e = , 1 1 0 0 1 2 0 0 0 0 0 0 e2V e1 = , e2V e2 = . Hom(2AA,AA) 0 0 A

A ⊕ Hom(2A ,A ) 0 0 2A Case 2. e V e = A A , e V e = , 1 1 0 0 1 2 0 0 2 0 0 0 e2V e1 = 0, , e2V e2 = . f0 0 0 A

Case 1. In this case, ﬁrst we compute f 0 f 0 0 0 ∈ e V e · e V e ⊆ e V e . 0 0 0 0 1 1 1 1 1 1 f 0 f 0 a + f · r 0 Let 0 0 = 0 for some a, r ∈ A. Now 0 0 0 0 0 0 f 0 f 0 f 0 f 0 0 = 2 0 0 = 2 0 0 = 0 0 0 0 0 0 0 0

a + f · r 0 2a 0 2 0 = , 0 0 0 0 so 2a = 0. Thus a = 0 or a = 2. Also f 0 f 0 0 2 f 0 0 2 0 2 0 0 = 0 = . 0 0 0 0 0 0 0 0 0 0 0 0 AN EXAMPLE OF OSOFSKY AND ESSENTIAL OVERRINGS 9

On the other hand, f 0 f 0 0 2 a + f · r 0 0 2 0 0 = 0 = 0 0 0 0 0 0 0 0 0 0 0 2a + 2r 0 2 0 2r = . 0 0 0 0 0 0

So 2r = 2, hence r = 1 or r = 3. Thus a + f0 · r = f0 or a + f0 · r = 2 + f0. So f 0 f 0 f 0 f 0 f 0 2 + f 0 0 0 = 0 or 0 0 = 0 . 0 0 0 0 0 0 0 0 0 0 0 0

Thus Case 1 can be divided into the following subcases: f 0 f 0 f 0 Subcase 1.1. 0 0 = 0 . 0 0 0 0 0 0 f 0 f 0 2 + f 0 Subcase 1.2. 0 0 = 0 . 0 0 0 0 0 0

f 0 f 0 f 0 For Subcase 1.1. 0 0 = 0 : 0 0 0 0 0 0 1 0 f 0 f 0 f 0 Step 1. 0 = 0 since 0 ∈ e V e . 0 0 0 0 0 0 0 0 1 1 0 2 f 0 0 2 Step 2. 0 = 0 because ∈ e V e and 0 0 0 0 0 0 1 2 f 0 0 ∈ e V e . 0 0 1 1 0 0 f 0 0 0 Step 3. 0 = . To show this, ﬁrst note that f0 0 0 0 f0 0 0 0 f0 0 0 0 f0 0 ∈ e2V e1 · e1V e1 ⊆ e2V e1. So = 0 or f0 0 0 0 f0 0 0 0 0 0 f 0 0 0 0 0 f 0 0 = . Assume that 0 = 0. Then f0 0 0 0 f0 0 f0 0 0 0 0 0 f 0 0 2 0 0 0 2 0 0 0 = = f0 0 0 0 0 0 f0 0 0 0 0 2 and 0 0 f 0 0 2 0 0 0 2 0 0 0 = = . f0 0 0 0 0 0 0 0 0 0 0 0 0 0 f 0 0 0 Thus we have a contradiction. Therefore 0 = . f0 0 0 0 f0 0 0 0 f 0 f 0 Step 4. 0 = 0 because 0 ∈ e V e . 0 1 0 0 0 0 1 1 Similarly, we get Steps 5 and 6. 1 0 0 0 f 0 0 0 Step 5. = 0. Step 6. 0 = 0. 0 0 f0 0 0 0 f0 0 10 GARY F. BIRKENMEIER, JAE KEOL PARK AND S. TARIQ RIZVI

0 2 0 0 0 2 0 0 Step 7. = 0. For this, note that ∈ 0 0 f0 0 0 0 f0 0 0 2 0 0 a + f0 · r 0 e1V e2·e2V e1 ⊆ e1V e1. Thus = for some a, r ∈ 0 0 f0 0 0 0 0 2 0 0 0 2 0 0 A. Therefore we see that 0 = 2 = 2 = 0 0 f0 0 0 0 f0 0 a + f · r 0 2a 0 2 0 = . So 2a = 0, hence a = 0 or a = 2. 0 0 0 0 0 2 0 0 0 2 0 2 0 0 Also we have that = = 0 and 0 0 f0 0 0 0 0 0 0 2 0 2 0 0 0 2 a + f · r 0 0 2 0 2a + 2r = 0 = = 0 0 f0 0 0 0 0 0 0 0 0 0 0 2r . Thus 2r = 0, so r = 0 or r = 2. Therefore a + f · r = 0 or a + f · r = 2. 0 0 0 0 0 2 0 0 2 0 Next suppose that = . Then 0 0 f0 0 0 0 0 2 0 0 f 0 0 2 0 0 2 0 0 = = 0 0 f0 0 0 0 0 0 f0 0 0 0 and 0 2 0 0 f 0 2 0 f 0 1 0 f 0 0 = 0 = 0 + 0 0 f0 0 0 0 0 0 0 0 0 0 0 0 1 0 f 0 f 0 f 0 2f 0 0 = 0 + 0 = 0 = 0. 0 0 0 0 0 0 0 0 0 0 0 2 0 0 So we have a contradiction. Therefore = 0. 0 0 f0 0 0 0 Finally noting that ∈ e2V e1, we get Steps 8 and 9. f0 0 0 0 0 0 0 0 0 0 0 0 Step 8. = 0. Step 9. = . f0 0 f0 0 0 1 f0 0 f0 0

By Step 1 through Step 9, there is a multiplication on V which extends the R-module multiplication of V over R such that a + f · r 2b a + f · r 2b a + f · r 2b a 2b 1 0 1 1 2 0 2 2 = 1 0 1 1 2 2 + f0 · s1 c1 f0 · s2 c2 f0 · s1 c1 0 c2 a 0 f · r 0 f · r 0 f · r 0 0 2b f · r 0 1 0 2 + 0 1 0 2 + 1 0 2 + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 f · r 0 0 0 f · r 0 0 2 + 0 2 + f0 · s1 0 0 0 0 c1 0 0 a 0 0 0 f · r 0 0 0 0 2b 0 0 1 + 0 1 + 1 + 0 0 f0 · s2 0 0 0 f0 · s2 0 0 0 f0 · s2 0 0 0 0 0 0 0 0 0 + = f0 · s1 0 f0 · s2 0 0 c1 f0 · s2 0 AN EXAMPLE OF OSOFSKY AND ESSENTIAL OVERRINGS 11

a a + f · r a + f · a r + f · r r 2a b + 2b r + 2b c 1 2 0 1 2 0 1 2 0 1 2 1 2 2 1 1 2 . f0 · s1a2 + f0 · s1r2 + f0 · c1s2 2s1b2 + c1c2

Let •1 denote this multiplication on V .

f 0 f 0 2 + f 0 For Subcase 1.2. 0 0 = 0 : 0 0 0 0 0 0 In this case, similar to Subcase 1.1, we have the following 1 0 f 0 f 0 0 2 f 0 Step 1. 0 = 0 . Step 2. 0 = 0. 0 0 0 0 0 0 0 0 0 0 0 0 f 0 0 0 0 0 f 0 Step 3. 0 = . Step 4. 0 = 0. f0 0 0 0 f0 0 0 1 0 0 1 0 0 0 f 0 0 0 Step 5. = 0. Step 6. 0 = 0. 0 0 f0 0 0 0 f0 0 0 2 0 0 0 0 0 0 Step 7. = 0. Step 8. = 0. 0 0 f0 0 f0 0 f0 0 0 0 0 0 0 0 Step 9. = . 0 1 f0 0 f0 0

Therefore, in this case, there exists a multiplication •2 on V which extends the R-module multiplication of V over R such that a1 + f0 · r1 2b1 a2 + f0 · r2 2b2 •2 = f0 · s1 c1 f0 · s2 c2

a a + 2r r + f · r a + f · a r + f · r r 2a b + 2r b + 2b c 1 2 1 2 0 1 2 0 1 2 0 1 2 1 2 1 2 1 2 . f0 · s1a2 + f0 · s1r2 + f0 · c1s2 2s1b2 + c1c2

Case 2. As in Case 1, we see that

f 0 f 0 f 0 f 0 f 0 2 + f 0 0 0 = 0 or 0 0 = 0 . 0 0 0 0 0 0 0 0 0 0 0 0

Thus we have two subcases, Subcase 2.1 and Subcase 2.2 as follows:

f 0 f 0 f 0 Subcase 2.1. 0 0 = 0 . 0 0 0 0 0 0 As in Subcase 1.1, we have the following. 1 0 f 0 f 0 0 2 f 0 Step 1. 0 = 0 . Step 2. 0 = 0. 0 0 0 0 0 0 0 0 0 0 0 0 f 0 2 0 0 0 f 0 Step 3. 0 = . Step 4. 0 = 0. f0 0 0 0 f0 0 0 1 0 0 1 0 0 0 2 0 f 0 0 0 Step 5. = . Step 6. 0 = 0. 0 0 f0 0 0 0 0 0 f0 0 0 2 0 0 0 0 0 0 Step 7. = 0. Step 8. = 0. 0 0 f0 0 f0 0 f0 0 12 GARY F. BIRKENMEIER, JAE KEOL PARK AND S. TARIQ RIZVI

0 0 0 0 2 0 Step 9. = . 0 1 f0 0 f0 0

Thus by Step 1 through Step 9, there is a multiplication •3 on V such that •3 extends the R-module multiplication of V over R: a1 + f0 · r1 2b1 a2 + f0 · r2 2b2 x y •3 = , f0 · s1 c1 f0 · s2 c2 z w

where

x = a1a2 + 2s1r2 + 2a1s2 + 2c1s2 + f0 · r1a2 + f0 · a1r2 + f0 · r1r2,

y = 2a1b2 + 2r1b2 + 2b1c2, z = f0 · s1a2 + f0 · c1s2 + f0 · s1r2, w = 2s1b2 + c1c2.

f 0 f 0 2 + f 0 Subcase 2.2. 0 0 = 0 . 0 0 0 0 0 0 As in Steps 1–9 of Subcase 1.1, we have the following. 1 0 f 0 f 0 0 2 f 0 Step 1. 0 = 0 . Step 2. 0 = 0. 0 0 0 0 0 0 0 0 0 0 0 0 f 0 2 0 0 0 f 0 Step 3. 0 = . Step 4. 0 = 0. f0 0 0 0 f0 0 0 1 0 0 1 0 0 0 2 0 f 0 0 0 Step 5. = . Step 6. 0 = 0. 0 0 f0 0 0 0 0 0 f0 0 0 2 0 0 0 0 0 0 Step 7. = 0. Step 8. = 0. 0 0 f0 0 f0 0 f0 0 0 0 0 0 2 0 Step 9. = . 0 1 f0 0 f0 0

Now by Step1 through Step 9, there is a multiplication •4 on V which extends the R-module multiplication of V over R: a1 + f0 · r1 2b1 a2 + f0 · r2 2b2 x y •4 = , f0 · s1 c1 f0 · s2 c2 z w where

x = a1a2 + 2r1r2 + 2s1r2 + 2a1s2 + 2c1s2 + f0 · r1a2 + f0 · a1r2 + f0 · r1r2,

y = 2a1b2 + 2r1b2 + 2b1c2, z = f0 · s1a2 + f0 · s1r2 + f0 · c1s2, w = 2s1b2 + c1c2.

The multiplications •1, •2, •3, and •4 are well deﬁned and they extend the R-module multiplication of V over R. Thus (V, +, •1), (V, +, •2), (V, +, •3), and (V, +, •4) are all possible compatible ring structures on V .

Deﬁne θ2 :(V, +, •2) → (V, +, •1) by a + f0 · r 2b a + 2r + f0 · r 2b θ2 = . f0 · s c f0 · s c AN EXAMPLE OF OSOFSKY AND ESSENTIAL OVERRINGS 13

Then we see that θ2 is a ring isomorphism. Also deﬁne θ3 :(V, +, •3) → (V, +, •1) and θ4 :(V, +, •4) → (V, +, •1) by

a + f0 · r 2b a + 2s + f0 · r 2b θ3 = , f0 · s c f0 · s c and a + f0 · r 2b a + 2r + 2s + f0 · r 2b θ4 = . f0 · s c f0 · s c ∼ ∼ Then we see that θ3 and θ4 are also ring isomorphisms. So (V, +, •1) = (V, +, •2) = ∼ (V, +, •3) = (V, +, •4). 5. Compatible ring structures on S: there are exactly four compatible ring structures on S such that (α) two compatible ring structures are QF and they are isomorphic; and (β) the other two compatible ring structures are not even right FI-extending and they are isomorphic. Proof of 5. By [BPR4, Theorem 10], there are exactly four ring multipli- a 2b a 2b cations on S which are: For 1 1 , 2 2 ∈ S, f0 · r1 c1 f0 · r2 c2 a1 2b1 a2 2b2 ◦(k,t) = f0 · r1 c1 f0 · r2 c2

a a − ta r + 2kb r + tc r 2a b + 2b c 1 2 1 2 1 2 1 2 1 2 1 2 , f0 · r1a2 + f0 · c1r2 2r1b2 + c1c2 where k ∈ A and t ∈ Soc(A). In this case, note that ◦(0,0) = ◦(2,0), ◦(1,0) = ◦(3,0), ◦(0,2) = ◦(2,2), and ◦(1,2) = ◦(3,2). Thus all possible ring multiplications ∼ are ◦(0,0), ◦(1,0), ◦(0,2), and ◦(1,2). Also, by [BPR4, Theorem 10], (S, +, ◦(0,2)) = ∼ (S, +, ◦(0,0)) and (S, +, ◦(1,0)) = (S, +, ◦(1,2)). Moreover, by [BPR4, Theorem 10] (S, +, ◦(1,0)) (hence (S, +, ◦(1,2))) is QF, while (S, +, ◦(0,0)) (hence (S, +, ◦(0,2))) is ∼ not even right FI-extending. Thus (S, +, ◦(1,0)) =6 (S, +, ◦(0,0)). 6. Compatible ring structures on U: there are exactly two compatible ring structures and they are mutually isomorphic. Proof of 6. Assume that U has a compatible ring structure. Then as in the 1 0 proof for the case of V , 1 = 1 because R ≤ess U . Also let e = ∈ R U R R R 1 0 0 0 0 and e = ∈ R. Then 1 = e +e , e2 = e , e = e2, and e e = e e = 0. 2 0 1 U 1 2 1 1 2 2 1 2 2 1 Thus U = e1Ue1 + e1Ue2 + e2Ue1 + e2Ue2. Also as in the proof for the case of V , 0 2A 0 0 1. e Ue = . 2. e Ue = . 1 2 0 0 2 2 0 A A ⊕ Hom(2A ,A ) 0 3. e Ue = A A , and 4. e Ue = 0. 1 1 0 0 2 1 14 GARY F. BIRKENMEIER, JAE KEOL PARK AND S. TARIQ RIZVI

f 0 f 0 To ﬁnd compatible ring structures on U, note that 0 0 ∈ 0 0 0 0 f 0 f 0 a + f · r 0 e Ue · e Ue ⊆ e Ue . Now take 0 0 = 0 . Then 1 1 1 1 1 1 0 0 0 0 0 0

0 2 f 0 0 2 f 0 f 0 0 2 = 0 = 0 0 = 0 0 0 0 0 0 0 0 0 0 0 0

f 0 f 0 0 2 a + f · r 0 0 2 0 2a + 2r 0 0 = 0 = . 0 0 0 0 0 0 0 0 0 0 0 0 So 2a + 2r = 2. On the other hand, 2 0 f 0 f 0 1 0 f 0 f 0 0 0 = 2 0 0 = 0 0 0 0 0 0 0 0 0 0 0 0

f 0 f 0 0 0 f 0 2 0 0 = 0 = 0. 0 0 0 0 0 0 0 0 Also 2 0 f 0 f 0 2 0 a + f · r 0 0 0 = 0 = 0 0 0 0 0 0 0 0 0 0 1 0 a + f · r 0 1 0 a 0 1 0 f · r 0 2 0 = 2 + 0 = 0 0 0 0 0 0 0 0 0 0 0 0 a 0 f · r 0 a + f · r 0 2a 0 2 + 0 − 2 0 = . 0 0 0 0 0 0 0 0

Thus 2a = 0, hence 2r = 2. Therefore a + f0 · r = f0 or a + f0 · r = 2 + f0. So f 0 f 0 f 0 f 0 f 0 2 + f 0 0 0 = 0 or 0 0 = 0 . 0 0 0 0 0 0 0 0 0 0 0 0

f 0 f 0 f 0 Case 1. 0 0 = 0 . 0 0 0 0 0 0 0 2 f 0 0 0 f 0 In this case, 0 ∈ e Ue · e Ue = 0. Also 0 ∈ 0 0 0 0 1 2 1 1 0 1 0 0 e2Ue2 · e1Ue1 = 0. Thus there is a multiplication on U such that U is a compatible ring structure under this multiplication, which is: a + f · r 2b a + f · r 2b 1 0 1 1 2 0 2 2 = 0 c1 0 c2

a a + f · a r + f · r a + f · r r 2a b + 2r b + 2b c 1 2 0 1 2 0 1 2 0 1 2 1 2 1 2 1 2 . 0 c1c2 We let }1 denote this multiplication.

f 0 f 0 2 + f 0 Case 2. 0 0 = 0 . 0 0 0 0 0 0 0 2 f 0 0 0 f 0 In this case also 0 = 0 and 0 = 0 as in Case 1. 0 0 0 0 0 1 0 0 Thus there is a multiplication on U such that U has a compatible ring structure AN EXAMPLE OF OSOFSKY AND ESSENTIAL OVERRINGS 15 under this multiplication, which is:

a + f · r 2b a + f · r 2b 1 0 1 1 2 0 2 2 = 0 c1 0 c2

a a + 2r r + f · a r + f · r a + f · r r 2a b + 2r b + 2b c 1 2 1 2 0 1 2 0 1 2 0 1 2 1 2 1 2 1 2 . 0 c1c2

We use }2 to denote this multiplication.

By the previous argument, (U, +, }1) and (U, +, }2) are all possible compatible ring structures on U. Deﬁne ψ :(U, +, }2) → (U, +, }1) by

a + f · r 2b a + 2r + f · r 2b ψ 0 = 0 . 0 c 0 c

∼ Then we can see that ψ is a ring isomorphism. Thus (U, +, }1) = (U, +, }2). 7. Compatible ring structures on T : there are exactly two compatible ring structures on T and they are isomorphic. Proof of 7. Assume that T has a compatible ring structure with R. As in 4, we see that A 0 0 0 (1) e T e = , (2) e T e = 0, and (3) e T e = . 1 1 0 0 2 1 2 2 0 A

0 A 0 1 0 2 0 3 Claim. e T e = or e T e = 0, , , . 1 2 0 0 1 2 0 2 0 0 0 2 x y x y Proof of Claim. First, note that if ∈ e T e , then = 0 z 1 2 0 z x y 0 y e = . From (1), (2), and (3), we have that 0 z 2 0 z

0 1 a 0 0 b 0 0 = + , 0 0 0 0 0 c 0 d

0 b where ∈ e T e since T = e T e + e T e + e T e + e T e and e T e = 0. 0 c 1 2 1 1 1 2 2 1 2 2 2 1 0 1 0 1 0 0 Thus a = 0, b = 1 and d = −c. So = + with 0 0 0 c 0 −c 0 1 ∈ e T e . Therefore we have the following cases. 0 c 1 2 0 1 0 A (α) c = 0. In this case, ∈ e T e . Thus ⊆ e T e , so 0 0 1 2 0 0 1 2 0 A e T e = since |e T e | = 4. 1 2 0 0 1 2 16 GARY F. BIRKENMEIER, JAE KEOL PARK AND S. TARIQ RIZVI

x 2y 0 1 x 2y (β) c = 1. There is ∈ R such that 0 6= = 0 z 0 1 0 z 0 z 0 z ∈ R since R ≤ess T . Hence 0 6= ∈ e R, a contradiction. So 0 z R R 0 z 1 this case cannot happen. 0 1 0 2 0 1 (γ) c = 2. Then ∈ e T e . Thus = 2 ∈ 0 2 1 2 0 0 0 2 0 3 0 1 0 1 e T e , = 3 ∈ e T e , 0 = 4 ∈ e T e . Thus e T e = 1 2 0 2 0 2 1 2 0 2 1 2 1 2 0 1 0 2 0 3 0, , , because |e T e | = 4. 0 2 0 0 0 2 1 2 0 1 x 2y (δ) c = 3. Then ∈ e T e . Thus there is ∈ R with 0 3 1 2 0 z 0 1 x 2y 0 z 0 z 0 6= = ∈ R. Hence 0 6= ∈ e R, which is a 0 3 0 z 0 3z 0 3z 1 contradiction. So this case cannot happen. Thus the proof of Claim is completed. By (1), (2), (3), and Claim, we get the following. A 0 0 A Case 1. e T e = , e T e = , e T e = 0, and 1 1 0 0 1 2 0 0 2 1 0 0 e T e = . 2 2 0 A

A 0 0 1 0 2 0 3 Case 2. e T e = , e T e = 0, , , , 1 1 0 0 1 2 0 2 0 0 0 2 0 0 e T e = 0, and e T e = . 2 1 2 2 0 A

1 0 0 1 0 1 Case 1. In this case, = since e ∈ e T e and 0 0 0 0 0 0 1 1 1 0 1 0 1 0 1 ∈ e T e . Also ∈ e T e · e T e = 0. Similarly, we have 0 0 1 2 0 0 0 0 1 2 1 2 0 0 0 1 that ∈ e · e T e = 0. 0 1 0 0 2 1 2

Therefore there exists a multiplication on T such that T has a compatible ring structure under this multiplication which is: a b a b a a a b + b c 1 1 2 2 = 1 2 1 2 1 2 . 0 c1 0 c2 0 c1c2

We let 1 denote this multiplication. We remark that the ring (T, +, 1) is the 2-by-2 upper triangular matrix ring over the ring A.

Case 2. We ﬁnd another compatible ring structure on T by the following steps. AN EXAMPLE OF OSOFSKY AND ESSENTIAL OVERRINGS 17

1 0 0 1 0 1 Step 1. = . 0 0 0 0 0 2 1 0 0 1 0 1 1 0 To show this, note that = since e = and 0 0 0 2 0 2 1 0 0 0 1 0 1 1 0 0 1 1 0 0 1 ∈ e T e . Thus = = + 0 2 1 2 0 2 0 0 0 2 0 0 0 0 1 0 0 0 1 0 0 1 = . 0 0 0 2 0 0 0 0 0 0 0 1 0 0 Step 2. = . 0 1 0 0 0 2 0 0 0 1 For this step, note also that ∈ e · e T e = 0. Thus 0 = 0 1 0 2 2 1 2 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 = + = + 0 1 0 2 0 1 0 0 0 1 0 2 0 1 0 0 0 0 0 0 0 1 0 0 . Therefore = . 0 2 0 1 0 0 0 2 0 1 0 1 0 2 Step 3. = . 0 0 0 0 0 0 0 1 0 1 For Step 3, also note that ∈ e T e · e T e = 0. Now 0 2 0 2 1 2 1 2 0 1 0 1 0 1 0 0 0 1 0 0 0 = = + + = 0 2 0 2 0 0 0 2 0 0 0 2 0 1 0 1 0 1 0 0 0 0 0 1 0 0 0 0 + + + = 0 0 0 0 0 0 0 2 0 2 0 0 0 2 0 2 0 1 0 1 0 2 0 0 0 1 0 1 0 1 0 2 + + = + 0 0 0 0 0 0 0 2 0 0 0 0 0 0 0 0 because 0 0 0 1 0 0 0 0 0 1 = + = 0 2 0 0 0 1 0 1 0 0 0 0 0 1 0 0 2 = 2 = 0 0 1 0 0 0 2 by Step 2. Therefore 0 1 0 1 0 2 = . 0 0 0 0 0 0

By Steps 1, 2, and 3 of Case 2, there is also a multiplication 2 on T such that T has a compatible ring structure under this multiplication which is: a1 b1 a2 b2 a1a2 a1b2 + 2b1b2 2 = . 0 c1 0 c2 0 c1c2 + 2a1b2 + 2c1b2

a b a b Now deﬁne θ :(T, +, ) → (T, +, ) by θ = . Then 1 2 0 c 0 2b + c we can see that θ is a ring isomorphism. Consequently, part (i) can be proved by 1 through 7. 18 GARY F. BIRKENMEIER, JAE KEOL PARK AND S. TARIQ RIZVI

(ii) This part can be routinely checked.

Theorem 2.3. Let R, V, S, U and T be as in Theorem 2.2. Then we have the following. (i) R is not right FI-extending.

(ii) (V, +, •1) (hence (V, +, •2), (V, +, •3) and (V, +, •4)) is right extending, but not right quasi-continuous.

(iii) (S, +, ◦(1,0)) (hence (S, +, ◦(1,2))) is right self-injective, while (S, +, ◦(0,0)) (hence (S, +, ◦(0,2))) is not right FI-extending. (iv) (U, +, }1) (hence (U, +, }2)) is right FI-extending, but not right extending. (v) (T, +, 1) (hence (T, +, 2)) is right FI-extending, but not right extending. 0 2A Proof. (i) Let I = . Then I R. But there is no idempotent e ∈ R 0 0 E ess such that IR ≤ eRR. Thus R is not right FI-extending. 1 0 0 0 (ii) Let ∆ := (V, +, • ). Recall that e = and e = ∈ ∆. 1 1 0 0 2 0 1 f 0 1 + f 0 Let g = 0 , g = 0 . Then e ∆ = g ∆ ⊕ g ∆ . We can 1 0 0 2 0 0 1 ∆ 1 ∆ 2 ∆ see that each gi∆∆ is uniform, hence it is extending. Also gi∆∆ is gj∆∆-injective for each i 6= j. Thus e1∆∆ is extending by [DHSW, Proposition 7.10, p.59]. Since e2∆∆ is uniform, e2∆∆ is extending.

By computations, all right ideals K of ∆ such that K ∩ e2∆ = 0 are: 0 2A 2A 0 1 + f 0 f 0 2A 2A 0, , , 0 ∆, 0 ∆, , 0 0 0 0 0 0 0 0 0 0

1 + f 2 2 + f 0 1 0 0 ∆, 0 ∆, ∆, 0 0 0 0 0 0 0 2 0 2A 2 2 1 + f 2 ∆, , ∆, 0 ∆, 0 2 0 2A 0 2 0 2 f 0 2 + f 0 1 0 0 ∆, 0 ∆, and ∆. f0 0 f0 0 f0 0 In these right ideals K, we can verify that closed right ideals are:

1 + f 0 f 0 1 0 f 0 1 0 0, 0 ∆, 0 ∆, ∆, 0 ∆, and ∆. 0 0 0 0 0 0 f0 0 f0 0

Also these are direct summands of ∆∆. Thus every closed right ideal K of ∆ such that K ∩ e2∆ = 0 is a direct summand of ∆∆. Also all the possible right ideals I of ∆ such that I ∩ e1∆ = 0 are: 0 0 0 0 0 0 0 2 , , ∆, ∆, 0 2 Hom(2AA,AA) 2A 0 1 0 2

f 0 0 2 0 2 0 ∆, ∆, and ∆. f0 0 0 1 f0 1 AN EXAMPLE OF OSOFSKY AND ESSENTIAL OVERRINGS 19

Among these, all of the closed right ideals are: 0 0 f 0 0 2 0 2 ∆, 0 ∆, ∆, and ∆, 0 1 f0 0 0 1 f0 1

which are direct summands of ∆∆. Thus every closed right ideal I of ∆ with I ∩ e1∆ = 0 is a direct summand of ∆∆. Consequently, by [DHSW, Lemma 7.9, p.59], ∆ is right extending. f 0 Next, we show that ∆ is not right quasi-continuous. For this, let v = 0 f0 0 2 ∈ ∆. Then v = v . Thus v∆∆ is a direct summand of ∆∆. Now e2∆∆ is a direct summand of ∆∆ and v∆ ∩ e2∆ = 0. But v∆ ⊕ e2∆ is not a direct summand of ∆∆. So ∆ is not right quasi-continuous. Thus by the proof of Theorem 2.2 (Step 4), (V, +, •2), (V, +, •3), and (V, +, •4) are not right quasi-continuous. ∼ (iii) By the proof of Theorem 2.2 (Step 5), (S, +, ◦(1,0))(= (S, +, ◦(1,2))) is right ∼ self-injective, but (S, +, ◦(0,0))(= (S, +, ◦(0,2))) cannot be right FI-extending.

(iv) Let Φ = (U, +, }1). Then f 0 1 + f 0 0 0 Φ = 0 Φ ⊕ 0 Φ ⊕ Φ . Φ 0 0 Φ 0 0 Φ 0 1 Φ

f 0 1 + f 0 0 0 We see that 0 Φ , 0 Φ and Φ are uniform, hence 0 0 Φ 0 0 Φ 0 1 Φ they are extending. Thus Φ is right FI-extending by [BMR, Theorem 1.3]. 2 2 Next, take K = Φ. Then the only possible e = e2 ∈ Φ with K ≤ eΦ 0 2 Φ Φ ess is 1. Thus if Φ is right extending, then KΦ ≤ ΦΦ. But this is absurd because f 0 0 Φ ∩ K = 0. Thus Φ is not right extending, so by the proof of Theorem 0 0 2.2 (Step 6)(U, +, }2) also is right FI-extending, but not right extending.

(v) Let Γ = (T, +, 1). Since A is self-injective, it is FI-extending. So Γ is right FI-extending from [BMR, Corollary 2.5]. Next, by [BPR3, Proposition 2.11(ii)], Γ is not right extending. Hence by the proof of Theorem 2.2 (Step 7), also (T, +, 2) is right FI-extending, but not right extending. Definition 2.4. ([BPR2, Definition 2.1]) Let K denote a class of rings and let R be a ring. Then an right essential overring S of R in a fixed injective hull of RR is called a K right ring hull of R if S is a minimal right essential overing of R with S ∈ K. For more details on ring hulls and their applications, see [BPR2]. From Theorems 2.2 and 2.3, and their proofs, we obtain the following properties of these ring hulls. Proposition 2.5. Let R, V, S, U and T be as in Theorem 2.2. Then we have the following.

(i) All right FI-extending right ring hulls of R are precisely: (V, +, •1), (V, +, •2), (V, +, •3), (V, +, •4), (S, +, ◦(1,0)), (S, +, ◦(1,2)), (U, +, }1), (U, +, }2), (T, +, 1), and (T, +, 2).

(ii) All right extending right ring hulls of R are precisely: (V, +, •1), (V, +, •2), (V, +, •3), (V, +, •4), (S, +, ◦(1,0)), and (S, +, ◦(1,2)). 20 GARY F. BIRKENMEIER, JAE KEOL PARK AND S. TARIQ RIZVI

(iii) All right quasi-continuous right ring hulls of R are precisely: (S, +, ◦(1,0)) and (S, +, ◦(1,2)).

(iv) All right continuous right ring hulls of R are precisely: (S, +, ◦(1,0)) and (S, +, ◦(1,2)).

(v) All right self-injective right ring hulls of R are precisely: (S, +, ◦(1,0)) and (S, +, ◦(1,2)).

3. Osofsky Compatibility As was mentioned in Section 2, Osofsky [O2] has shown that no right injective 2 hull of the ring Z4 Z4 can have a compatible ring structure. 0 Z4 The following lemma from [BPR3, Corollary 2] extends the above example. Lemma 3.1. ([BPR3, Corollary 2.2]) Let A be a commutative local QF-ring with A Soc(A) J(A) 6= 0. Then R = is not a right Osofsky compatible A-algebra. 0 A An open question was posed in [BPR3, Question (ii)] whether this result can be extended to the case when the commutative QF-ring A is not necessarily local. In the following, we give an answer to the question in the aﬃrmative by characterizing a class of algebras which are not right Osofsky compatible. Theorem 3.2. Let A be a commutative QF-ring and let A Soc(A) R = . 0 A Then the following are equivalent. (i) J(A) 6= 0.

(ii) No injective hull of RR has a compatible A-algebra structure (i.e., R is not a right Osofsky compatible A-algebra).

n Proof. Assume that J(A) 6= 0. Then A = (⊕i=1Ai) ⊕ B (ring direct sum), where n ≥ 1, each Ai is local with J(Ai) 6= 0, and J(B) = 0. Let

n n A1 Soc(A1) (⊕i=2Ai) ⊕ B (⊕i=2Soc(Ai)) ⊕ B Λ = and Ω = n . 0 A1 0 (⊕i=2Ai) ⊕ B

Then R = Λ ⊕ Ω and E(RR) = E(ΛR) ⊕ E(ΩR). Then E1 = E(ΛR) = E(ΛΛ) and E2 = E(ΩR) = E(ΩΩ). By Lemma 3.1, E1 = E(ΛΛ) does not have a compatible A1-algebra structure. Assume to the contrary that E(RR) has a compatible A-algebra structure (E(RR), +, •), where • extends the R-module multiplication of E(RR) over R. For any x, y ∈ E1, we have that (x, 0) • (y, 0) = (µ(x, y), ν(x, y)) for some µ(x, y) ∈ E1 and ν(x, y) ∈ E2. Then ν(x, y) is additive in each of the variables. Consider any (r1, r2) ∈ R = Λ ⊕ Ω. Then [(x, 0) • (y, 0)] • (r1, r2) = (µ(x, y), ν(x, y)) • (r1, r2) = (µ(x, y)r1, ν(x, y)r2). Also, we see that (x, 0) • [(y, 0) • (r1, r2)] = (x, 0) • (yr1, 0) = (µ(x, yr1), ν(x, yr1)). Hence ν(x, yr1) = ν(x, y)r2. By taking r1 as unity in R1 and r2 = 0, we get ν = 0. Thus (x, 0) • (y, 0) = (µ(x, y), 0) AN EXAMPLE OF OSOFSKY AND ESSENTIAL OVERRINGS 21 for x, y ∈ E1. Let x ◦ y = µ(x, y) for x, y ∈ E1. Then we can see that (E1, +, ◦) is a compatible A1-algebra structure on E1 for which the multiplication ◦ extends the Λ-module multiplication. But this is a contradiction by Lemma 3.1. Thus E(RR) has no compatible ring structure. 0 0 Let E be any injective hull of RR. Assume to the contrary that that E 0 has a compatible A-algebra structure (E , +, }). There is an R-isomorphism f : 0 E(RR) → E which extends the identity map on RR. For v, w ∈ E(RR) deﬁne −1 v w = f (f(v) } f(w)). Then we see that (E(RR), +, ) is a compatible A- algebra structure on E(RR), which is a contradiction. Consequently, no injective hull of RR has a compatible A-algebra structure. Conversely, if J(A) = 0, then A is semisimple Artinian, so R is right nonsingu- lar. Thus E(RR) = Q(R), so every injective hull of RR has a compatible A-algebra structure.

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Department of Mathematics, University of Louisiana at Lafayette, Lafayette, Louisiana 70504-1010, U. S. A. E-mail address: [email protected]

Department of Mathematics, Busan National University, Busan 609-735, South Korea E-mail address: [email protected]

Department of Mathematics, Ohio State University at Lima, Lima, OH 45804- 3576, U. S. A. E-mail address: [email protected]