1. Introduction and Preliminaries
Let
S be a monoid. We call a nonempty set
A a right
S-act over
S if we have a mapping
given by
such that for all
and
We refer the reader to [
1,
2,
3] for all undefined terms concerning acts over monoids.
Mahmoudi and Renshaw, in [
3], introduced coessential epimorphisms and covers of cyclic
S-acts. They provided a characterization of coessential epimorphisms using congruence classes and left unitary submonoids. Furthermore, they defined various types of covers such as strongly flat covers, Condition
covers, and projective covers. They also established necessary and sufficient conditions for cyclic
S-acts to possess Condition
covers, strongly flat covers, and projective covers.
In 2010, Khosravi et al. investigated the covers of Condition (P) and strongly flat covers for arbitrary
S-acts over monoids in [
4]. Much like a perfect monoid, an
-perfect monoid was identified by condition (A) possessing a strongly flat cover for every cyclic right
S-act, and a
-perfect monoid was defined by condition (A) and having having a Condition (
P) cover for each cyclic right
S-act. Additionally, the authors introduced a new discription for perfect monoids, defining them as monoids where each strongly flat right
S-act has a projective cover.
Bailey and Renshaw introduced articles on covers of
S-acts over monoids in 2013 and 2014, respectively. They considered free, torsion-free, divisible, and injective covers in [
5]. As we all know, Enochs’s celebrated flat cover conjecture was ultimately proven in [
6] in 2001. And in [
1], they attempted to mimic some of Enochs’s work within the category of
S-acts over monoids, primarily concentrating on strongly flat acts. Of course, the covers of [
3] are slightly different from Enochs’ notion of covers.
Subsequently, several additional papers concerning covers have appeared (e.g., [
7,
8,
9,
10]). Recently, Irannezhad and Madanshekaf [
11] researched covers of
S-acts over monoids with Condition (
), and Qiao et al. introduced Condition (
) covers in a similar way in [
12]. Now, we restrict our attention to these notions.
Let
S be a monoid. Recall [
3] that a right
S-act
together with an
S-epimorphism
is a cover of a right
S-act
if for any proper subact
of
the restriction
is not an
S-epimorphism. Moreover, we call an
S-epimorphism
coessential in [
3] if
is an
S-epimorphism for every
S-act
and every
S-map
, then
g is an
S-epimorphism. And we define an act
, along with an
S-epimorphism
, as an
X-cover of
if
satisfies property
X and
f is coessential. One can readily observe that an
S-morphism
being a cover of
A is equivalent to
f being a coessential epimorphism.
Let
be a class of right
S-acts closed under isomorphisms, meaning if
and
, then
. Now let us explore another concept of covers (Enochs’ notion). let
A be a right
S-act. Recall [
1] that an
S-act
is called an
-precover of
A if there exists an
S-homomorphism
such that, for every
S-homomorphism
:
, for
, there is a homomorphism
f:
with
. Alternatively, the following diagram
commutes. We define
as an
-cover of
A if the precover satisfies the condition that for any endomorphism
such that
,
f must be an automorphism.
Recall [
2] that a right
S-act
is called strongly flat if it is both pullback flat and equalizer flat. It is a commonly acknowledged fact that an
S-act
A is strongly flat if, and only if, it satisfies both Condition (
P) and Condition (
E). Now, a right
S-act
is said to satisfy Condition
if
Recall that a right
S-act
satisfies Condition
if
In this paper, our primary focus lies on finitely generated
S-acts, particularly on the notions of strong flatness and Condition (
P), and we find valuable characterizations.
Let
S be a monoid. By [
3], it is said to be right reversible if there exist
such that
for all
. And it is said to be left collapsible if there is
such that
for all
. Let
T be a submonoid of
S.
T is called a left unitary submonoid if and only if whenever
then
.
In [
3], Mahmoudi M. and Renshaw J. solved a problem concerning the covers of cyclic
S-acts over monoids. However, the problem of whether all
S-acts have covers remains open. Recall [
2] that every
S-act has a unique decomposition into indecomposable subacts. If a finitely generated
S-act satisfies Condition (
P), then it is a coproduct of cyclic subacts. Based on the above facts, in this paper, we consider the covers of finitely generated
S-acts and coproducts of
S-acts and then study Condition
covers, strongly flat covers, and projective covers. This research brings us closer to understanding the covers of any
S-acts. Therefore, this study of the covers of finitely generated
S-acts plays an important role.
The main results here are descriptions of finitely generated acts and coproducts of
S-acts in
Section 2, and we provide necessary and sufficient conditions for finitely generated
S-acts to have covers in
Section 3. Next, in
Section 4, we concentrate on
-covers, strongly flat covers, and projective covers, and the main conclusions extend some known results in [
3]. In
Section 5, our exploration delves into Enochs’ notion of covers of acts over monoids, with a specific emphasis on
-precovers.
Throughout this paper, S denotes a monoid, denotes a class of right S-acts closed under isomorphisms, is said that A is a n-generated right S-act, and . And will denote a coproduct of cyclic subacts for all . We focus solely on the right S-acts, often just called S-acts.
2. Finite Generation and Coproducts
This section aims to describe the finite generation and coproduct of
S-acts. A subset
U of an
S-act
A is a generating set for
A if, for any
, there exist
,
such that
. In other words,
U is a set of generating elements for
if
, where
. An
S-act
A is said to be finitely generated (resp. cyclic) if it has a finite (resp. one-element) generating set. Further details may be found, for example, in [
2].
Proposition 1. Let S be a monoid and be S-act, is a subact of A. For every , is finitely generated, then A is finitely generated.
Proof. Let . Then, . □
Lemma 1. Let S be a monoid and an S-act, are subacts of A for . And for , suppose that is either finitely generated or empty. If A is finitely generated, then for all is finitely generated.
Proof. It is clear by Lemma 5.3 of [
13]. □
Corollary 1. Let S be a monoid. S-act is finitely generated if and only if S-act is finitely generated.
It is important for monoids to consider the covers of finitely generated acts. Now, we first aim to investigate the cover of coproducts of acts.
Proposition 2. Suppose that is a family of S-morphisms, where . Let , , and satisfies . Then, is a cover of A if and only if is a cover of for every .
Proof. Sufficiency. Since , is a family of epimorphisms, it is clear that is epimorphic. Assume that f is not coessential, that is, there exists a proper subact of B such that is epimorphic. Therefore, there is such that is a proper subact of . Thus, is not epimorphism since is a coessential epimorphism, so there exists such that for all . However, by the surjectivity of , we obtain that for some . Thus, there exists such that , . Then which contradicts .
Necessity. Let be a coessential epimorphism. Then, we obtain that is a family of epimorphisms. To show that is coessential for all , assume that there exists a map such that is not coessential, where . Then is an epimorphism for some proper subact of . Thus, is an epimorphism. But is a proper subact of and by assumption is not epimorphic, a contradiction. So, is a family of coessential epimorphisms. □
Now, we concentrate on covers of finitely generated S-acts, and we begin with some notations. For an element a of a right S-act A, we denote by the set , the submonoid of right identities of a, and it is clear that is a left unitary submonoid of S. We denote by the set . In particular, , the right annihilator congruence of a. It is clear that is isomorphic to under the S-isomorphism , where denote the class of with respect to an equivalence relation .
Lemma 2 ([
2], Proporsition III.13.14)
. Let be a finitely generated S-act. If satisfies Condition , then is a coproduct of cyclic subacts. Let S be a monoid and , where are right S-acts. Then, A is strongly flat (resp. satisfies Condition ) if and only if is strongly flat (resp. satisfies Condition ) for every . The following results turn out to be very helpful in the following section.
Proposition 3. If the finitely generated S-act is strongly flat (resp. satisfis Conditon (P)), then is a left collapsible(resp. right reversible) submonoid of S.
Proof. Let A satisfy Condition . By Lemma 2, satisfies Condition for . Thus, is right reversible. And, in a similar way, we obtain that is a left collapsible submonoid of S if A is strongly flat. □
Proposition 4. Let T be a submonoid of a monoid S and A a right S-act, . If , then .
Proof. Since , for any and , then . Thus, , that is, . Therefore, . It is clear that . □
Lemma 3. Let S be a monoid and P a left collapsible(resp. right reversible) submonoid of S, is n-generated. If , then is left collapsible (resp. right reversible) and is strongly flat(resp. satisfies Condition (P)).
Proof. Since , by Proposition 4, is left collapsible (resp. right reversible). Notice that in either case, P is right reversible and so defined by if and only if there exist with . Notice that . In fact, if then there exist with and so . Conversely, if , then there are with and from . So and hence . It is clear that satisfies Condition . Then, satisfies Condition . If in addition P is left collapsible, then is strongly flat. □
It is clear that any cover of a cyclic right
S-act is cyclic by Lemma 2.3 of [
3]. The following theorem concerns the covers of
n-generated right
S-acts.
Theorem 1. Let be a cover of . Then, is n-generated if and only if is n-generated.
Proof. Necessity. Assume that B is n-generated, is a generating set of B. Since is an epimorphism, A is generated by n elements. Now, we are ready to prove n is the smallest. Suppose that , . Then, , . Therefore, , which contradicts that B is n-generated. Thus, A is n-generated.
Sufficiency. Suppose . Since f is an epimorphism, there exists such that Take , clearly and . By the definition of cover, we have . Therefore, B is generated by n elements. Now, we are ready to prove a result that n is the smallest. If , and , we have that A is generated by , a contradiction. □
Let be n-generated and . The following Proposition 5 shows that is a cover of A implies that is a family of covers of .
Proposition 5. Let . Let be a family of S-morphisms, and satisfies . Then, is a cover of A implies is a family of covers of .
Proof. If for any , it is clear by Proposition 2. Otherwise, suppose . Since f is an epimorphism, for , there exists such that . If , we have for some . But , a contradiction. Therefore, , we obtain that . By Proposition 2, is a cover of A implies that is a cover of . □
Conversely, if B is n-generated and A is a disjoint union of n subacts, the argument always holds in Proposition 6.
Proposition 6. Suppose that , , is a family of S-morphisms, and satisfies . Then is a cover of A if and only if is a family of covers of .
Proof. Let be a cover of A. If there exist such that , that is, , where , then is an epimorphism, a contradiction. So, f is one-to-one. Assume . Then, , and which contradicts . Hence, or the element in has no image under the action of f. If the latter, then is an epimorphism, a contradiction. Therefore, . Conversely, let be a family of covers of , we obtain the same result . By Proposition 2, is a cover of A if and only if is a family of covers of . □
Remark 1. From the several propositions above, it is shown that if A and B are the union (coproduct) of n S-acts and , respectively, then is a cover of A if and only if is a family of covers of for .
Let
S be a monoid and
X an
S-act. Recall [
2] that
X is Noetherian if every congruence on
X is finitely generated. Additionally, a monoid
S is called Noetherian if it is Noetherian as an
S-act over itself.
Proposition 7. Let S be a Noetherian monoid and be a cover. Then, A is Noetherian if and only if B is Noetherian.
Proof. Let
be a cover of
A. If
B is Noetherian. By Lemma 7.6 of [
5], then
B is finitely generated, and so is
A by Theorem 1. All finitely generated
S-acts over a Noetherian monoid are Noetherian, thus
A is Noetherian. Conversely, it is obvious. □
3. Covers of Finitely Generated Acts
Let
S be a monoid and
right congruences on
S. In Mahmoudi and Renshaw [
3], it is proved that a cyclic act
has a cover
if and only if
and for all
,
. The next propositions are needed to obtain the result.
Proposition 8. Let S be a monoid. Then, is isomorphic to a finitely generated subact of if and only if there exist , such that for any .
Proof. Let be an S-monomorphism and . For , there is such that , . Since f is a monomorphism, then . Thus, we have for some and . And so for any . Let , . Then, . In addition, suppose that , and we have since f is monomorphic. Therefore, .
Conversely, the mapping defined by is well-defined, and f is an S-monomorphism. □
Lemma 4. Let S be a monoid and . Consider the S-monomorphism given by , . Then, f is onto if and only if for every .
Proof. Since f is an S-epimorphism. For any , there exist and such that . And since , it follows that , namely , a contradiction. Thus and so , hence .
Conversely, since , there exists such that for every . So . It is easy to see that f is an epimorphism. □
Lemma 5. Let S be a monoid, and . If is a coessential epimorphism, then there exists such that , is isomorphic, where . And given by is a coessential S-epimorphism. In particular, .
Proof. Since is a coessential epimorphism, it follows that A and B are n-generated. For any , there exist and such that . Suppose that given by is a monomorphism whose composite with f is clearly onto. Since B is a cover of A, and so , , is a coessential S-epimorphism. It then follows that . □
We now present a fundamental theorem that yields the main result of this section.
Theorem 2. Let S be a monoid and a n-generated S-act. The map given by is a coessential epimorphism if and only if , , and for all , .
Proof. Let for , and . Then, because f is well-defined. Hence, . For , there exist and such that , but , we obtain , a contradiction. Therefore, and so . Since is a coessential epimorphism, by Lemma 5, there is a subact such that and is a coessential epimorphism. So we have for some . And since is isomorphic by Lemma 5, it follows that , thus as required.
Conversely, if the given conditions hold, then clearly f is well-defined. Let B be an S-subact of and suppose that is onto. Then, for , there exists such that , and we obtain , so . However, , and hence . By assumption, there exists such that , in the sense that for an arbitrary . So , that is, , and f is a cover. □
If
, we can easily obtain the Theorem 2.7 in [
3].
Theorem 3. Let S be a monoid and a n-generated S-act. If the natural map is a coessential epimorphism, where . Then, , is a subgroup of S.
Proof. Since is a coessential epimorphism. Then, for an arbitrary element , . And . We can verify that for . Moreover, , it is easy to see that and so . Consequently, is a subgroup of S. □
Proposition 9. Let S be a monoid. Then, the map is a coessential epimorphism, where , if and only if , is a subgroup of S and for .
Proof. Let , be a coessential epimorphism. Let for and . We have since f is well-defined. By Theorem 3, is a subgroup of S.
Conversely, suppose that is a subgroup of S. For any , there exists such that . Thus, . In addition, since for , we can verify that f is a coessential epimorphism by Theorem 2. □
Proposition 10. Let S be a right simple semigroup with a 1 adjoined and , . If for and . Then given by is a coessential epimorphism.
Proof. Since , we can verify that , where T is a right simple semigroup. So for every . Then for , . It is easy to see that f is a coessential epimorphism since . □
Remark 2. It follows from the above that covers of finitely generated S-acts need not be unique. If S is a group then every n-generated S-act has as a cover, where . And so proper n-generated S-acts do not have unique covers. In fact, if the map is onto, then is trivially a cover of .
4. -Generated Flat Covers
In Mahmoudi and Renshaw [
3], it is proved that an equivalent characterization of a cyclic act having flat properties covers, such as strongly flat, Condition
, and projective. The next theorems investigate the flatness covers of
n-generated
S-acts.
Theorem 4. Let S be a monoid. Then, the n-generated S-act has a strongly flat cover B if and only if and contains a left collapsible submonoid R such that for all , .
Proof. Suppose that A has a strongly flat cover B. By Theorem 1, . Then, by Theorem 2, we can assume that and that for all , . Moreover, since B is strongly flat, by Lemma 2, and R is left collapsible by Proposition 3.
Conversely, suppose that and R is a left collapsible submonoid of such that for all , . Define , then and further is strongly flat by Lemma 3. Define a map given by and note that f is a well-defined S-epimorphism. To see this, first notice that if f is well-defined, then it is clearly an S-map which it is on. Since , it is easy to see that , but and hence . And for any , is not necessary to consider because . Further, for all , so f is coessential by Theorem 2. □
We can easily obtain the Theorem 3.2 of [
3] if
in Theorem 4.
Example 1. Let S be a left cancellative monoid. Then, n-generated S-act has a strongly flat cover if and only if , , is a subgroup of S, and in this case, is a strongly flat cover of A, where . First, notice that the only strongly flat cover of a n-generated S-act (assuming that it has one) is then . Notice also that not all finitely generated S-acts need have a strongly flat cover (see, for example, Remark 3.6 in [3]). Theorem 5. Let S be a monoid. Then, the n-generated S-act has a -cover B if and only if and contains a right reversible submonoid R such that for all , .
Proof. Suppose that A has a -cover, . Then, by Theorem 2, we can assume that and that for all , . Moreover, since B satisfies Condition , by Lemma 2, and R is right reversible by Proposition 3.
Conversely, suppose that and R is a right reversible submonoid of such that for all , . Define , then and further satisfies Condition by Lemma 3. Define a map given by and note that f is a well-defined S-epimorphism. Since , it follows that , but and hence . And for any , is not necessary to consider because . Further, since for all , , then f is coessential by Theorem 2. □
Similarly, if
, we can easily obtain Theorem 4.2 of [
3].
Proposition 11. Let S be a group. Then, every n-generated S-act has at least two -covers but only has a unique strongly flat cover.
Proof. Let . Since S is a group, then all S-acts satisfy condition and so in particular A is a -cover of itself. But is a subgroup of S; thus, is a cover of A by Proposition 9, where , while the latter part follows immediately from the Example 1. □
According to Proposition 11, we have an obvious example, if under multiplication, then the only subgroup is . Then, and N are -covers of . However, has a strongly flat cover N if .
Lemma 6 ([
14], Corollary 3.8)
. A right S-act is projective if and only if , where for idempotents , . Theorem 6. Let S be a monoid and a n-generated S-act. Then, the following are equivalent:
- (1)
A has a projective cover;
- (2)
There exists a submonoid R of which has a left zero element and for all , .
- (3)
The submonoid of S has a minimal right ideal generated by an idempotent.
Proof. (1)⇒(2) Let be a projective cover of A. Then by Theorem 2, take , for all , . Moreover by Lemma 6 there exists with and thus obviously and . If , then and so is a left zero element of .
(2)⇒(3) Assume that is a left zero element of R and consider the right ideal of . If this is not minimal, then there exists with . By assumption, there exists with , and since is a left unitary submonoid, then . Since z is a left zero element of R, , and so , which is a contradiction. Therefore, the submonoid of is a minimal right ideal generated by an idempotent.
(3)⇒(1) Suppose that with is a minimal right ideal of . Clearly is projective. Define the map by . Then f is well-defined since if , and , then and . In addition, it is also coessential since if , then by the minimality of , and so for some . But then and so . Hence, the result follows by Theorem 2. □
If
in Theorem 6, we can obtain the Theorem 5.2 in [
3].
5. -Precovers
In this section, we study Enochs’ notion of cover [
1] in the category of acts over monoids and focus on
-precovers and
-covers, where
is a class of
S-acts closed under isomorphisms. For example, we denote the class of all projective
S-acts by
. Clearly the concept of cover by Enochs’ notion is slightly different from that of the coessential cover.
Proposition 12. Let S be a monoid and let satisfy the property that for -act-S. If A has an -precover, then has an -precover.
Proof. Let be -precover of A. Define to be the obvious induced map. Let and let . Now, by the hypothesis, , so, since C is an -precover of A, there exists such that . So define to be the induced map, and clearly . Thus, has an -precover. □
Proposition 13. Let A be an S-act and a class of S-acts. Any retract of any act satisfying satisfies . Then, the -precover of A, if it exists, is the -precover of the retract of A.
Proof. Please see the following diagram
Let
be the
-precover and
B the retract of
A. Then
and
satisfy
. For
S-map
,
, by the
-precover property, there exists an
S-map
such that
. We claim that
Hence,
is the
-precover of
B. □
Proposition 14. Let be a class of acts and . Consider the following commutative diagram of S-acts:If is an -precover of S-act A, then is also a -precover of A. Proof. Let be an S-homomorphism for . By the -precover property, there exists an S-map such that . Hence, from . So is a -precover of A. □
Theorem 7. Let be a class of S-acts and be a monomorphism. Consider the following commutative diagram of S-acts:If is an -precover of B, then is an -precover of A. Proof. For every S-map , where . Since is an -precover of B, there exists an S-map with . Then, since f is a monomorphism, and we are done. □
Dually, if is an S-epimorphism, we obtain the following Theorem 8.
Theorem 8. Let S be a monoid and the class of all projective S-acts. Let be an S-epimorphism. Consider the following commutative diagram of S-acts:If is a -precover of A, then is a -precover of B. Proof. Let be -precover of A and an S-epimorphism. To show that is a -precover of B, we assume that any S-map , where . By assumption, there exists such that . Since is a -precover of A, we obtain that for some . So . Therefore, is a -precover of B. □
Remark 3. It is clear that if is an -cover of A, then is an -cover of B in Theorem 8. A question that could be brought up is whether Theorem 8 is valid for free S-acts. It is easy to answer this question positively.
Theorem 9. Let S be a monoid and a class of S-acts. If is an -precover of B and is an -precover of A. Then, C is an -precover of B.
Proof. Suppose that is an -precover of B and is an -precover of A. For any S-map , for , there is an S-map with since is an -precover of B. Similarly, because is an -precover of A, we have for some S-map . Thus, we can verify that . So is an -precover of B, that is, C is an -precover of B. □
Proposition 15. Let S be a monoid and B a right S-act. Let be a class of S-acts closed under factor acts. If is a -precover for B, then is a -precover for B, where is a congruence on A.
Proof. Let be a congruence on A contained in and . First note that . By the Homomorphism Theorem for acts, there exists a homomorphism , defined by , with , where . For each morphism with , there exists a morphism such that . Thus , that is, is a -precover for B. □
Proposition 16. Let be a class of S-acts which is closed under subacts. And suppose that all subacts of B have -covers. If and is an -precover and C is a proper subact of B. Then, there exists an S-subact of A and a homomorphism such that it is an -cover of C.
Proof. By assumption, we note that
. Thus,
is an
-precover of
C. Let
be an
-cover of
C. Then, we have the commutative diagram
such that
and
, and so
. Since
h is an
-cover of
C, it follows that
is an automorphism of
. Thus,
is monomorphism. Take
as an
S-subact of
A, then
is an
-cover of
C. □