Semigroup Forum
Springer-Verlag New York Inc.
Version of November 17, 1997
Flat Acts That Satisfy Condition (P )
James Renshaw and Akbar Golchin
Communicated by John M. Howie
1.
Introduction
We deal in this paper with what is generally referred to as homological classification
of monoids by flatness properties. We have the following sequence of properties,
arranged in strictly decreasing order of strength that an act over a monoid may or
may not possess.
projective ⇒ strongly flat ⇒ condition (P ) ⇒ flat ⇒weakly flat ⇒torsion free
Many papers have appeared recently (for example [4], [5], [6], [8], [14]) investigating
the conditions on a monoid which are necessary and sufficient to make various of
the above flatness properties coincide, either for all right acts, or for all right acts
of a certain type.
There are still however a number of open problems and in particular the problem of
when (weak) flatness implies condition (P ). To date most of the definitive results
of this problem have been found when restricting attention to certain classes of
monoids. In this article, we continue this investigation and extend some results
of [8] and many of the main results of recent papers on this subject appear as
corollaries.
We start in section two with basic definitions and results which will be needed
later. Section three extends one of the main results in [8] to certain types of right
subelementary monoid. In the fourth section, we give a characterization of which
monoids have the property that every e ∈ E(S)\{1} is right zero and use this to
give a characterization of eventually regular monoids for which all flat (cyclic) right
acts satisfy condition (P ). We also use this characterization to deduce some of
the main results from the literature. In section five we consider left PSF monoids
and extend some of the results in [3] and [14].
2.
Preliminaries
Throughout this paper S will denote a monoid. We refer the reader to [11] for
basic definitions and terminology relating to semigroups and acts over monoids.
A right S -act A is called flat (resp. (principally) weakly flat) if the tensor functor
A ⊗ − preserves all monomorphisms (resp. all embeddings of (principal) left ideals
of S into S ). A is said to satisfy condition (P ) if whenever a, a0 ∈ A, u, v ∈ S
and au = a0 v , there exist a00 ∈ A, s, t ∈ S such that a = a00 s, a0 = a00 t, su = tv .
A is said to be torsion free if as = a0 s with a, a0 ∈ A, s ∈ S right cancellative
implies a = a0 .
We frequently deal in this paper with cyclic acts (that is acts of the form S/ρ
where ρ is a right congruence on S ) and so we give characterizations of the various
flatness concepts for such acts which will be used most often in this paper.
2
Renshaw and Golchin
Lemma 2.1.
on S .
[4, Lemma 2.1] Let S be a monoid and let ρ be a right congruence
1. S/ρ is projective if and only if there exists e2 = e ∈ S such that e ρ 1 and
u ρ v implies eu = ev for all u, v ∈ S .
2. S/ρ is strongly flat if and only if for all u, v ∈ S with u ρ v there exists
s ∈ S such that su = sv and s ρ 1.
3. S/ρ satisfies condition (P ) if and only if for all u, v ∈ S with u ρ v there
exist s, t ∈ S such that su = tv and s ρ 1 ρ t.
4. S/ρ is flat if and only if for any left congruence λ on S and any u, v ∈ S
if u(ρ ∨ λ)v then there exist s, t ∈ S with suλtv such that s(ρ ∨ λu)1 and
t(ρ ∨ λv)1.
5. S/ρ is weakly flat if and only if for all u, v ∈ S with u ρ v there exist s, t ∈ S
such that su = tv, s(ρ ∨ ∆u)1 and t(ρ ∨ ∆v)1
Note that in (4) above, λu denotes the left congruence on S defined by x(λu)y if
(xu)λ(yu) and in (5) above ∆ denotes the equality relation on S .
For a monoid S and x, y ∈ S we shall denote by ρ(x, y), the smallest right
congruence on S containing (x, y).
Lemma 2.2.
([4, Lemma 3.1]) Let S be any monoid such that every flat cyclic
right S -act satisfies condition (P ). Then every e ∈ E(S)\{1} is a right zero
element of S .
The converse of the above result is false (see [14, comments after Lemma 4.1]),
but if we restrict attention to monocyclic acts (i.e. acts of the form S/ρ(x, y))
then the converse is true (see [6, Theorem 4.3]). That it is also true for regular
monoids, follows from Theorem 4.7 below. Clearly then, monoids for which every
e ∈ E(S)\{1} is a right zero element play an important role in this theory and will
be considered further in section 3. First we mention some results and definitions
that will be required later.
Lemma 2.3.
[4, Lemma 3.3] Let S be any monoid and let ρ be a right congruence on S such that S/ρ is weakly flat. If e and f are right zero elements of
S and e ρ f then e = f .
A monoid S is called left PP if every principal left ideal of S is projective. It can
be shown that S is left PP if and only if for every x ∈ S , there exists e2 = e ∈ S
such that ex = x and ux = vx implies ue = ve. Every regular and every right
cancellative monoid is left PP.
Lemma 2.4.
([1, Proposition 1.2]) Let S be a monoid. Then a right S -act A
is weakly flat if and only if A is principally weakly flat and for all left ideals I and
J of S , AI ∩ AJ = A(I ∩ J).
Renshaw and Golchin
Lemma 2.5.
equivalent:
3
[7, Theorem 4] For an arbitrary monoid S the following are
1. all right S -acts are weakly flat;
2. all finitely generated right S -acts are weakly flat;
3. all cyclic right S -acts are weakly flat;
4. S is a regular monoid and for any x, y ∈ S there is an element z ∈ Sx ∩ Sy
such that (x, z) ∈ ρ(x, y).
If S is a monoid then an element x ∈ S is called eventually regular if ∃n ∈ N such
that xn is regular. S is called eventually regular if all its elements are eventually
regular. Clearly regular monoids and periodic monoids are eventually regular.
On the other hand, a subset of a monoid which contains no eventually regular
elements will be called a regular-free subset. Notice that regular-free semigroups
are equivalent to idempotent free semigroups and so must consist of elements of
infinite order.
An element x is right nil if x 6= 1 and xn is a right zero element for some n ∈ N .
A monoid S is called right elementary if S = G ∪˙ N where G is a group and N
is the set of all right nil elements of S . Notice that we include the case when N
is empty. It is fairly easy to see that if N is not empty then it is an ideal of S .
Notice also that G is a subgroup of S .
A monoid S is called right subelementary if S = C ∪˙ N where C is the set of
all right cancellative elements of S and N is the set of all right nil elements of S
(again allowing the case when N is empty). Notice that C is a submonoid of S .
Theorem 2.6.
[8, Theorem 3 and Theorem 4] If S is a right elementary monoid
then all weakly flat cyclic right S -acts satisfy condition (P ).
If S is a periodic monoid such that all flat cyclic right S -acts satisfy condition
(P ) then S is right elementary.
Consequently, if S is periodic then the following statements are equivalent :
1. all weakly flat cyclic right S -acts satisfy condition (P ),
2. all flat cyclic right S -acts satisfy condition (P ),
3. S is right elementary.
We extend the above theorem in the next section by considering right subelementary monoids.
3.
Right Subelementary Monoids
Lemma 3.1.
Let S = C ∪ N be a monoid such that C is the set of right
cancellative elements of S and N is the set of right nil elements of S . Then
C ∩ N = Ø and N is either empty or an ideal of S .
4
Renshaw and Golchin
Proof.
Let x ∈ N and y ∈ S . If xy ∈ C then x ∈ C , a contradiction, and so
xy ∈ N .
Suppose then that yx ∈ C . Notice that yx 6= 1 since if yx = 1 and n ∈ N was
such that xn is right zero and no such smaller n has this property, then for n > 1
we have xn = yxn = xn−1 giving a contradiction, and for n = 1, x = yx = 1,
another contradiction. Since xy ∈ N then there exists m ∈ N with (xy)m a
right zero and m the smallest such positive integer. Hence, (xy)m+1 = (xy)m
and so (yx)m+2 = y(xy)m+1x = y(xy)mx = (yx)m+1 and therefore yx = 1 - a
contradiction as required. Hence, in this case yx ∈ N .
We aim to consider cyclic acts over a right subelementary monoid but first we need
a few technical lemmas.
Lemma 3.2.
Let S = C ∪˙ N be a right subelementary monoid. Suppose that
ρ is a right congruence on S such that S/ρ is weakly flat.
1. If x ∈ C, y ∈ N are such that x ρ y , then S/ρ is projective.
2. If x, y ∈ C , then x ρ y if and only if there exist s, t ∈ S with sx = ty and
s ρ 1 ρ t.
3. If S/ρ is not projective and a, b ∈ C are such that a ρ 1 ρ b, then there
exist s, t ∈ C with s ρ 1 ρ t and sab = tba.
4. If S/ρ is not projective and a, b ∈ C and c, x ∈ S are such that a ρ b(∆x)c,
then there exist s, t ∈ C with s ρ 1 ρ t and sax = tcx (i.e. sa(∆x)tc).
5. If S/ρ is not projective and if there exist s1 , s2 , t1 , t2 ∈ C and a, b, c ∈ S
such that s1 ρ 1 ρ t1 , s2 ρ 1 ρ t2 and such that s1 a = t1 b, s2 b = t2 c, then
there exist s, t ∈ C with s ρ 1 ρ t and sa = tc.
(Note that if S is a right subelementary monoid, then from Lemma 2.1, if S/ρ is
projective then either ρ = ∆ or there exists e2 = e(∈ N) such that e ρ 1 ∈ C .
Hence if ρ 6= ∆ then the converse of (1) above is true for right subelementary
monoids.)
Proof.
1. We know that there exist s, t ∈ S such that sx = ty and such that s(ρ ∨
∆x)1(ρ ∨ ∆y)t. Since y ∈ N then sx = ty ∈ N and so s ∈ N . Hence there
exists n ∈ N such that sn is right zero. Now x ∈ C means that ∆x = ∆
and so it follows that sn ρ 1.
Now let u, v ∈ S be such that u ρ v . Then sn u ρ u ρ v ρ sn v . Since sn u
and sn v are both right zero, then by Lemma 2.3, sn u = sn v and the result
follows from Lemma 2.1(1).
Renshaw and Golchin
5
2. If x ρ y then we know that there exist s, t ∈ S such that sx = ty and such
that s(ρ ∨ ∆x)1(ρ ∨ ∆y)t. But x, y ∈ C means that ∆x = ∆, ∆y = ∆ and
so s ρ 1 ρ t as required.
The converse is straightforward.
3. Notice that ab ρ ba (ρ 1) and so from (2), there is s, t ∈ S such that
s ρ 1 ρ t and such that sab = tba. Finally, s, t ∈ C by (1).
4. Since a ρ b then by (2), there exist s, t ∈ S with sa = tb and s ρ 1 ρ t.
Therefore sax = tbx = tcx. Finally, s, t ∈ C by (1).
5. Notice that, t1 ρ s2 ρ 1 and so from (3), there exist s0 , t0 ∈ C with s0 ρ 1 ρ t0
and s0 s2 t1 = t0 t1 s2 . Hence s0 s2 s1 a = s0 s2 t1 b = t0 t1 s2 b = t0 t1 t2 c. Now put
s = s0 s2 s1 and t = t0 t1 t2 .
Lemma 3.3.
Let S = C ∪˙ N be a right subelementary monoid and suppose
that in addition, ∀a ∈ C, ∀b ∈ N, b ∈ Sab. Let ρ be a right congruence on S such
that S/ρ is weakly flat but not projective and let u, v ∈ N . Then u ρ v if and
only if there exist x, y ∈ C with x ρ 1 ρ y and xu = yv . Moreover, ρ|N ⊆ L.
Proof.
Let u ρ v so that there exist s, t ∈ S with s(ρ ∨ ∆u)1(ρ ∨ ∆v)t and
su = tv . Hence we have
s = s0 ρ t0 (∆u)s1 ρ . . . sn ρ tn (∆u)sn+1 = 1
(∗)
for some si , ti ∈ S .
We show that there exist a, b ∈ C with a ρ 1 ρ b and asu = bu and we do this by
considering two cases :
Case 1. All the si , ti ∈ C .
First, if n + 1 = 0 (in other words if s = 1) then put a = b = 1. Otherwise,
n + 1 ≥ 1 and we use an inductive argument on n. If n + 1 = 1 then we have
s = s0 ρ t0 (∆u)1.
From Lemma 3.2 (4), there exist a, b ∈ C with a ρ 1 ρ b and asu = as0 u = bu as
required.
Now, if n + 1 > 1 then from s = s0 ρ t0 (∆u)s1 and Lemma 3.2 (4), we
have a0 , b0 ∈ C with a0 ρ 1 ρ b0 and a0 su = a0 s0 u = b0 s1 u. By induction
on s1 ρ . . . sn ρ tn (∆u)1, there exist a1 , b1 ∈ C with a1 ρ 1 ρ b1 and a1 s1 u = b1 u.
The result now follows from Lemma 3.2 (5).
Case 2. At least one of the si or ti ∈ N . Since sn+1 = 1 ∈ C then working
backwards from the right hand end of (∗) we see that there are two possibilities
that can arise :
1. si ρ ti for some i with si ∈ N and ti ∈ C . This case is impossible since we
are assuming that S/ρ is not projective.
6
Renshaw and Golchin
2. ti (∆u)si+1 for some i with ti ∈ N and si+1 ∈ C . In this case, since
u ∈ Ssi+1 u then there exists r ∈ S with u = rsi+1 u = rti u and since
rti ∈ N then we can deduce that u is right zero. Consequently, we can take
a = b = 1.
In a similar way, there exist c, d ∈ C with c ρ 1 ρ d and ctv = dv .
Now by Lemma 3.2 (3), there exist g, h ∈ C with g ρ 1 ρ h and gac = hca.
Consequently, on putting x = hcb, y = gad we see that
xu = (hcb)u = hcasu = gacsu = gactv = (gad)v = yv
and x = (hcb) ρ 1 ρ (gad) = y as required.
Conversely, if x ρ 1 ρ y and xu = yv then u ρ xu = yv ρ v as required.
Suppose that (u, v) ∈ ρ|N . Then by the previous part, there exist x, y ∈ C with
x ρ 1 ρ y and xu = yv . Notice also that by assumption, there exist z, w ∈ S with
u = zxu, v = wyv . Hence u = zxu = zyv and v = wyv = wxu and so (u, v) ∈ L
as claimed.
Notice that from Lemmas 3.2 and 3.3, we have
Corollary 3.4.
Let S = G ∪˙ N be a right elementary monoid and suppose
that ρ is a right congruence on S such that S/ρ is weakly flat but not projective.
Then ρ ⊆ L.
The following is straightforward.
Lemma 3.5.
Let S be a monoid and suppose that N is a right nil ideal of S .
Then u ∈ N is regular if and only if u is right zero.
Lemma 3.6.
Let S = C ∪˙ N be a right subelementary monoid such that
∀a ∈ C, b ∈ N, b ∈ Sab and let ρ be a congruence on S such that S/ρ is weakly
flat. Suppose that u ∈ N, v ∈ S and suppose that u is right zero. Then u ρ v if
and only if either u = v or there exists a right zero e ∈ S with e ρ 1 and such
that u = ev .
Proof.
If the given condition holds then it is clear that u ρ v .
Suppose then that u 6= v and consider the following two cases :
1. v ∈ C . By Lemma 3.2(1), S/ρ is projective and so there exists e2 = e ∈ S
such that e ρ 1 and eu = ev . Now e 6= 1 and so e ∈ N and so by Lemma 3.1
and Lemma 3.5 e is right zero.
2. v ∈ N . If S/ρ is projective then we proceed as in the first case. Otherwise,
by Lemma 3.3 there exist x, y ∈ C such that x ρ 1 ρ y and xu = yv . Now by
assumption there exists z ∈ S such that v = zyv and so u = zxu = zyv = v .
We can now deduce one of our main theorems in this section.
Renshaw and Golchin
7
Theorem 3.7.
Let S = C ∪˙ N be a right subelementary monoid and suppose
that in addition, ∀a ∈ C, ∀b ∈ N, b ∈ Sab. Let ρ be a right congruence on S such
that S/ρ is weakly flat. Then S/ρ satisfies condition (P ).
Proof.
Let u ρ v . By Lemma 3.2 (1 & 2), we need only consider the case where
u, v ∈ N . Moreover, we can assume that S/ρ is not projective. The result then
follows by Lemma 3.3.
It is fairly clear that if C is a group or if N contains all the right zeros of S then
the condition “b ∈ Sab” is satisfied and so we can deduce as corollaries [8, Theorem
3] (see Theorem 2.6) and part of [4, Theorem 3.6] (see Theorem 4.5 below).
One obvious question that arises here is, “Is the condition ‘b ∈ Sab’ necessary for
such a monoid to have the property that all weakly flat cyclic right acts satisfy
condition (P) ?”. Certainly if the element a ∈ C has finite order then a must be
invertible and so the condition will be satisfied. Hence this property will be true
for periodic monoids. In fact in this case, the right cancellative monoid C will be
a (periodic) group.
The converse of Theorem 3.7 is untrue. In [5, Proposition 3.1], Bulman-Fleming
and Kilp showed that the monoid S given by the presentation
< x, y|xy = x = yx > ∪{1}
was such that every weakly flat S -act satisfies condition (P ) but it contains
elements that are neither cancellative nor nil.
However, let us examine the question of which monoids have the property that all
their (weakly) flat cyclic right acts satisfy condition (P ).
4.
Flatness and Condition (P )
Theorem 4.1.
Let S be a monoid. Then every e ∈ E(S)\{1} is right zero if
and only if S is a disjoint union of a group, a right nil set and a regular-free set.
Proof.
First, if S is a disjoint union of a group, a right nil set and a regularfree set, then the only non-identity idempotents will have to be right nil and hence
right zero.
Conversely, suppose that every e ∈ E(S)\{1} is right zero. Suppose that x ∈ S
is eventually regular. Then there exists n ∈ N and x0 ∈ S such that xn x0 xn = xn .
Since xn x0 is idempotent then either xn x0 = 1 and so x is right invertible or xn x0
(and hence xn = xn x0 xn ) is right zero giving that x is right nil. It is then easy to
see that the right invertible elements form a group and so S is a disjoint union as
claimed.
Notice that in the previous theorem, S = G ∪˙ N ∪˙ F where G is a group, N
contains all the right nil elements of S and F is the set of all regular-free elements
of S . It is also the case that N ∪˙ F is an ideal of S , that F is closed under
the taking of powers and that for this type of monoid, the condition that F is
8
Renshaw and Golchin
a regular-free subset is equivalent to the condition that it is an idempotent-free
subset.
It is unknown whether G ∪˙ N is a submonoid of S .
If S = G ∪˙ N ∪˙ F is right reversible then it can contain at most one right zero
element which would then be a zero of S and so N is either empty or nil (every
element has a power equal to 0). Conversely, if N is nil then S must be right
reversible.
There is an alternative decomoposition for the monoids in Theorem 4.1 that will
be useful.
Corollary 4.2.
Let S be a monoid. Then every e ∈ E(S)\{1} is right zero if
and only if S = C ∪˙ N ∪˙ F 0 where C is right cancellative, N is as before and F 0
is the subset of F which contains no right cancellative elements.
From Lemma 2.2 we can deduce
Corollary 4.3.
Let S be a monoid such that all flat cyclic right S -acts satisfy
condition (P ). Then S is a disjoint union of a group, a right nil set and a regularfree set.
It is known that the converse to this corollary is false (see [14, comments after
Lemma 4.1] or [5, Propositions 3.3 & 3.4]).
The following is straightforward.
Lemma 4.4.
S is a left PP monoid such that every e ∈ E(S)\{1} is right zero
if and only if S is right subelementary and every element in the right nil part is
right zero.
Proof.
Suppose that S is a left PP monoid such that every e ∈ E(S)\{1} is
right zero and let x ∈ S . Since S is left PP then there exists e2 = e ∈ S such that
ex = x and ux = vx implies ue = ve. Now if e = 1 then x is right cancellative
otherwise e is a right zero and so therefore is x.
Conversely, if S is right subelementary and every element in the right nil part is
right zero then by Corollary 4.2 every e ∈ E(S)\{1} is right zero. If x ∈ S then
define e = 1 if x is right cancellative and e = x if x is right zero. Then in either
case, e2 = e, ex = x and ux = vx implies ue = ve for all u, v ∈ S and so S is left
PP.
From Lemma 2.2, Theorem 3.7 and Lemma 4.4 we can deduce
Theorem 4.5.
([4, Theorem 3.6]) S is left PP and every (weakly) flat cyclic
right S -act satisfies condition (P ) if and only if every element of S is either right
zero or right cancellative.
Clearly, from Theorem 4.1 we have
Corollary 4.6.
Let S be an eventually regular monoid.
E(S)\{1} is right zero if and only if S is right elementary.
Then every e ∈
Consequently we can deduce from Lemma 2.2 and Theorem 2.6 the following
extension to [8, Theorem 4] (which also provides a converse for Lemma 2.2 in
the case S is eventually regular).
Renshaw and Golchin
Theorem 4.7.
equivalent :
9
If S is an eventually regular monoid then the following are
1. all weakly flat cyclic right S -acts satisfy condition (P ),
2. all flat cyclic right S -acts satisfy condition (P ),
3. every e ∈ E(S)\{1} is right zero,
4. S is right elementary.
We can now give an alternative proof of the main theorem in [4].
Theorem 4.8.
are equivalent
[4, Theorem 3.4] For any monoid S the following statements
1. S is right nil.
2. Every weakly flat cyclic right S -act is projective.
3. Every weakly flat cyclic right S -act is strongly flat.
4. Every flat cyclic right S -act is projective.
5. Every flat cyclic right S -act is strongly flat.
Proof.
Note that (2) ⇒ (4) ⇒ (5) are obvious, as is (2) ⇒ (3) ⇒ (5).
(1) ⇒ (2) If ρ = ∆ then S/ρ = S is clearly projective. Otherwise, by Lemma 3.3,
if S/ρ is not projective then ρ = ∆ giving a contradiction.
(5) ⇒ (1) By corollary 4.3, S = G ∪˙ N ∪˙ F . But by [3, Proposition 2.10], every
element of S is aperiodic (xn = xn+1 for some n ∈ N ). Hence F = Ø and G = {1}
as required.
If a monoid S has the property that all flat right S -acts satisfy condition (P ),
then it is known that |E(S)| = 1 ([3, Theorem 2.4]) and so we see that
Corollary 4.9.
Let S be a monoid such that all flat right S -acts satisfy condition (P ). Then S is a disjoint union of a group and a regular-free subsemigroup.
Again, the converse is false as can be seen from [5, Proposition 3.3].
The following appeared as one of the main results in [12] although it also follows
implicitly from [4, Corollary 3.2] and [3, Theorem 2.4].
Theorem 4.10.
only if S = {1}.
[12, Theorem 2.1] All flat right S -acts are strongly flat if and
Proof.
If S = {1} then it is clear that all flat S -acts are strongly flat.
For the converse, notice that from Corollary 4.9, S is a disjoint union of a group
and a regular-free subsemigroup but from Theorem 4.8, S is also a right nil monoid
and so S = {1}.
The following appeared as the main theorem in [2].
10
Renshaw and Golchin
Theorem 4.11.
[2, Theorem 2] Let S be any monoid. Then the following
statements are equivalent:
1. S is right cancellative.
2. S is left PP and every weakly flat right S -act satisfies condition (P ).
3. S is left PP and every flat right S -act satisfies condition (P ).
Proof.
(1) ⇒ (2) Let A be a weakly flat S -act and suppose that au = a0 v
0
with a, a ∈ A, u, v ∈ S . By Lemma 2.4, there exists a00 ∈ A and w ∈ Su ∩ Sv
such that au = a0 v = a00 w . But then w = su = tv for some s, t ∈ S and so
au = a00 su and a0 v = a00 tv . Since A is torsion free and u, v are right cancellative,
then a = a00 s, a0 = a00 t as required and A satisfies condition (P ).
(2) ⇒ (3) obvious
(3) ⇒ (1) Straightforward from the definition of left PP.
Theorem 4.12.
Suppose that S is a monoid such that every e ∈ E(S)\{1} is
right zero. Then all cyclic right S -acts are weakly flat if and only if S is a group
or a 0-group.
Proof.
Suppose that all cyclic right S -acts are weakly flat. Then it is known
(Lemma 2.5) that S is regular and for all x, y ∈ S there exists z ∈ Sx ∩ Sy such
that (z, x) ∈ ρ(x, y). In particular, S is right reversible. But by Theorem 4.7,
S is also right elementary and it is easy to see that the right nil elements are all
right zero. However, right reversible monoids have at most one right zero and so
S is either a group or a 0-group.
Conversely, suppose that S is a group or a 0-group and let x, y ∈ S . If y = 0
then z = y ∈ Sx ∩ Sy . If y 6= 0 then notice that z = x = xy −1y ∈ Sx ∩ Sy . In
both cases (z, x) ∈ ρ(x, y) and so all cyclic right S -acts are weakly flat.
Notice that if all cyclic right S -acts satisfy condition (P ) then they are all weakly
flat. Also, all flat cyclic right S acts will then satisfy condition (P ) and so
from Lemma 2.2 every e ∈ E(S)\{1} is right zero. Hence S is a group or a
0-group and so we have the ”only if” part of [13, Theorem 2.1] (the ”if” part being
straightforward).
5.
Left PSF monoids
If S is a monoid then an element x ∈ S is called right semi-cancellative if whenever
sx = tx with s, t ∈ S then there exists r ∈ S with x = rx and sr = tr .
A monoid S is called left PSF if all its principal left ideals are strongly flat. Every
left PP monoid is clearly left PSF and it was proved in ([14, Lemma 3.3]) that S
is left PSF if and only if all elements of S are right semi-cancellative.
Recall that if all flat cyclic right S -acts satisfy condition (P ), then S = C ∪˙ N ∪˙ F 0
where C contains all the right cancellative elements of S , N contains all the right
nil elements of S and F 0 the regular-free which are not right cancellative.
Renshaw and Golchin
11
Lemma 5.1.
Let S = C ∪˙ N ∪˙ F 0 be a left PSF monoid. Then for all
x ∈ N ∪˙ F 0 , either x is right zero or there exists r ∈ F 0 such that x = rx and
r 6= x.
Proof.
If x 6∈ C then there exists a, b ∈ S with ax = bx and a 6= b. Since x
is right semi-cancellative then there exists r ∈ S with x = rx and ar = br . Now,
r 6∈ C because a 6= b and if r ∈ N then there exists k ∈ N with r k right zero from
which we deduce that x is right zero.
If r = x then r ∈ E(S), a contradiction.
Corollary 5.2.
Let S be a monoid. Then S is left PSF and right subelementary if and only if every element of S is either right cancellative or right zero.
Consequently, for such a monoid, all (weakly) flat cyclic right S -acts satisfy condition (P ).
Proposition 5.3.
[14, Propositions 2.1 & 2.3] If S is a monoid such that all
flat right S -acts satisfy condition (P ) then for every infinite sequence (x0 , x1 , . . .)
with xi = xi+1 xi , xi ∈ S, i = 0, 1, . . ., there exists a positive integer n such that
xn = xn+1 = . . . = 1.
The following extension to Theorem 4.11 is now easy to prove.
Theorem 5.4.
Let S be a left PSF monoid. Then all flat right S -acts satisfy
condition (P ) if and only if S is right cancellative.
Proof.
We see from Corollary 4.9 that S = C ∪˙ F 0 with C right cancellative
0
and F the set of regular free elements of S that are not right cancellative. We
claim that F 0 = Ø. If not then there exists x0 ∈ F 0 and so by Lemma 5.1 there
exists x1 ∈ F 0 with x0 = x1 x0 and x0 6= x1 . In a similar way, there exists
x2 ∈ F 0 with x1 = x2 x1 and x2 6= x1 . Continuing in this fashion we get an infinite
sequence (x0 , x1 , . . .) with xi = xi+1 xi , xi ∈ F 0 and xi 6= xi+1 , i = 0, 1, . . .. But
this contradicts Proposition 5.3.
The converse is straightforward.
We turn now to considering left PSF monoids over which all (weakly) flat cyclic
right acts satisfy condition (P ).
By a proper right ideal I of a monoid S , we mean one such that I 6= S . Note
that this is slightly different from the definition given in [11].
The following, extends part of [14, Theorem 4.3] to the non right reversible case.
Theorem 5.5.
Let S be a left PSF monoid. Then the following are equivalent
1. S = C or S = C ∪ {0}, with C right cancellative,
2. for every proper right ideal J of S with |J| > 1, there exists j ∈ J\Jj .
12
Renshaw and Golchin
Proof.
(1) ⇒ (2). Let J be a proper right ideal of S with |J| > 1 and suppose
that j ∈ Jj for all j ∈ J . Then there exists x ∈ J with x ∈
/ {0, 1} and so x = yx
for some y ∈ J . But x is right cancellative and so J = S , a contradiction.
(2) ⇒ (1). We can assume, without loss of generality, that S 6= {1}. Let
0 6= x0 ∈ S be such that x0 is not right cancellative. Then there exist a, b ∈ S
such that ax0 = bx0 but a 6= b. Since x0 is right semi-cancellative, then there
exists x1 ∈ S such that x0 = x1 x0 and ax1 = bx1 and so x1 is not right
cancellative. Continuing in this fashion, we construct a sequence (x0 , x1 , . . .) such
S
that xi = xi+1 xi and xi is not right cancellative for i = 0, 1, . . .. Let J = ∞
i=0 xi S
and notice that J is a proper right ideal of S . We claim that |J| > 1. Otherwise,
if J = {x0 } then x0 is a left zero of S and if I is the right ideal consisting of all
left zeros of S , notice that I 6= S since S 6= {1}. Hence, by our condition, |I| = 1
from which we easily deduce that x0 = 0, a contradiction and so |J| > 1. But
for all s ∈ S , xi s = xi+1 xi s ∈ Jxi s which contradicts our condition. Hence the
result.
Lemma 5.6.
Let S be a left PSF monoid. Then for each x0 ∈ S with x0 not
right cancellative, there exists a sequence (x0 , x1 , . . .), xi = xi+1 xi , and xi not right
cancellative, i = 0, 1, . . ..
Proof.
If x0 is not right cancellative, then there exists a, b ∈ S with ax0 = bx0
and a 6= b. But x0 is right semi-cancellative and so there exists x1 ∈ S with
x0 = x1 x0 and ax1 = bx1 . Since a 6= b then x1 is not right cancellative.
Continuing in this fashion we generate the required sequence.
Lemma 5.7.
Let S be a monoid. If for every proper right ideal J of S with
|J| > 1, there exists j ∈ J\Jj , then for every infinite sequence (x0 , x1 , . . .) with
xi = xi+1 xi , i = 0, 1, . . ., there exists n ∈ N with xn ∈ E(S).
Proof.
Suppose (x0 , x1 , . . .) is an infinite sequence with xi = xi+1 xi , and
S
/ E(S), i = 0, 1, . . .. Let J = ∞
xi ∈
i=0 xi S and notice that J is a proper right
ideal of S with |J| > 1 and for every j ∈ J, j ∈ Jj . This is a contradiction as
required.
Lemma 5.8.
Let S be a left PSF monoid with the property that for every
infinite sequence (x0 , x1 , . . .) with xi = xi+1 xi , i = 0, 1, . . ., there exists n ∈ N
with xn ∈ E(S). Then for every element x ∈ S either x is right cancellative or
there exists e ∈ E(S)\{1} such that x = ex.
Proof.
Let x0 ∈ S and suppose that x0 is not right cancellative. Then by
Lemma 5.6 there exists a sequence (x, x1 , . . .) with xi = xi+1 xi and xi not right
cancellative i = 0, 1, . . .. Hence there exists n ∈ N with xn ∈ E(S) and since
x0 = xn x0 , the result follows.
The following theorem generalizes [4, Theorem 3.6].
Renshaw and Golchin
Theorem 5.9.
13
Let S be a monoid. Then the following are equivalent,
1. S is right subelementary with the right nil elements all right zero;
2. S is right subelementary and left PSF;
3. S is left PSF, for any infinite sequence (x0 , x1 , . . .) with xi = xi+1 xi , i =
0, 1, . . ., there exists n ∈ N with xn ∈ E(S) and every weakly flat cyclic right
S -act satisfies condition (P );
4. S is left PSF, for any infinite sequence (x0 , x1 , . . .) with xi = xi+1 xi , i =
0, 1, . . ., there exists n ∈ N with xn ∈ E(S) and every flat cyclic right S -act
satisfies condition (P );
5. S is left PSF, for every infinite sequence (x0 , x1 , . . .) with xi = xi+1 xi , i =
0, 1, . . ., there exists n ∈ N with xn ∈ E(S) and every e ∈ E(S)\{1} is right
zero;
6. S is left PP and every weakly flat cyclic right S -act satisfies condition (P );
7. S is left PP and every flat cyclic right S -act satisfies condition (P ).
Proof.
(1) and (2) are equivalent by Corollary 5.2.
(1), (6) and (7) are equivalent by Theorem 4.5.
(1) ⇒ (3) Straightforward from Theorem 4.5
(3) ⇒ (4) Obvious.
(4) ⇒ (5) Lemma 2.2.
(5) ⇒ (1) Lemma 5.8.
Proposition 5.10.
ideal of S . Then
[5, Proposition 2.2] Let S be a monoid and let J be a right
1. S/J has property (P ) if and only if J = S and S is right reversible, or
|J| = 1.
2. S/J is flat (or equivalently weakly flat) if and only if S is right reversible
and j ∈ Jj for all j ∈ J .
Lemma 5.11.
Let S be a right reversible monoid. If S is left PSF and all
flat cyclic right S -acts satisfy condition (P ) then for every infinite sequence
(x0 , x1 , . . .) with xi = xi+1 xi , i = 0, 1, . . . there exists n ∈ N with xn ∈ E(S).
Proof.
Suppose that there exists an infinite sequence (x0 , x1 , . . .) with xi =
S
xi+1 xi , i = 0, 1, . . . and such that each xi 6∈ E(S). Let J = ∞
i=0 xi S and note
that J is a proper right ideal of S such that j ∈ Jj for all j ∈ J . Hence
from Proposition 5.10(2) we see that S/J is flat and so from Proposition 5.10(1),
|J| = 1, a contradiction.
From Theorem 5.9 and Lemma 5.11, we can immediately deduce part of [14,
Theorem 4.3].
14
Renshaw and Golchin
Corollary 5.12.
Let S be a right reversible monoid. Then S is left PSF and all
flat cyclic right S acts satisfy condition (P ) if and only if S = C or S = C ∪ {0}
where C is a right cancellative monoid.
If a monoid S is such that all flat cyclic acts have property (P ), then we have
indicated that it has a structure of the form S = G ∪˙ N ∪˙ F . This has proved
useful in that some of the main results in the literature can be deduced as fairly
simple consequences of this structure. However, it is clear that more details of
the structure of the regular-free part will be needed if a full classification of these
monoids is to follow from these techniques.
Acknowledgement
The authors would like to thank the referee for some useful comments and S
Bulman-Fleming for pointing out an error in an earlier draft of this paper. The
second named author would like to express his thanks to the Iranian Ministry of
Culture and Higher Education for supporting him in his research.
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Faculty of Mathematical Studies
University of Southampton
Southampton
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ENGLAND