The categories of left and right modules of a semigroup

for all

s, t \in S

. Often we shall write

s x

for

s . x

if there is no ambiguity. A left

S

-set (or a left

S

-module) is a pair

(X, a_{X})

where

X

is a set and

a_{X}

is an action of

S

on the left of

X

. A [left]

S

map

α : X \to Y

from a left

S

-set

X

to a left

S

-set

Y

(or a left

S

-morphism) is a map

α : X \to Y

that satisfy the condition

We denote by

M o d S

the category with

S

-modules as objects and

S

-maps as morphisms. There is a functor

U : M o d S \to S e t

sending every

X \in 𝖛 M o d S

to the underlying set

U (X)

and every morphism

f : X \to Y \in M o d S

to the map

U (f)

determined by

f

. In the sequal, we refer to

U

as the foregetful functor for left-modules. A subobject relation on

𝖛 M o d S

is deﬁned as follows: For

X \in 𝖛 M o d S

let

a_{X} : S \times U (X) \to U (X)

denote the action of

S

U (X)

given by the module structure on

X

. For

X, Y \in 𝖛 M o d S

write

The relation on

𝖛 M o d S

deﬁned above is a subobject relation on

M o d S

. In fact, it is easy to check that the

J = ȷ Y X

is a morphism in

M o d S

if and only if

X

and

Y

satisfy Equation (10). The class of all these morphisms form a choice of subobjects in

M o d S

(see Equation 2.1.2). Notice that, if

X

and

Y

are submodules of

Z \in M o d S

then it follows from Equation (10) that

X \subseteq Y

if and only if

U (X) \subseteq U (Y)

It is clear that the foregetful functor

U : M o d S \to S e t

is faithful and preserve images. It follows that

(M o d S, U)

satisﬁes axiom (Sb:b) of Deﬁnition 3.1. If

X, Y \in 𝖛 M o d S

and

x \in U (X) \cap U (Y)

, then

D = {s x : s \in S^{1}

is a lefmodule such that

x \in D \subseteq X

and

x \in D \subseteq Y

. Therefore

(M o d S, U)

satisﬁes axiom (Sb:c) also. If

𝒞 \subseteq M o d S

is a small subcategory of

M o d S

with imagages and if

U_{0} = U ∣ 𝒞

then

(𝒞, U_{0})

clearly satisfy axioms (Sb:b) and (Sb:c). Further, if the partially ordered set

𝖛 𝒞

of leftmodules has a largest member

Z

, then it follows from Equation (10) that the functor

U_{0}

is an embedding. In this case

𝒞

is an

𝒮

-category with respect to

U_{0}

. For convenience of using these facts in the sequal we state as a propositionm below.

Proposition 4.1. Let $M o d S$ be the category with subobjects in which objects are left $S$ modules, morphisms are left module morphisms and the subobject relation is deﬁned by Equation (10). The category $M o d S$ has images and the foregetful functor $U : M o d S \to S e t$ is faithful and preserve subobjects. Moreover, the pair $(M o d S, U)$ satisfy axioms (Sb:b) and (Sb:c) of Deﬁnition 3.1. If $𝒞 \subseteq M o d S$ is a small subcategory of $M o d S$ with images and if $U_{0} = U ∣ 𝒞$ then $(𝒞, U_{0})$ satisfy axioms (Sb:b) and (Sb:c). Further, if there is a leftmodule $Z \in 𝖛 M o d S$ which is an upperbound of partially ordered subset $𝖛 𝒞$ of $𝖛 M o d S$ then, by Equation (10), the functor $U_{0}$ is an embedding. In this case $𝒞$ is an $𝒮$ -category with respect to $U_{0}$ .

Proof. The ﬁrst part of the statement is routine to verify. Equation (10) shows that the natural forgetful functor preserves subobjects. To show that $M o d S$ has images, let $f : X \to Y \in M o d S$ . Then $U (f)$ is a mapping from $U (X)$ to $U (Y)$ and $h = U {(f)}^{\circ}$ is a surjective map of $U (X)$ onto $Y' = cod h = im U (f) \subseteq U (Y)$ . If $s \in S$ and $y \in Y'$ there is $x \in X$ with

s y = s (h (x)) = s (f (x) = f (s x) \in Y' .

Hence there is a module $X' \subseteq Y$ with $U (X') = Y'$ and a morphism $f^{\circ} : X \to X'$ with $U (f^{\circ}) = h$ . Since $h$ is surjjecive, $f^{\circ}$ is an epimorphism. Also

U (f) = h ȷ U (Y) Y' = U (f^{\circ}) ȷ U (Y) U (X') = U (f^{\circ} ȷ Y X') .

Since $U$ is faithful, $f = f^{\circ} ȷ Y X'$ . Thus $f = f^{\circ} ȷ Y X'$ is a canonical factorization of $f$ . Let $f = g ȷ Y X ”$ be any canonical factorization of $f$ . Then

U {(f)}^{\circ} ȷ U (Y) U (X') = U (f) = U (g) ȷ U (Y) U (X ”) .

Since the left-hand side of equation above is the image factoprization of $U (f)$ in $S e t$ , we have $U (X') \subseteq U (X ”)$ . Since $X'$ and $X ”$ are submodules of $Y$ we must have $X' \subseteq X ”$ . Therefore $X' = im f$ by the deﬁnition of images (see Equation 2.2). Moreover, we have

U (im f) = U (X') = im U (f) .

If $𝒞 \subseteq M o d S$ is a small subcategory with images then it is clear that $𝒞$ satisfy axioms (Sb:b) and (Sb:c) with respect to $U_{0} = U ∣ 𝒞$ . Furthermore if there exists a largest module $Z$ for the partially ordered set $𝖛 𝒞$ then by Equation (10) $𝖛 U_{0}$ is an orderembedding of $𝖛 𝒞$ into $𝖛 S e t$ and so, $(𝒞, U_{0})$ satisﬁes axiom (Sb:a). Therefore $𝒞$ is ab $𝒮$ -category withrespect to $U_{0}$ . This completes the proof. $□$

Evidently

S^{1}

(see Cliﬀord and Preston [1961], page 4), the semigroup obtained by adjoining identity to

S

is a left [as well as right]

S

-module where the action is the multiplication by elements of

S

on the left [right] of elements in

S^{1}

. Similarly any left ideal

Λ

[righti deal

I

] of

S

is a left [right]

S

-module and is a proper submodule of

S^{1}

. Let

𝕃 (S)

denote the full subcategory of

M o d S

whose objects are left ideals of

S^{1}

; that is, if

X \in 𝖛 𝕃 (S)

then either

X = S^{1}

X

is a left ideal in

S

. Morphisms in this category are called right translations. It may be observed that, while

S^{1}

is a faithful left (as well as right) module, a left [right] ideal may not be faithful as a module. For

a \in S

, the map

ρ_{a} : s \mapsto s a

is a morphism in

𝕃 (S)

from

S^{1}

onto the principal left ideal

L (a) = S^{1} a

. This map

ρ_{a}

is called the principal right translation of

S

determined by

a

Since, for every

X \in 𝖛 𝕃 (S)

, the action of

S

X

is the restriction of the product in

S

S \times U (X)

𝖛 U

is an order embedding of

𝖛 𝕃 (S)

into

𝖛 S e t

. Thus the natural foregetful functor

U : 𝕃 (S) \to S e t

is an embedding of

𝕃 (S)

into

S e t

. Therefore the pair

(𝕃 (S), U)

satisﬁes axiom (Sb:a) of Deﬁnition 3.1. Axioms (Sb:b) and (Sb:c) hold by Proposition 4.1. Thus we have:

In the following, for brevity, we shall avoid explicit reference to the forgetful functor

U

as far as possible. Instead, we shall assume that the given

𝒮

-category has been identiﬁed with its image in

S e t

(see Proposition 3.3) so that

U

reduces to the inclusion.

We can dualise the deﬁnitions and results above. Thus a right

S

module is a set on which

S

acts on the right in the usual sense. A morphism

f : X \to Y

of right modules is a mapping that preserve rightr actions. There is a category

M o d S

in which objects are right

S

-modules and morphisms are morphisms of right modules. We can deﬁne a subobject relation in

M o d S

by dualising Equation (10). The category

M o d S

with subobjects so deﬁned has images and the foregetful functor

U : M o d S \to S e t

(that forgets right action) preserve subobjects and images of morphisms.

Remark 4.1: Notice that a left $S$ -module $X$ uniquely determines a dual (contravariant) representation $λ : s \mapsto λ_{s}$ where $λ_{s}$ denote the map on the set $U (X)$ deﬁned by

λ_{s} (x) = s . x for all x \in X .

Conversely every contravariant representation of $S$ by functions on the set $X$ uniquely determines a left $S$ -module structure on $X$ . Let $X$ and $Y$ be left $S$ -modules giving representations $ρ$ and $λ$ respectively. A map $α : X \to Y$ is a morphism of $S$ -modules if and only if $α$ commutes with $ρ$ and $λ$ :

λ \circ α = α \circ ρ .

Thus in the following we may replace modules by representations and mprphisms by maps commuting with representations. We shall say a representation $ρ$ (by functions on a set $X$ ) or the associated module $M_{ρ}$ is faithful if the function $ρ : S \to 𝒯 X$ is injective. Dual remarks are valid for right modules of $S$ and representationys by functions on a set.

For each

a \in S

L (a) = S^{1} a

and

R (a) = a S^{1}

are principal left and right ideals of

S

respectively. Notice that the natural left action of

S

L (a)

can be extended to an action of

S^{1}

so that

L (a)

is a left

S^{1}

-set which is generated by

a

and thus

L (a)

is a cyclic

S^{1}

-set. [see E. Krishnan, 2000, §2.5,2.6]. A morphism

ρ : L (a) \to L (b)

of left ideals is a mapping which satisﬁes the condition: there exists

s \in S^{1}

such that

u ρ = u s

for all

u \in L (a)

. Thus

ρ

is the restriction of the right translation

ρ_{s} : x \mapsto x s

L (a)

and is a morphism of the corresponding left

S^{1}

-set. We shall refer to a triplet

(a, s, b) \in S \times S^{1} \times S

as left-admissible triple in

S

if it satisfy the condition

is a morphism in

𝕃 (S)

induced by the principasl right translation

ρ_{s}

. Conversely every morphism

ρ : L (a) \to L (b) \in 𝕃 (S)

between principal ideals in

S

indused by a principal right translation

ρ_{s}

determines a left admissible triple

(a, s, b)

such that

ρ = ρ (a, s, b)

. Notice that these representations of certain morphisms in

𝕃 (S)

as left-admissible triples are not, in general, unique. In fact, we can see that

Proposition 4.3 (The category of principal left ideals). To every semigroup $S$ there corresponds a category $𝕃 0 (S)$ speciﬁed as follows:

i)

𝖛 𝕃 0 (S) = {L (a) : a \in S}

ii)

𝕃 0 (S) (L (a), L (b)) = {ρ (a, s, b) = ρ_{s} ∣ L (a) : s \in S^{1} with a s \in L (b)}

iii)

Composition in

𝕃 0 (S)

is given by the rule

ρ (a, s, b) ρ (c, t, d) = \{\begin{matrix} ρ (a, s t, d) & if L (b) = L (c); \\ undeﬁned & if L (b) \neq L (c) . \end{matrix}

iv)

ρ (a, s, b) = ȷ L (b) L (a)

if and only if

ρ (a, s, b) = ρ (a, 1, b)

. Moreover

{ρ (a, 1, b) : a \in L (b)}

is a choice of subobjects in

𝕃 0 (S)

and the foregetful functor

U : 𝕃 0 (S) \to S e t

is an embedding category

𝕃 0 (S)

(with choice of subobjects above) into

S e t

v)

ρ (a, s, b)

is a morphism in

𝕃 0 (S)

, then

im ρ (a, s, b) = L (a s) and ρ (a, s, b) = ρ (a, s, a s)) ρ (a s, 1, b)

(14)

gives the image-factorization of $ρ (a, s, b)$ in $𝕃 0 (S)$ .

Furthermore, $𝕃 0 (S)$ is an $𝒮$ -category is an $𝒮$ -subcategory of $𝕃 (S)$ .

Proof. By deﬁnition, for each left admissible triplet $(a, s, b)$ , the map $ρ (a, s, b) = ρ_{s} ∣ L (a)$ is a morphism in $𝕃 (S)$ . Hence

𝕃 0 (S) = \cup {𝕃 0 (S) (L (a), L (b)) : a, b \in S}

is a subset of the set of morphisms of $𝕃 (S)$ . Also morphisms $ρ (a, s, b)$ and $ρ (c, t, d)$ are composable iﬀ $L (b) = L (c)$ . In this case, the triplet $(a, s t, d)$ is left admissible and we haveith respect to

\begin{array}{l} ρ (a, s, b) ρ (c, t, d) & = (ρ_{s} ∣ L (a)) (ρ_{t} ∣ L (c)) \\ = ρ_{s} ρ_{t} ∣ L (a) \\ = ρ_{s t} ∣ L (a) \\ = ρ (a, s t, d) . \end{array}

Therefore $𝕃 0 (S)$ is closed with respect to composition of maps. It follows that the composition deﬁned by the rule given in iii) coincides with composition of the corresponding maps. Now, for any $a \in S$ , $(a, 1, a)$ is left-admissible and $ρ (a, 1, a) = 1_{L (a)}$ . Hence $𝕃 0 (S)$ is the morphism set of a subcategory of $𝕃 (S)$ . Clearly, the vertex set of $𝕃 0 (S)$ is the set of all principal left ideals of $S$ .

It is obvious that $L (a) \subseteq L (b)$ if and only if $(a, 1, b)$ is an admissible triple and $ρ (a, 1, b) = ȷ L (b) L (a)$ . Therefore inclusions between vertices of $𝕃 0 (S)$ are morphisms of the form $ρ (a, 1, b)$ with $a, b \in S$ . It is easy to verify that the set ${ρ (a, 1, b) : a \in L (b)}$ is a choice of subobjects for $𝕃 0 (S)$ . Further, for any $ρ = ρ (a, s, b)$ , we can see that the function $ρ$ maps the set $L (a)$ onto the set $L (a s)$ . Therefore $L (a s)$ is the image of $ρ$ in $𝕃 0 (S)$ as well as in $𝕃 (S)$ . Since $𝕃 (S)$ is an $𝒮$ -category, by Corollary 3.2 $𝕃 0 (S)$ is an $𝒮$ -subcategory of $𝕃 (S)$ . $□$

Let

ϕ : S \to T

be a homomorphism of semigroups. If

(a, s, b)

is a left admissible triple in

S

, by Equation (11)

a s = t b

for some

t \in S^{1}

. Then

a ϕ, b ϕ \in T

s ϕ, t ϕ \in T^{1}

and

(a ϕ) (s ϕ) = (t ϕ) (b ϕ)

. Therefore

(a ϕ, s ϕ, b ϕ)

is an admissible triple in

T

. We see that the equations

for each

a \in S

and admissible triple

(a, s, b)

S

is an inclusion-preserving functor

𝕃 0 (ϕ) : 𝕃 0 (S) \to 𝕃 0 (T)

. Furthermore the assignments

is a functor

𝕃_{0}

from the category

𝔖

of semigroups to the category

𝔖 𝔠 𝔞 𝔱

of small

𝒮

-categories.

Dually there is a full subcategory

ℝ (S) \subseteq M o d S

of all right ideals of

S^{1}

in which morphisms are called left-translations. The morphism

ℝ (S)

is called the principal left translation of

S

determined by

a

Dually we can deﬁne the category

ℝ 0 (S)

of principal right ideals. A morphism in

ℝ 0 (S)

from

R (a)

R (b)

, denoted by

λ (a, s, b) : R (a) \to R (b)

is a left translation

λ_{s} ∣ R (a)

with

s \in S^{1}

and

s a \in R (b)

. We shall say that a triplet

(a, s, b) \in S \times S^{1} \times S

satisfying the condition

s a \in R (b)

is right admissible. Notice that

(a, s, b)

is right admissible if and only if

s a = b t

for some

t \in S^{1}

. Moreover all deﬁnition and results for

𝕃 0 (S)

holds for

ℝ 0 (S)

with appropriate modiﬁcations. Proof of the following statement (the dual of Proposition 4.3) is obtained by dualising (that is, replacing left ideals by right ideals, left translations by right translations, etc.) the proof of Proposition 4.3.

Proposition 4.4 (The category of principal right ideals). To every semigroup $S$ there corresponds a category $ℝ 0 (S)$ speciﬁed as follows:

i)

𝖛 ℝ 0 (S) = {R (a) : a \in S}

ii)

ℝ 0 (S) (R (a), R (b)) = {λ (a, s, b) = λ_{s} ∣ R (a) : s \in S^{1} with s a \in R (b)}

iii)

Composition in

ℝ 0 (S)

is given by the rule

λ (a, s, b) λ (c, t, d) = \{\begin{matrix} λ (a, t s, d) & if R (b) = R (c); \\ undeﬁned & if R (b) \neq R (c) . \end{matrix}

iv)

λ (a, s, b) = ȷ R (b) R (a)

if and only if

λ (a, s, b) = λ (a, 1, b)

. Moreover

{λ (a, 1, b) : a \in R (b)}

is a choice of subjects in

ℝ 0 (S)

and

ℝ 0 (S)

is a category with subobjects in which the set-theoretic inclusions among vertexes in

ℝ 0 (S)

is the subobject relation.

v)

λ (a, s, b)

is a morphism in

ℝ 0 (S)

, then

im λ (a, s, b) = R (s a) and λ (a, s, b) = λ (a, s, s a)) λ (s a, 1, b)

gives the image-factorization of $λ (a, s, b)$ in $ℝ 0 (S)$ . Furthermore, $ℝ 0 (S)$ is an $𝒮$ -category and is a subcategory of $ℝ (S)$ .

The proposition Proposition 4.3 shows that there are close relations between the structure of the category

𝕃 0 (S)

and the semigroup

S

. For example, the following observations are useful:

Corollary 4.5. Let $S$ be a semoigroup and let $ρ = ρ (a, s, b) : L (a) \to L (b)$ be a morphism in $𝕃 0 (S)$ . Then we have

(a): $ρ$ is surjective if and only if $a s ℒ b$ ;
(b): $ρ$ is a split injection if and only if there exists $s' \in S^{1}$ such that $(b, s', a)$ is left admissible and $a s s' = a$ .
(c): $ρ$ is a split surjection if and only if if there exists $s' \in S^{1}$ such that $(b, s', a)$ is left admissible and $b s' s = b$ .
(d): $ρ$ is an isomorphism if only if $a ℛ a s ℒ b$ .

Statements dual to the above are valid for morphisms in $ℝ 0 (S)$ .

Proof. The statement (a) is an immediate consequence of the deﬁnition of $ρ (a, s, b)$ . To prove (b), assume that an element $s' \in S^{1}$ exists satisfying the requirements. Then $σ : ρ (b, s', a)$ is a morphism in $𝕃 0 (S)$ such that

ρ σ = ρ (a, s, b) ρ (b, s', a) = ρ (a, s s', a)

by Proposition 4.3. Since $a s s' = a$ , by (13) we have $ρ (a, s s', a) = ρ (a, 1, a) = 1_{L (a)}$ . Therefore $ρ$ is a split monomorphism. Conversely if $ρ = ρ (a, s, b)$ is a split monomorphism, then there is $σ = ρ (b, s', a) : L (b) \to L (a)$ such that $ρ σ = 1_{L (a)}$ . $(b, s', a)$ is left admissible and the fact that $ρ$ is a split monomorphism gives

ρ σ = ρ (a, s, b) ρ (b, s', a) = ρ (a, s s', a) = ρ (b, 1, b)

and so, $a s s' = a$ . Consequently, the element $s'$ above satisﬁes the requirements.

To prove (c), assume that $ρ = ρ (a, s, b) : L (a) \to L (b)$ is a split epimorphoism so that $σ ρ = 1_{L (b)}$ for some morphism $σ = ρ (b, s', a)$ . Then by the deﬁnition of composition in $𝕃 0 (S)$ $σ ρ = ρ (b, s' s, b)$ and by (13), we have

σ ρ = ρ (b, s' s, b) = ρ (b, 1, b)

This gives $b s' s = b$ . On the other hand, if $s'$ exist satisfying the requirements in (c), then $σ = ρ (b, s', a)$ is a morphism in $𝕃 0 (S)$ from $L (b)$ to $L (a)$ and since $b s' s = b$ , by (13),

σ ρ = ρ (b, s' s, b) ρ (b, 1, b) = 1_{L (b)} .

Hence $ρ$ is a split epimorphism. The statement (d) follows from (b) and (c). $□$

Remark 4.2: Notice that $𝕃 0 (S)$ is a proper subcategory of the category $M o d S$ of all left $S^{1}$ -systems. In general, $𝕃 0 (S)$ may not be a full subcategory of $M o d S$ . Another category which may be of interest in this connection is the full subcategory $𝕃 (S)$ of $M o d S$ whose objects are left ideals of $S$ . We have

𝕃 0 (S) \subseteq 𝕃 (S) \subseteq M o d S .

Suppose that $ρ : L (a) \to L (b)$ is a morphism in $𝕃 (S)$ . If $a$ is a regular element in $S$ there is an idempotent $e ℒ a$ . Let $s = e ρ$ . Then $s = e s \in L (b)$ and $(e, s, b)$ is anadmissible triple. By Equations (12) and (13) $ρ = ρ (e, s, b) = ρ (a, s, b)$ . It follows that the inclusion $𝕃 0 (S) \subseteq 𝕃 (S)$ is fully-faithful if $S$ is regular. Dual remarks hold for the category $ℝ 0 (S)$ of all principal right ideals, the category $ℝ (S)$ of all right ideals and the category $M o d S$ of all right $S^{1}$ -systems.

4. The categories of left and right modules of a semigroup