Classes of modules closed under projective covers

Theorem

For a ring R, the following properties are equivalent:

1. for each projective left R-module P and every K ∈ R -Mod : If there exists an epimorphism P ↠ K , then there exists a monomorphism K ↣ P ,

2. R is an artinian principal ideal ring,

3. R -hcov = R -qcov and R is left perfect,

4. R -quot ⊆ R -her ,

5. R -her ⊆ R -quot ;

Definition 2.1

Given a class of left R -modules C , we say that it is closed under projective covers whenever, for any M ∈ C such that its projective cover P ( M ) exists, then P ( M ) ∈ C . Also, we will denote by R -cov the class of module classes closed by projective covers.

Example 1

We give some examples of module classes closed under projective covers.

(1) Let R = Z and C = { M ∈ Z -Mod ∣ M is not a left free module. } . Then, C ∈ R -cov , although no module in C has a projective cover. Indeed, as Z is a PID, being a left projective module is equivalent to being a left free module. In contrast, in Z -Mod, the only modules that have a projective cover are the ones that are already free ([8, Example 5.1.2(4)]).

(2) Let C = { M ∣ M is a left finitely generated module } . Then, C ∈ R -cov .

Let n ∈ N and M = ∑ i = 1 n R x i be a left finitely generated module. If P ( M ) does not exist, there is nothing to prove. If P ( M ) exist, we have the following diagram:

which can be lifted by the epimorphism φ : R ( n ) ↠ P ( M ) and we are done.

(3) Let R be a ring, J = Rad ( R ) its Jacobson radical, e ∈ R such that e 2 = e and

It is clear that C ∈ R -cov since Re is the projective cover of Re/Je.

Definition 2.2

Let A be a set of closure properties and C be a class of left R -modules. We denote by ξ A ( C ) the least class of left modules containing C and being closed under the properties of A .

Proposition 2.1

If C ⊆ R -Mod , then

1. ξ ≤ ( C ) = { N ∣ t h e r e e x i s t s a m o n o m o r p h i s m N ↣ C w i t h C ∈ C } .

2. ξ P ( C ) = C ∪ { P ( C ) ∣ C ∈ C s u c h t h a t P ( C ) e x i s t s } .

3. ξ ↠ ( C ) = { M ∣ t h e r e e x i s t s a n e p i m o r p h i s m C ↠ M w i t h C ∈ C } .

Example 2

Let R = Z p 3 for some prime number p . It is clear that R is a commutative uniserial QF ring (thus, perfect) with a finite chain of ideals:

Thus, we have that ( p 2 ) is a simple left R -module. Consider C = { M ∣ M ≅ ( p 2 ) } then, we have

with ( p 2 ) being a simple left R -module. In contrast, as the projective cover P ( ( p 2 ) ) is a direct summand of R , which is an indecomposable R -module, then P ( ( p 2 ) ) ≅ R . Hence,

Therefore, ξ ↠ ξ P ≠ ξ P ξ ↠ as we claim.

Proposition 2.2

Let R be a ring. If C ⊆ R -Mod is a class of left R-modules such that for all C ∈ C : P ( C ) exists. Then, C ∈ R-cov implies that ξ ↠ ( C ) ∈ R-cov .

Proof

Let M ∈ ξ ↠ ( C ) . We can assume that P ( M ) exists. As M ∈ ξ ↠ ( C ) , then there exist C ∈ C and an epimorphism C ↠ M . Since C ∈ C , P ( C ) exists and belongs to C . Now, we have the following diagram:

where p 1 , p 2 are the projective covers of C and M , respectively, and the morphism φ : P ( C ) ⟶ P ( M ) exists since P ( C ) is projective. As ker p 2 ≪ P ( M ) , then Im φ = P ( M ) , i.e., φ is an epimorphism, therefore P ( M ) ∈ ξ ↠ ( C ) .□

Lemma 2.1

Let R be a ring and C ⊆ R -Mod a class of left R-modules such that for every C ∈ C , P ( C ) exists. Then:

ξ ↠ ( ξ P ( C ) ) = { M ∣ t h e r e e x i s t s a n e p i m o r p h i s m P ( C ) ↠ M w i t h C ∈ C } .

Proof

Let K = { M ∣ there exists an epimorphism P ( C ) ↠ M with C ∈ C } now, given M ∈ ξ P ( C ) , we have two cases: M ∈ C or M = P ( C ) for some C ∈ C . In the first case, we take p : P ( M ) ↠ M , and in the second case, 1 M : M = P ( C ) ↠ M , then, in any case, M ∈ K . Thus, ξ P ( C ) ⊆ K .

Now, K is closed under quotients. Indeed, given N ∈ K , M a left R -module and an epimorphism N ↠ M , then by definition, there exists P ( C ) ↠ N for some C ∈ C , so there exists P ( C ) ↠ M , then M ∈ K .

Finally, consider a class D containing ξ P ( C ) and being closed under quotients. If M ∈ K , there exists an epimorphism f : P ( C ) ↠ M for some C ∈ C . Now, since C ⊆ ξ P ( C ) ⊆ D ∈ R-cov , P ( C ) ∈ D . Hence, M ∈ D . Therefore, K is the smallest class containing ξ P ( C ) and that is closed under quotients, in other words, K = ξ ↠ ( ξ P ( C ) ) .□

Proposition 2.3

If C is class of left R-modules such that for every C ∈ C , P ( C ) exists. Then, ξ ↠ ( ξ P ( C ) ) = ξ { ↠ , P } ( C ) .

Proof

We note that, if K = { M ∣ , there exists an epimorphism P ( C ) ↠ M with C ∈ C } , then by the previous lemma and Proposition 2.1:

1. C ⊆ ξ P ( C ) ⊆ K .

2. K ∈ R-quot .

3. K = ξ ↠ ( ξ P ( C ) ) ∈ R-cov .

Then, it is sufficient to prove that if D is a class such that C ⊆ D and D ∈ R- q c o v , then K ⊆ D . Indeed, given a left R -module M ∈ K , by Lemma 2.1, there exists an epimorphism P ( C ) ↠ M for some C ∈ C . Thus, as C ∈ C ⊆ D and D ∈ R- q c o v , then M ∈ D . Therefore, ξ { ↠ , P } ( C ) = K = ξ ↠ ( ξ P ( C ) ) .□

Observation 1

As proved in the above proposition, generating the least class closed under quotients and left projective covers containing C is obtained by first generating the least class of left modules containing C closed under left projective covers and then consider the least class of left modules closed under quotients containing the former class. However, as Example 2 points out, we cannot obtain the least class of left R -modules closed under quotients and left projective covers by reversing the process described above.

Example 3

Let R = Z p 3 and C = { M ∣ M ≅ ( p 2 ) } as in Example 2. Then:

as ( p 2 ) being a simple left R -module. On the contrary

Therefore, ξ ≤ ξ P ≠ ξ P ξ ≤ as we claim.

Example 4

Let R = a ( x , y ) 0 a ∣ a , x , y ∈ Z 2 be a formal triangular matrix ring; thus, R is a commutative local artinian ring as we can see in the lattice of ideals of R , regarding that J denotes the Jacobson’s radical of R .

Where

and

As the ring R is local, all the left simple R -modules are isomorphic. Then S 1 ≅ R ∕ J . Since J is superflous in R, then P ( S 1 ) ≅ R .

As R being artinian, P ( J ) exists; however, if we suppose that P ( J ) ≅ R or P ( J ) ≅ S 1 , we obtain

Both cases are contradictions since J = Soc ( R ) = S 1 ⊕ S 2 ⊕ S 3 ; we can conclude that P ( J ) ∉ ξ ≤ ( R ) . Then, if C = { M ∣ M ≅ R } , we have

but it is clear that P ( J ) ∈ ξ { ≤ , P } ( C ) but P ( J ) ∉ ξ ≤ ( ξ P ( C ) ) . Consequently, ξ { ≤ , P } ≠ ξ ≤ ξ P as we claim.

Definition 2.3

Let C be a class of left R -modules. We define

1. η 0 ( C ) = C ,

2. η ( C ) = ξ P ( ξ ≤ ( C ) ) ,

3. η n + 1 ( C ) = η η n ( C ) ,

4. η ∞ ( C ) = ⋃ n ∈ N η n ( C ) .

Observation 2

Note that from Definition 2.3, given n ∈ N :

Note that, for each n ∈ N , C ⊆ η n ( C ) , since C ⊆ ξ ≤ ( C ) ⊆ η ( C ) (every R -module is a submodule of itself); moreover, η n ( C ) ⊆ ξ ≤ ( η n ( C ) ) ⊆ η n + 1 ( C ) for each n ∈ N .

Example 5

Let R be a left hereditary ring, and C = { M ∣ M is a left projective R -module } . Then, η ∞ ( C ) = ξ ≤ ( C ) .

Proof

As M is a left projective module, we have that P ( M ) = M for each M ∈ C , and every submodule C of an element in C is also left projective, we have that

It is clear that ξ ≤ ( ξ ≤ ( C ) ) = ξ ≤ ( C ) , then η 2 ( C ) = ξ ≤ ( C ) = η ( C ) . Therefore,

Example 6

Let R be a principal ideal domain (PID) and consider C = { R } . Then, η ∞ ( R ) = ξ ≤ ( R ) .

Proof

Since R is a PID, every ideal of R is free as left R -module and consequently left projective, then { P ( C ) ∣ C ∈ ξ ≤ ( R ) } = ξ ≤ ( R ) . Then, as in the previous example:

In particular, for the PID R = Z , we have that η ∞ ( Z ) = { n Z ∣ n ∈ Z } and for R = K [ x ] where K is a field, η ∞ ( K [ x ] ) = { ( 0 ) } ∪ { ( x k ) ∣ k ∈ N ∪ { 0 } } .□

Example 7

If R is a left artinian, left hereditary, and left uniserial ring and let C be the class of all left simple R -modules, then η ∞ ( C ) = C ∪ ξ ≤ ( R ) .

Proof

Since R is a left artinian ring, the left projective cover P ( S ) exists for each S ∈ C . Moreover, for each S ∈ C , we have that P ( S ) = Re for some idempotent element e of R and consequently, P ( S ) ≅ R for each left simple R -module as R being left uniserial.

Note that ξ ≤ ( C ) = { 0 } ∪ C and { P ( C ) ∣ C ∈ ξ ≤ ( C ) } = { 0 } ∪ { M ∣ M ≅ R } using the previous argument. Hence,

Now, it is clear that M ∈ ξ ≤ ( η ( C ) ) = C ∪ ξ ≤ ( R ) . Hence,

as R being left hereditary. Then

Finally, note that

At this stage, it is clear that η 3 ( C ) = η 2 ( C ) . Therefore,

Lemma 2.2

If C is a class of left R-modules, then η ∞ ( C ) = ξ { P , ≤ } ( C ) .

Proof

Since C ⊆ η n ( C ) for each n ∈ N , C ⊆ η ∞ ( C ) . We will prove that η ∞ ( C ) is closed under submodules and projective covers.

Let M ∈ η ∞ ( C ) and K ↣ M a monomorphism, by definition, M ∈ η n ( C ) for some n ∈ N . Then, K ∈ ξ ≤ ( η n ( C ) ) ⊆ η n + 1 ( C ) ⊆ η ∞ ( C ) . On the contrary, if P ( M ) exists, then P ( M ) ∈ η n + 1 ( C ) as M ∈ ξ ≤ ( η n ( C ) ) , thus P ( M ) ∈ η ∞ ( C ) and consequently, η ∞ ( C ) ∈ R- h c o v .

Finally, we will prove by induction that η ∞ ( C ) is the smallest class closed under submodules and projective covers containing C .

Let D be a class of left R -modules closed under submodules and projective covers containing C . For n = 0 we have that η 0 ( C ) = C ⊆ D . On the contrary, suppose that η n ( C ) ⊆ D and take M ∈ η n + 1 ( C ) . By Observation 2, we have either of the following two cases: M ∈ ξ ≤ ( η n ( C ) ) or M ∈ { P ( N ) ∣ N ∈ ξ ≤ ( η n ( C ) ) such that P ( N ) exists } , we now apply the inductive hypothesis since D is a class closed under submodules and projective covers. As we can see, in any case M ∈ D , therefore η ∞ ( C ) ⊆ D .□

Proposition 2.4

Let R be a ring such that R-qcov ⊆ R -hcov . Then, for every left projective R-module P and K ∈ R -Mod :

If there exists a monomorphism K ↣ P , there exists an epimorphism P ↠ K .

Proof

Let P be a left projective module, K ∈ R -Mod and a monomorphism K ↣ P . Now, consider C = { M ∣ M ≅ P } ; note that C satisfies the hypothesis of Lemma 2.1 then P ∈ ξ ↠ , P ( C ) and since ξ ↠ , P ( C ) ∈ R -qcov   ⊆ R -hcov so, K ∈ ξ ↠ , P ( C ) , and there exists an epimorphism P ↠ K .□

Theorem 2.1

Let R be a left perfect ring. The following statements are equivalent:

• (1) R- q c o v ⊆ R- h c o v .

• (2) For each left projective R-module P and K ∈ R -Mod :

• If there exists a monomorphism K ↣ P , then there exists an epimorphism P ↠ K .

Proof

(1) ⇒ (2) See Proposition 2.4.

(2) ⇒ (1) Consider C ∈ R- q c o v , M ∈ C and a monomorphism N ↣ M . We are going to prove that C it is closed under submodules. As R is a left perfect ring, then there exists a left projective cover P ( M ) and we have the following diagram:

where L is the pullback for the epimorphism and monomorphism. Now, since M ∈ C we have that P ( M ) ∈ C , then for the monomorphism L ↣ P ( M ) , by hypothesis, there exists an epimorphism P ( M ) ↠ L . Hence L ∈ C . Finally, as C is closed under quotients it follows that N ∈ C . Therefore, C ∈ R- h e r .□

Theorem

Let R be a left hereditary ring and M ∈ R -Mod . If there exists M ↣ F with F being a free R-module, then

Example 8

Let R = Z and consider C = { M ∣ M ≅ Z p ∞ } .

Let P be a left projective module, K ∈ R -Mod and a monomorphism K ↣ P . Since Z is a PID, every left projective module is free, then by Kaplansky’s theorem:

Now, since Z is a PID, we have that I x = ( 0 ) or I x ≅ Z , i.e., K ≅ Z ( Y ) for some Y ⊆ X ; consequently, K is covered by P = Z ( X ) and (2) of the previous theorem is fulfilled. On the contrary, as Z p ∞ is not a free module and its isomorphic to all its quotients, we have that C ∈ R -qcov , but Z p is a proper submodule of C , thus R -qcov ⊈ R -hcov , as we claim.

Proposition 2.5

Let R be a ring in which the following property holds for every left projective module P and K ∈ R -Mod : If there exists an epimorphism P ↠ K , then there exists a monomorphism K ↣ P . Then, R is a QF ring.

Proof

Let M ∈ R -Mod , then there exist a free R -module F and an epimorphism F ↠ M . As F is a left projective module, by the hyphotesis, then there exists a monomorphism M ↣ F , so every left R -module embeds in a free module. Therefore, R is a Q F -ring.□

Proposition 2.6

Let R be a ring in which the following property holds for every left projective module P and K ∈ R -Mod : If there exists an epimorphism P ↠ K , then there exists a monomorphism K ↣ P . Thus, R- h c o v ⊆ R- q c o v .

Proof

Take C ∈ R- h c o v , M ∈ C and an epimorphism M ↠ N , since R is a Q F ring, there exists P ( M ) , moreover, P ( M ) ∈ C (since the class is closed under projective covers), and we also have

Given that P ( M ) is left projective, then by hypothesis, there exists a monomorphism N ↣ P ( M ) and since C is closed under submodules, N ∈ C .□

Theorem 2.2

Let R be a left hereditary ring. The following statements are equivalent:

• (1) R- h c o v ⊆ R- q c o v .

• (2) For every left projective module P and K ∈ R -Mod : If there exists an epimorphism P ↠ K , then there exists a monomorphism K ↣ P .

Proof

( 1 ) ⇒ ( 2 ) Let P a left projective R -module, K ∈ R -Mod and an epimorphism P ↠ K . If we consider C = { M ∣ M ≅ P } , clearly ξ P ( C ) = C . Since R is a left hereditary ring, ξ ≤ ( C ) = { M ∣ M ≅ L with 0 ≤ L ≤ P } . Now, according to Definition 2.3, we have that η ( C ) = ξ P ( ξ ≤ ( C ) ) = ξ ≤ ( C ) , and η 2 ( C ) = η ( η ( C ) ) = ξ ≤ ( C ) , then by Lemma 2.2, we obtain that ξ { ≤ , P } ( C ) = ξ ≤ ( C ) . Finally, we can see that ξ ≤ ( C ) = ξ { ≤ , P } ( C ) ∈ R -hcov   ⊆ R -qcov by the hyphotesis, then as K is a quotient of P , K ∈ ξ ≤ , P ( C ) . Therefore, exists a monomorphism K ↣ P .

( 2 ) ⇒ ( 1 ) See Proposition 2.6.□

Theorem 2.3

For a left perfect ring R, the following conditions are equivalent:

• (1) R- q c o v = R- h c o v .

• (2) For each left projective R-module P and K ∈ R -Mod , we have:

• there exists an epimorphism P ↠ K if and only if there exists a monomorphism K ↣ P .

Proof

(2) ⇒ (1) It follows immediately from Theorem 2.1 and Proposition 2.6.

(1) ⇒ (2) Suppose that R- q c o v = R- h c o v , let P be a left projective R -module and K ∈ R -Mod .

( ⇒ ) If there exists an epimorphism P ↠ K , then K ∈ ξ { ↠ , P } ( P ) .

We will see that ξ ≤ ( P ) ∈ R -cov . For this, we take L ∈ ξ ≤ ( P ) , then by Proposition 2.1, there exists a monomorphism L ↣ P , and since R is left perfect, there exists P ( L ) , and as R -qcov ⊆ R -hcov , by Theorem 2.1, then there exists an epimorphism P ↠ L and the following diagram commutes:

Since P ( L ) is left projective, it follows that P ( L ) is a direct summand of P , and consequently, there is a monomorphism P ( L ) ↣ P ; in other words, the projective cover P ( L ) ∈ ξ ≤ ( P ) i.e., ξ ≤ ( P ) ∈ R -cov then ξ { ≤ , P } ( P ) ⊆ ξ ≤ ( P ) . Now, by hypothesis and [6, Remark 2.3]: ξ { ≤ , P } ( P ) = ξ { ↠ , P } ( P ) and since K ∈ ξ { ↠ , P } ( P ) it follows that K ∈ ξ { ≤ , P } ( P ) . Hence, there exists a monomorphism K ↣ P .

( ⇐ ) By Theorem 2.1.□

Theorem 3.1

The following propositions are equivalent:

1. R is an artinian principal ideal ring.

2. R is a left principal ideal ring and QF.

3. The injective hull and the projective cover of each (left or right) finitely generated R-module are isomorphic.

4. For each left R -module M , R Soc ( M ) ≅ M J M and for each rigth R-module N, Soc R ( N ) ≅ N J N where J denotes the Jacobson radical of the ring R.

5. For every ideal I of R , R ∕ I is Q F .

6. R- h e r = R-quot .

Proof

See [6, Theorem 4.1]□

Theorem 3.2

For a ring R, the following properties are equivalent:

• (1) For each projective left R-module P and every K ∈ R -Mod :

• if there exists an epimorphism P ↠ K , then there exists a monomorphism K ↣ P .

• (2) R is an artinian principal ideal ring.

• (3) R- h c o v = R- q c o v and R is left perfect.

• (4) R- q u o t ⊆ R- h e r .

• (5) R- h e r ⊆ R- q u o t .

Proof

(1) ⇒ (2) By Proposition 2.5, R is Q F .

On the contrary, let I be an ideal of R , P a left projective R ∕ I -module and for a K ∈ R ∕ I -Mod an epimorphism β : P ↠ K .

Note that I P = I K = 0 as left R ∕ I -modules, so for each r ∈ R and m ∈ P ( m ∈ K ) , we define the action r m ≔ ( r + I ) m , then P and K are left R -modules and β can be regarded as an epimorphism of left R -modules, then for (1) there exists a monomorphism α : K ↣ P of left R -modules.

Note that α ( ( r + I ) m ) = α ( r m ) = r α ( m ) = ( r + I ) α ( m ) , then α is a monomorphism of left R ∕ I -modules, i.e., R ∕ I is QF for each ideal I of R , then by Theorem 3.1, R is an artinian principal ideal ring.

(2) ⇒ (3) See [5, Theorem 38].

(3) ⇒ (1) See Theorem 2.3.

(5) ⇔ (2) and (4) ⇔ (2) See [6, Theorem 4.2].□

Corollary 3.1

Let R be a left hereditary ring. The following statements are equivalent:

1. R is a QF ring.

2. R is an artinian principal ideal ring.

3. R -hcov   ⊆ R -qcov .

Proof

(1) ⇒ (2) Is clear.

(2) ⇒ (3) It follows from Theorem 3.2.

(3) ⇒ (1) See Theorem 2.3.□

Lemma 3.1

If R - q c o v ⊆ R - h c o v , then for each left R-module M the following statements are equivalent.

1. M is left noetherian.

2. M is finitely generated.

Proof

( 1 ) ⇒ ( 2 ) It is clear.

( 2 ) ⇒ ( 1 ) Let C be the class of left finitely generated R -modules, then C ∈ R- q c o v ⊆ R - h c o v as we see in Example 1.2. Now, given M ∈ C and N a submodule of M , we have that N ∈ C , then N is finitely generated, so M is left noetherian.□

Corollary 3.2

If R- q c o v ⊆ R -hcov , then R is a left principal ideal ring.

Proof

Since R is left projective, by Theorem 2.1 for every ideal I of R there exists an epimorphism R ↠ I , I is cyclic and R is a left principal ideal ring.□

Corollary 3.3

If R- q c o v ⊆ R - h c o v , then for every simple left R-module S, P ( S ) is left noetherian when exists.

Proof

Let S be a simple left R -module, and suppose that P ( S ) exists. Then, we have the following commutative diagram:

and h ( R ) + ker g = P ( S ) hence h ( R ) = P ( S ) , then P(S) is cyclic, so by Lemma 3.1, P ( S ) is left noetherian.□

Theorem 3.3

For a left self-injective ring R, the following statements are equivalent:

1. R -qcov   ⊆ R -hcov .

2. R is an artinian principal ideal ring.

Proof

(1) ⇒ (2) As R is left self-injective and left prinicipal ideal ring and thus left noetherian, then R is a Q F . Therefore, by [10, Proposition 25.4.6.B (c)], R is an artinian principal ideal ring.

(2) ⇒ (1) It follows from Theorem 3.2.□

Lemma 3.2

[12, Lemma 2] Let R be a ring and I an ideal of R. If M is a quasi-injective R ∕ I -module, then M, regarded as R-module, is also quasi-injective. Also, if M is a quasi-injective R-module such that I M = 0 , then R ∕ I -module M is also quasi-injective.

Lemma 3.3

[12, Lemma 3] Let R be a ring. Then, every left cyclic R-module is quasi-injective if and only if every left cyclic R ∕ K -module is R ∕ K -quasi-injective for each two-sided ideal K of R .