arXiv:0910.1995v1 [math.RA] 11 Oct 2009
Remarks On ℵ0-Injectivity E. Momtahan [email protected] Department of Mathematics, Yasouj University, ...

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Remarks On ℵ0-Injectivity E. Momtahan [email protected] Department of Mathematics, Yasouj University, Yasouj, Iran

Abstract We continue our review of the literature via countable injectivity.

2000 Mathematics Subject Classifications: 16D90; 16D50.

1

Introduction

In this article, by R we mean an associative ring with identity and M is a unitary R-module. An R-module M is said to be ℵ0 -injective (f-injective) if every (module) ¯ I = f , where homorphism f ∈ HomR (I, M), there exists f¯ ∈ HomR (R, M), such that f| I is any countably generated (finitely generated) right ideal of R. A ring R is said to be right ℵ0 -self-injective (f-injective) if it is ℵ0 -injective (f-injective) as a right R-module. In this note, by C(X) we mean the ring of all real valued continuous over a completely regular space (or equivalently a Tychonoff space). The reader is referred to [8] for undefined notations and definitions. A module is called extending (ℵ0 -extending) if every submodule (countably generated submodule) is a direct summand. This paper is a continuation of [15] and [16], in which the author studied ℵ0 injectivity of modules and rings. In [15], many well-known observations on injective modules have been seen to be true if we replace injectivity by ℵ0 -injectivity. Among them, for example, one can quote, this one: a ring R is semisimple artinian if and only every R-module (or every countably generated R-module) is ℵ0 -injective. Using this result, one can prove the following proposition which is well-known if we replace injectivity by ℵ0 -injectivity. The equality of the first two parts of the following result is due to B. Osofsky (see [14, page 114, excersise 11]). 1

Proposition 1.1. The following assertions are equivalent: 1. R is semisimple Artinian 2. the intersection of any two injective R-modules is injective. 3. the intersection of any two ℵ0 -injective R-modules is ℵ0 -injective. Proof. The proof which is given for (1) ⇔ (3) works also for (1) ⇔ (2). (1) ⇒ (3) is obvious. Suppose that the intersection of any two ℵ0 -injective R-modules is ℵ0 injective. Now suppose that R were not semisimple Artinian. In [15], it has been observed that a ring is semi-simple Artinian if and only if every module is ℵ0 -injective. Hence by our assumption there is an R-module M which is not ℵ0 -injective. Set E = E(M) ⊕ E(E(M)/M). Then E is injective. Let x belong to E(M) \ M. Set y = (x, x + M) ∈ E. Then (M, 0) is essential in (E(M), 0) and E((M, 0) + yR) ⊆ E and the intersection of those two injectives is M which is not ℵ0 -injective. A contradiction.

Remark 1.2. It is well-known that a ring is von Neumann regular if and only if every module over R is f-injective. Now the same method used in the above result shows that (i) a ring R is von Neumann regular if and only if the intersection of any two f-injective module is f-injective. It is also well-known that a ring R is (left and right) artinian serial with Jac(R)2 = 0 if and only if every module over R is extending (see [6, 13.5]). Again, the same line of proof shows that a ring R is (left and right) artinian serial if and only if the intersection of any two extending modules is extending. However, if every finitely generated (or even cyclic) R-module is ℵ0 -injective, we do not need to access the semisimplicity of R (contrary to Osofsky’s famous result which asserts that a ring R is semisimple artinian if and only if every cyclic Rmodule is injective). In fact, a ring is right ℵ0 -self-injective regular if and only if every cyclic R-module is ℵ0 -injective. In [16], it has been found out that the wellknown result of Y. Utumi that every right (or left) self-injective ring module its Jacobson radical is von Neumann regular, is no longer true if we replace injectivity by ℵ0 -injectivity. A commutative ℵ0 -self-injective ring with many additional properties has been constructed that is not von Neumann regular modulo its Jacobson radical. The next proposition will help us in the sequel. It is well-known for injectivity (see [21]).

2

Proposition 1.3. Let M be an R-module and I ⊆ Ann(M). If M is an ℵ0 -injective R-module, then M is an ℵ0 -injective R/I-module. The converse is true provided that R is a right fully idempotent ring. Proof. Let M be an ℵ0 -injective R-module, and f : A/I −→ M be an R/Ihomomorphism, where A/I is a countably generated right ideal. Thus we may put P A = B + I, where B = ∞ k=1 ak R and then let f (ak + I) = mk , where mk ∈ M. Defining g : B −→ M by g(ak ) = mk = f (ak + I), by our hypothesis g can be extended to R. Suppose now M is ℵ0 -injective as an R/I-module, and let f : A −→ M be an R-homomorphism, where A is a countably generated right ideal. First we claim that T T Pk=n f (A I) = 0: for every x ∈ A I we have x = k=1 ak bk where ai and bi belong to T A I and therefore k=n k=n X X f (x) = f ( ak bk ) = f (ak )bk k=1

k=1

T but f (ak )bk ∈ MI = 0 so f (A I) = 0. In as much as A + I/I ∼ = A/A I and T f (A I) = 0 we infer that f¯ : A + I/I −→ M is an R/I-module homomorphism so f¯ : R/I −→ M, i.e., f¯(a + I) = (a + I)(r + I) = ar + I for some r ∈ R, i.e. f (a) = ar. T

2

ℵ0-Ikeda-Nakayama rings

In [11, Theorem 1], it has been proved that a ring R is right f-injective if and only T if (i)l(I1 I2 ) = l(I1 ) + l(I2 ), where I1 and I2 are finitely generated left ideals of R and (ii) r(l(I)) where I is a principal ideal right ideal of R. Rings satisfy in (i), for each pair of right ideals are called Ikeda-Nakayama (IK-)rings. Such rings have been extensively studied (see for example [4]). IK-rings have been shown to be a special class of quasi-continuous rings (see [4], where it has been observed for the first time). Furthermore, a theory of IK-modules is shown in [20],. It is easy to see that every right self-injective module is a right IK-ring. However, the converse is not true. This is because, the ring of integers, Z, is an IK-ring that is not self-injective, since Z is not a divisible abelian group. Here we consider ℵ0 -IK rings and study their behaviour in Boolean rings and rings of real valued continuous functions over a Tychonoff space. T Definition 2.1. A ring is called a right ℵ0 -IK ring if l(I1 I2 ) = l(I1 ) + l(I2 ), where I1 and I2 are countably generated left ideals of R. We begin with a lemma which is useful in the sequel: 3

Lemma 2.2. If R is a commutative ℵ0 -selfinjective ring, then R is an ℵ0 -IK ring. T Proof. Obviously l(I1 ) + l(I2 ) ⊆ l(I1 I2 ). Suppose on the other hand that a ∈ T l(I2 I2 ), we can define a homomorphism α : I1 + I2 −→ R as: (

b b ∈ I1 , (1 + a)b b ∈ I2 . T Since these two expressions coincide on I1 I2 , by ℵ0 -injectivity of R, ∃c ∈ R such that α(b) = cb for b ∈ I1 . Thus, we have cb = b, i.e., (c − 1)b = 0. Consequently we write a = (c − 1) + (1 + a − c)(c − 1 ∈ l(I1 ) and (1 + a − c) ∈ l(I2 )), this proves the lemma. α(b) =

For the sequel we need the following result which is an slight modification of Theorem 8 in [20]. Lemma 2.3. Let MR be a right R-module and S = End(M). Then the following are equivalent: 1. M is ℵ0 -π-injective (ℵ0 -quasi continuous) 2. For any two countably generated submodules A and B of MR , S = lS (A)+lS (B). Using these results we have the following result in C(X). In the following a space X is called extremally disconnected if open sets have open closure. By a basically disconnected space we mean a space in which co-zero sets (i.e. the complement of zero sets) have open closure. Theorem 2.4. Let X be a completely regular space. Then the following are equivalent: 1. for any two countably generated ideals A and B of C(X),with A Ann(A) + Ann(B) = C(X)

T

B = (0);

2. X is basically disconnected. Proof. (1)⇒ (2) By Lemma 1.2, we observe that C(X) is ℵ0 -continuous. Hence it is an ℵ0 -extending ring. By [3], Theorem 3.3, X is basically disconnected. (2)⇒ (1) Since X is basically disconnected, again by [3], Theorem 3.3, we observe that C(X) is ℵ0 -extending. Let he1 i and he2 i be two ideals which are summands of a commutative ring R. It is well-known that he1 i + he2 i = he1 + e2 − e1 e2 i and 4

T he1 i he2 i = he1 e2 i. Hence C(X) satisfies summand sum property (SSP) and summand intersection property. This implies that C(X) is an ℵ0 -extending ring if and only if C(X) is an ℵ0 -quasi continuous ring.

Theorem 2.5. Let X be a completely regular space. Then the following are equivalent: 1. for any two ideals A and B of C(X) with A

T

B = (0),

Ann(A) + Ann(B) = C(X) 2. X is an extremally disconnected space, Proof. (1)⇒(2) By [20], Theorem 8, C(X) is a quasi-continuous ring and hence extending. By [3], Theorem 3.5, we deduce that X is extremally disconnected. (2)⇒(1): By [3], Theorem 3.5. when X is extremally disconnected, then C(X) is an extending ring. But as we stated in the proof of the above theorem, C(X) has always summand sum property. Therefore it is quasi-continuous when it is extending. Now by [20, Theorem 8] the implication follows. This completes the proof. Remark 2.6. It is natural to speculate on the space X if C(X) satisfies the following stronger conditions: (i) for any two ideals A and B of C(X), Ann(A) + Ann(B) = C(X), and (ii) for any two countably generated ideals A and B of C(X), Ann(A) + Ann(B) = C(X). We conjecture that in the first case X is an extremally P -space (and hence C(X) is self-injective regular in this case) and in the second case, X would be a P -space. We leave these questions open for those who are interested in the theory of rings of continuous functions. In [7], it has been shown that C(X) is ℵ0 -self-injective if and only if C(X) is regular. Hence if our conjecture (i.e., conjecture (ii)) is true, we would have: C(X) is ℵ0 -IK if and only if C(X) is ℵ0 -self-injective. The following result is due to O. A. S. Karamzadeh and A. A. Koochakpour [12]. S Let A and B be two subsets of the ring R. We say A B is an orthogonal set if S ∀x, y ∈ A B, x 6= y, then xy = 0. We say that an element x separates A from B if xa2 = a for all a ∈ A and xB = 0. If there exists such an x, then we say that A has a left separation from B. Lemma 2.7. For a strongly regular ring R the following are equivalent: 1. R is ℵ0 -self-injective 5

S T 2. If S T is a countable orthogonal set in R with S T = ∅, then S has a left separation from T . Andrew B. Carson [5], has shown that if R is an ℵ0 -complete Boolean ring (Boolean joins of countably many elements always exist), then R is ℵ0 -self-injective. He also showed that there are ℵ0 -self-injective Boolean rings which are not ℵ0 -complete. In spite of this fact, we show that in a Boolean ring the following fact holds: Proposition 2.8. Let R be a Boolean ring, then the following are equivalent: 1. R is an ℵ0 -IN-ring. 2. for any two ideals A and B of R with A 3. R is ℵ0 -self-injective.

T

B = (0), Ann(A) + Ann(B) = R.

Proof. (1)⇒(2) It is always true. (2)⇒(3): We want to show that R is ℵ0 -self-injective ring. According to Lemma 2.7, it is enough to show that each two disjoint orthogonal countable subsets of R can S T be separated by an element of R. Let S T be an orthogonal set with S T = ∅. Since R is an ℵ0 -IN ring and Ann(B) = Ann(hBi), where B ⊆ R and by hBi we mean the ideal generated by the set B, we have \ Ann(hSi hT i) = Ann(hSi) + Ann(hT i)

T It is then evident that R = Ann(hSi hT i) = Ann(hSi)+Ann(hT i), i.e. 1 = x+y, where xS = 0 and yT = 0. Now we see that 1 − x = y separates S from T , because yT = 0 and yS = (1 − x)S = S i.e. ya = (1 − x)a = a − xa = a. But R is a Boolean ring, hence ya2 = a. (3)⇒(1) follows from Lemma 2.1.

Along this line, we observe that a Boolean ring is an IK-ring if, and only if, it is self-injective. We need the following lemma which has been proved in [13]: Lemma 2.9. For a strongly regular ring R the following are equivalent: 1. R is self-injective S T 2. If S T is an orthogonal set in R with S T = ∅, then S has a left separation from T . 6

Proposition 2.10. Let R be a Boolean ring, then the following are equivalent: 1. R is an IN-ring, 2. for any two ideals A and B of R with A 3. R is self-injective.

T

B = (0), Ann(A) + Ann(B) = R.

Proof. Use Lemma 2.9 and follow exactly the proof of Proposition 2.8.

3

P -ℵ0-injectivity

P In [9], C. Faith has shown that a -injective R-module M with endomorphism ring S is characterized by the ascending chain condition on the lattice of S-submodules which are annihilators of subsets of R ([9, Prposition 3.3]). If E(R) denotes the injective hull of RR , and if M = E(R), this condition implies the ascending chain condition on annihilator right ideals (=right annulets) of R, and in case M = E(R) = R, this condition is equivalent to a.c.c. on right annulets ([9, Corollary 3.4 and Theorem 3.5]). Now we observe that some parts of these results by C. Faith in [9] can be P applied to answer a similar question concerning -ℵ0 -injectivity. Let MR be a left R-module and S = End(M). Following Faith’s nomenclature, for any subset X of M, X ⊥ = AnnR (X) = {r ∈ R | Xr = 0} is a right ideal of R. The set of such right ideals is denoted by A. By Aℵ0 , we mean all countably generated members of A. And by Bℵ0 , we mean the set of all annihilators of members of Aℵ0 in M. Since I −→ I ⊥ is an injective and order-inverting map between Aℵ0 and Bℵ0 , one satisfies the ascending chain condition if, and only if, the other one satisfies the descending chain condition. Proposition 3.1. If MR is an R-module, then Aℵ0 satisfies the ascending chain condition if and only if for each countably generated right ideal I of R there exists a finitely generated subideal I1 such that I ⊥ = I1⊥ . Proof. Assume a.c.c. for Aℵ0 or equivalently, the d.c.c. for Bℵ0 . Let I be a countably generated right ideal of R, and let I1 be a finitely generated subideal of I, such that 7

I1⊥ be a minimal element of the set {K ⊥ : K is a finitely generated subideal of I} in M. If x ∈ I, then I1 + xR is also a finitely generated subideal of I, and hence satisfies (I1 + xR)⊥ ⊆ I1⊥ . By the choice of I1 , necessarily (I1 + xR)⊥ = I1⊥ , so I1⊥ x = 0. Since this is true for all x ∈ I, then I1⊥ I = 0, whence, I1⊥ ⊆ I ⊥ . Moreover, I1 ⊆ I implies I ⊥ ⊆ I1⊥ , consequently I1⊥ = I ⊥ follows. Conversely, let I1 ⊆ I2 ⊆ · · · ⊆ In ⊆ · · · be a chain of countably generated right ideals of R lying in Aℵ0 , let Xi = Ii⊥ , i = 1, 2, · · · , be the corresponding elements of S Aℵ0 , and suppose also I = ∞ n=1 In . Now, let J be the finitely generated subideal if ⊥ ⊥ I such that I = J . Since J is finitely generated, there is an integer q such that J ⊆ Ik , k ≥ q. Thus, Ik⊥ ⊆ J ⊥ , k ≥ q. Moreover, ⊥

⊥

J =I =

∞ \

In⊥ ,

n=1

Consequently, Ik⊥ = J ⊥ , for k ≥ q. Then, Ik = (Ik⊥ )⊥ = Iq , k ≥ q, and the result follows.

Corollary 3.2. A ring R satisfies the a.c.c. on Bℵ0 if and only if each countably generated right ideal I contains a finitely generated ideal I1 , such that I ⊥ = I1⊥ . Proposition 3.3. The following conditions on an ℵ0 -injective module M are equivalent: 1. M (N) is ℵ0 -injective, 2. R satisfies the a.c.c. on the ideals in Aℵ0 , P 3. M is -ℵ0 -injective.

Proof. (1)⇒(2) Let I1 ⊆ I2 ⊆ · · · ⊆ Im ⊆ · · · be a strictly ascending chain of right S ⊥ ideals in Aℵ0 , let I = ∞ n=1 In , and let xn , be an element of In (taken in M) not ⊥ in In+1 , n = 1, 2, · · · . If r ∈ I, then xn r = 0 for all k ≥ q. Therefore the element r ′ = (x1 r, · · · , xn r, · · · ) lies in M (N) , even though x = (x1 , · · · , xn , · · · ) lies in M N . Let f denote a map defined by f (r) = r ′ for all r ∈ I. Assuming that M (N) is injective, there exists, by Baer’s criterion, an element y = (y1 , · · · , ym , 0, · · · ) ∈ M (R) such that f (r) = yr = (y1 r, · · · , ym r, 0, · · · ) = (x1 r, · · · , xm r, · · · ) 8

for all r ∈ I. But this implies that xt r = 0 for all t > m, for all r ∈ I, that is, ⊥ xt ∈ I ⊥ ⊆ It+1 , this contradicts the choice of xt . (2)⇒(3). Let I be a right ideal of R, and I1 = r1 R + · · · + rn R be the finitely generated subideal given by the above proposition such that I ⊥ = I1⊥ . Let f : I −→ M (Λ) be any map. Since M Λ is injective (it is direct product of injective modules), there exists an element p ∈ M Λ such that f (r) = pr for all r ∈ I. Since f (ri ) = pri ∈ M (Λ) , i = 1, · · · , n, there exists an element p′ ∈ M (Λ) such that pa ri = p′a ri for all a ∈ Λ, i = 1, · · · , n, where ga is the a coordinate of any g ∈ M Λ . Since r1 , · · · , rn generate I1 , this implies that pr = p′ r for all r ∈ I1 , hence (pa − p′a ) ∈ I1⊥ for all a ∈ Λ. Since I1⊥ = I ⊥ , it follows that for all x ∈ I and for all a ∈ A, that is px = p′ x, for all x ∈ I. Thus, f (x) = p′ x for all x ∈ I, with p′ ∈ M (Λ) , so M (Λ) is ℵ0 -injective by Baer’s criterion. Along this line and as an application of Proposition 1.3, we consider the following theorem which is well-known when we replace ℵ0 -injectivity by injectivity. We need the following lemma. P Lemma 3.4. Suppose R is a regular ring and M is an R-module. Then M is -ℵ0 injective if and only if R/A is an artinian ring where A = Ann(M). Proof. A standard proof given in [18], works here with only a slight modification. Theorem 3.5. Let R be a regular ring which is an algebra over a field F . Let M be an R-module with dimF M ≤ ℵ0 , and let I = Ann(M). Then the following are equivalent: 1. R/I is Artinian 2. M is

P -ℵ0 -injective

3. M is ℵ0 -injective.

Proof. Using the proof of the above theorem (see [18]) and apply Proposition 1.3.

4

On ℵ0-quasi injectivity

It is well-known that ℵ0 -injectivity is a Morita property for regular rings ([10, Chapter 14]). In fact, by a theorem of D. Handelman we have: Let R be a right ℵ0 -self-injective regular ring. If M is a finitely generated projective right R-module, then T = End(M) 9

is a right ℵ0 -self-injective regular ring. We say that a module is ℵ0 -quasi injective if every R-homomorphism f : B −→ M extends to M, when B is a countably generated submodule of M. By the proof of Handelman’s theorem we infer that: Corollary 4.1. Let R be an ℵ0 -selfinjective regular ring, and P is a finitely generated projective module, then P is ℵ0 -quasi injective. As we saw over ℵ0 -self-injective regular rings, every finitely generated free module is ℵ0 -quasi injective. This is a good motivation to study ℵ0 -quasi injectivity. J. Ahsan [1] introduced the concept of qc-ring. Here we will present results on ℵ0 -qc ring. Definition 4.2. A ring R is called right ℵ0 -qc ring if every R-homomorphic image of R as a right R-module is ℵ0 -quasi injective. The next result is an extension of a theorem of Nicholson and Yousif [17]. Although it is stated for quasi injective module, with only a slight modification, it is true for ℵ0 -quasi injective modules. Recall that a module M is Dedekind-finite if M ⊕N ∼ =M then N = 0. Proposition 4.3. Let M be a quasi injective module. Then M is Dedekind-finite if, and only if, M is co-hopfian. Proof. It is well-known that M is Dedekind-finite if, and only if, End(M) is a Dedekind-finite ring. Now let f : M −→ M be an R-homomorphism. If f is a monomorphism, then there exits g : f (M) −→ M such that g = f −1 , but f (M) is a submodule of M, so there exists g ′ : M −→ M such that g ′f = 1 (by quasi injectivity of M). But End(M) is Dedekind-finite so f g ′ = 1, whence, f is an epimorphism. Conversely, let f g = 1M . This implies that g is monic (g(m) = 0 ⇒ f g(m) = 1(m) = m = 0). By M is co-hofian, i.e., g is an epimorphism, i.e., ∃h such that gh = hg = 1. It follows that h = f , and gf = 1. Lemma 4.4. Let R be a ring, and I an ideal of R. If M is an ℵ0 -quasi injective R/I-module, then M is also ℵ0 -quasi injective as an R-module. Also, if M is an ℵ0 -quasi injective R-module, then M is also ℵ0 -quasi injective as an R/I-module Proof. The relation (∗): m(r + I) = mr(m ∈ M and r ∈ R) is used in each case to define M as a module over R or R/I, where M is given as a module over R/I or R. Consider that if N = hx1 , x2 , · · · i is countably generated as a an R/I-module, then for all x ∈ N there exist i1 , i2 , · · · , ik such that x = xi1 (ri1 + I) + · · · + xin (rin + I). 10

By (∗), x = xi1 ri1 + · · · + xin rin , this means that N is also countably generated as an R-module. It is easy to see that the concepts ”submodule” and ”homomorphism” coincide over each ring. Hence any diagram 0 −→ A −→ M ↓ M over any ring is also a diagram over the other ring. Thus M is ℵ0 -quasi injective over R if, and only if, M is ℵ0 -quasi injective over R/I module. The result follows. Lemma 4.5. Let R be a ring and K an arbitrary two sided ideal of R. Then R is a right ℵ0 -qc ring if, and only if, R/K is a right ℵ0 -qc ring. Proof. Suppose R is a right ℵ0 -qc ring and K is a two sided ideal of R. We show that R/K is a right ℵ0 -qc ring. Let I/K be any right ideal of R/K with I ⊆ R and consider the right R/K-module R/K/I/K ∼ = R/I as an R-module with K ⊆ I. Then K annihilates the R-module R/I, and therefore, R/I may be regarded as an R/K-module. Furthermore, R is a right ℵ0 -qc ring by hypothesis. Hence, R/I is R-ℵ0 -quasi-injective. Hence, by the above lemma, R/I, as an R/K-module, is R/Kℵ0 -injective. We have shown that any cyclic R/K-module is R/K-ℵ0-quasi-injective. Therefore R/K is a right ℵ-qc ring. Theorem 4.6. Let R be a commutative ring. Then R is an ℵ0 -qc ring if, and only if, every factor ring of R is an ℵ0 -self-injective ring. Proof. If I is an ideal of an ℵ0 -qc ring R, then R/I is an ℵ0 -qc ring by lemma 3.5. Therefore R/I is an ℵ0 -self-injective ring (it is evident that since RR is generated by the identity, we may infer that any homomorphism from countably generated ideal of R into RR can be extended to an endomorphism). Hence RR is ℵ0 -injective, and R is a cyclic R-module. Then, M ∼ = R/I for some ideal I of R. By assumption, R/I is R/I-ℵ0-quasi-injective. Hence R/I is R-ℵ0 -quasi-injective, and therefore R is an ℵ0 -qc ring.

5

Examples

In this section we provide some new examples of ℵ0 -self-injective (regular) rings which are perhaps of some interest for their own right. 11

Example 5.1. Let F be a field and G a group. If char(F) = 0, then the following are equivalent: 1. F [G] is ℵ0 -self-injective; 2. G is finite; 3. F [G] is self-injective. Proof. (1)⇒(2): Suppose F [G] is ℵ0 -self-injective, then it is p-injective. By a result of Farkas ([18]), G is locally finite. Now by a result of Villamayor-Connell [18], page 69, Theorem 1.5, F [G] is a regular ring. The rest of proof is the same as Renault’s Theorem ([18], Theorem 2.8). The remaining implications are well-known and can be found in [18]. Example 5.2. Let X be a Tychonoff space. In [7], it has been shown that C(X) is ℵ0 -self-injective if, and only if, C(X)/CF (X) is ℵ0 -self-injective if, and only if, X is a P-space and therefore, if, and only if, C(X) is regular. Example 5.3. Let A be a σ-Algebra. In [2], it has been observed that rings of all real valued A-measurable functions are ℵ0 -self-injective. Example 5.4. Let X be Tychonoff space. By D(X), we mean the lattice of conS tinuous functions f on X with values in the extended real numbers R {±∞}, for which f −1 R is a dense subset of X. In general, under pointwise addition and multiplication, D(X) is not a ring. However, when X is a quasi-F space, then D(X) is a ring. A Tychonoff space X is called a quasi F-space if every dense co-zero set S in X is C ∗ -embedded. Since D(X) ∼ = D(βX), we may without loss of generality suppose M(X,A) that X is a compact space. Then by a result due to Hager (see [2]), D(X) ∼ = N , for some certain N. This then implies that D(X) is also an ℵ0 -self-injective regular ring, for X a quasi-F space (see [2]) The following example is a slight modification of a theorem by F. L. Sandomierski (see [19]). Example 5.5. Let M be a right R-module and a direct sum of countably many nonzero submodules {Mn | n ∈ N} and S = End(M). Then S M is not an ℵ0 -injective S-module.

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Proof. Let ei : MR −→ Mi be the i-th projection of the module MR onto the submodule Mi , then {ei }i∈N is a countable set of orthogonal idempotents of S. Let S A be the ideal of S generated by {ei }i∈N . Since Mi 6= 0 for each i ∈ I, choose 0 6= xi ∈ Mi . Clearly there is an S-homomorphism f :S A −→S A such that ei f = xi = ei xi . If f were extendable to a homomorphism from S to S M, then it would be given by some element of M. However, for any element x ∈ M, ei x = 0 for all but finitely many i ∈ N, so f is not extendable to S S, so S M is not ℵ0 -injective.

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