Ann associated structure of a module

              Rev. Acad. Canar. Cienc., XXV, 9-22 (2013) (publicado en marzo de 2015)
AN ASSOCIATED STRUCTURE OF A MODULE
Sandip Jana & Supriyo Mazumder
Department of Pure Mathematics
University of Calcutta
35, Ballygunge Circular Road
Kolkata-700019, India
sjpml2@yahoo.co.in I supriyo88@gmail.com
Abstract
In this paper we have generalised the concept of a module in the sense that
every module can be embedded into this new structure, which we name as 'quasi
module', and every quasi module contains a module. In fact, we have replaced the
group structure of a module by a semigroup structure and invited a partial order
which has a significant role in formulating this new structure; it is this partial order
which is the prime key in relating a quasi module with a module. After discussing
severa! examples we have introduced the concept of order-morphism between two
quasi modules, discussed its various properties and finally proved an isomorphism
theorem regarding this order-morphism.
AMS Classification: 08A99, 13C99, 06F99
Key words : Module; quasi module; order-morphism; order-isomorphism.
1 Introduction
For any topological module M over a topological unitary ring R, the collection 'ef(M)
of all nonempty compact subsets of M is closed under usual addition of two sets and
multiplication of a set by any element of R. Also for any r, s E R and any A, B E 'ef(M)
with A<:;; B we have (r + s)A <:;; rA + sA and rA <:;; rB. Moreover, if e be the additive
identity in M then A - A = {e} iff A is a singleton set. Thus { {m} : m E M} is the
collection of all invertible elements of 'ef(M), {e} acting as the additive identity in 'ef(M) .
These singletons are the minimal elements of 'ef(M) with respect to the usual set-inclusion
as partial order. Now this collection of all minimal elements of 'ef(M) can be identified with
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the module M through the map { m} f-----7 m ( m E M). This makes a useful connection
between the hyperspace éef'(M) and its generating module M. The above facts are not just
a speciality of the hyperspace éef'(M); we have axiomatised these facts and introduced the
concept of a quasi module, as explained below.
In this paper we have generalised the concept of a module in the sense that every
module can be embedded into t his new structure, which we name as 'quasi module', and
every quasi module contains a module. In fact, we have replaced the group structure of a
module by a semigroup structure and invited a part ial order within t his structure which
has a significant role in formulating this new structure; it is this partial order which is the
prime key in relating a quasi module with a module. This partía! order is made compatible
with the semigroup operation and externa! composit ion (which is multiplication by an
unitary ring, in this case), while formulating the axiom for quasi module. A number of
examples have been discussed and it has been shown that every module over an unitary
ring can be embedded into a quasi module and every quasi module contains a module as a
sub-structure.
In section 3 we have introduced the concept of an order-morphism between two quasi
modules over a common unitary ring. Sorne of its propert ies have been discussed. Section 4
deals with the arbitrary product of quasi modules. We have shown that Cartesian product
of any family of quasi modules is again a quasi module. After defining the kernel of an
order-morphism we have proved that kernel of any order-morphism is a quasi module.
In the last section we have discussed an order-isomorphism theorem. For doing this we
have introduced first the concept of congruence in a quasi module and then constructed a
quotient structure which has been finally settled as a quasi module.
2 Quasi Module
Definition 2.1. Let (X,::::;) be a partially ordered set, '+' be a binary operation on X
and '·': R x X ---+ X be another composition [R being a unitary ring] . If the operations
and partía! order satisfy the following axioms then (X, +, ·,::::;) is called a quasi module (in
short qmod) over R .
A1 : (X ,+) is a commutative semigroup with ident ity e.
A2 : x::=;y(x, y E X )=>x+ z :::; y +z, r - x::=;r - y, VzEX,VrER.
A3 : (i) r · (x + y) = r · x + r ·y,
(ii) r· (s · x) = (rs) · x,
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(iii) (r + s) · x ::::; r · x + s · x,
(iv) 1 · x = x, ' l ' being the multiplicative identity of R
(v) O· x = (} and r · (} = (} (r E R)
V'x,y E X, V'r,s E R.
A4 : x + ( - 1) · x = (} if and only if x E X o : = { z E X : y 'Í z, V' y E X "- { z}}
A5 : For each x E X , :J y E X 0 such that y ::::; x.
The elements of the set X0 (which are evidently t he minimal elements of X with respect
to the partial order '::::; ') are called 'one arder' elements of X and, by axiom A4 , these are
the only invertible elements of X, the inverse of x(E X0 ) being (-1) · x, usually written
as '-x'. Also for any x, y E Xo and V'r E R we have, by axiom A4, X - X= e, y - y = (}
=;.. (x +y) - (x +y)=(} and rx - rx = r(x - x) = (} and hence rx, x + y E X 0 . Moreover,
for r, s E R and x E X0 we have (r + s)x::::; rx + sx =;.. (r + s)x = rx + sx (-.· rx + sx is of
order one). Thus we have the following result.
Result 2.2. For any quasi module X over an unitary ring R, the set X0 of all one arder
elements of X is a module over R.
Above result shows that every quasi module contains a module. It is now a routine work
to verify that, for a topological module Mover a topological unitary ring R, the collection
Cef'(M) of ali nonempty compact subsets of M forms a quasi module over R with usual set­inclusion
as partía! order and the relevent operations defined as follows : for A, B E "ef'(M)
and r E R, A+ B :={a+ b: a E A, b E B} and rA := {ra : a E A}. The identity element
of Cef'(M) is {e}, where (}is the ident ity element of M; the set of ali one order elements of
Cef'(M) is given by ["ef'(M)]o = { {m} : m E M}. If we identify { {m} : m E M} with M
we can say that, the topological module M is embedded into the quasi module Cef'(M). We
construct below an example which shows that every module (not necessarily topological)
over an unitary ring can be embedded into a quasi module over the same ring.
Example 2.3. Let M be a module over an unitary ring R. Let M := MU{w} (w tJ_ M).
Define '+ ', '·' and the partial order ''.Sp' as follows :
(i) The operation '+' between any two elements of M is same as in the module M and
x + w := w and w + x := w, Y x E M.
(ii) The operation '.' when applied on R x Mis same as in the module M and r · w := w,
if r( =/= O) E R and O· w := (}, (} being the identity element in M.
(iii) x '.Sp w, V'x E M and x '.Sp x, V'x E M.
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(i) The operation '+' between any two elements of M is same as in the module M and
x + w := w and w + x := w, V x E M.
(ii) The operation ' ·' when applied on R x Mis same as in the module M and r · w := w,
if r(f O) E R and O· w := B, () being the identity element in M.
(iii) x '.Sp w, Vx E M and x '.Sp x, Vx E M.
We show that ( M, +, ·, :SP) is a quasi module over R.
A1 : Clearly ( M, +) is a commutative semigroup with identity () and ''.Sp' is a partial order
inM.
A2 : Let y E M and r ( f O) E R . Then V x E M , x '.Sp w ~ x + y '.SP w + y = w and
r · x = rx '.Sp r · w = w. Also for any x E M, x '.SP x ~ x +y '.SP x +y and r · x '.Sp r · x.
Since O ·y = B, Vy E M so O · x '.SP O · z whenever x '.SP z ( x, z E iJ).
A3 : (i) For r f O and x E M we have r · (x + w) = r · w = w = r · x + r · w and
O· (x + w) =()=O· x + O· w
(ii) If rr' f O then r · (r' · w) = w = (rr') · w, otherwise r · (r' · w) = () = (rr') · w
(iii) If r + r' = O but not both O then (r + r') · w = () '.Sp w = r · w + r' · w
(iv) lR · x = x, V x E M, lR being the multiplicative identity of R.
(v) O· x = B, V x E M
The remaining cases follow immediately from the fact that M is a module over R.
A4 : Here [M] 0 = M. Since w + (- lR) · w = w f () and m + (- lR) · m = m - m = B,
V m E M we have x + ( - lR) · x = () iff x E M = [ MJo.
A5 : For each x E M , x '.Sp x and for w we have m '.Sp w, V m E M
Thus it follows that ( M , +, ·, :SP ) is a quasi module over R, where M is the set of ali
one order elements of M.
This example shows that every module is contained in a quasi module.
In this example if we consider M = <C, the vector space of ali complex numbers as a
module over the unitary ring íZ then the extended complex plane <C00 :=<CU { oo} becomes
a quasi module over Z, provided we define O.oo = O and z < oo, V z E <C.
Example 2.4. Let íZ be the ring of integers and z+ := {n E íZ : n ~ O} . Then under the
usual addition, z+ is a commutative semigroup with the identity O. Also it is a partially
ordered set with respect to the usual order (:S) of integers. If we define the ring multipli­cation
'·' : ÍZ X z+ --+ z+ by (m, n) >----+ lmln, then it is a routine WOrk to verify that
(z+, +, ·, :S) is a quasi module over Z. Here the set of ali one order elements is given by
[z+]o = {O}.
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Example 2.5. Let z+[x] be the set of all polynomials with coefficients taken from z+ :=
{ n E íZ. : n ?'. O}. Then with respect to the usual addition ( +) of polynomials it is a
commutative semigroup with the identity, viz. 'zero polynomial' O(x). Let us define the
ringmultiplication '·': íZ.xíZ.+[x]-----+ z+[x] by (m,ao+a1x+·· ·+anxn) >------+ lml(ao+a1x+
... + anxn). Again if f(x), g(x) E z+[x], where f(x) := ªº + a¡X + ... + anXn (an =!=O) and
g(x) := b0 + b1x + · · · + bmxm (bm =/=O) then we define J(x) ~ g(x) if degf(x) S degg(x)
and a; s b;, Vi = O, 1, ... , n ( = deg f (x)). Then clearly '~' is a partial order in z+[x]. It
now follows that (Z+[x], +, ·, ~) is a quasi module over the unitary ring íZ.; the set of all
one order elements of z+[x] is given by [z+[xJ] 0 = { O(x) }.
Example 2.6. Let Q+ := { ~ : p, q E z+, q =/= O, gcd(p, q) = 1} i.e. the set of all non­negative
rational numbers. Then with respect to the usual addition of rationals, Q+ be­comes
a commutative semigroup with zero (O) as the identity element. We define the ring
multiplication '8' by elements of the ring íZ. by, (r, ~) >------+ lrl~ (r E íZ.). The partial order
'S' on Q+ is defined as ~ S ;¡;- <* p1 S p2 and q1 = Q2. We now show that under these
operations and partial order (Q+, +, 8, s) becomes a qmod over the unitary ring íZ.. First
of all, it is clear that 'S' is truly a partial order on Q+. We are only to show the following
to establish this example of qmod.
A2 : If x, y E Q+ with x S y then for any z E Q+ we have z + x S z +y and for any r E íZ.
we have lrlx S lrly i. e. r 8 x S r 8 y.
A3 : (i) r 8 (x +y)= lrl(x +y)= lrlx + lrly = r 8 x + r 8 y, Vr E íZ., V x, y E Q+.
(ii) ri 8 (r2 8x) = lr1lhlx = hr2lx = (r1r2) 8 x, Vr1 , r2 E íZ., Vx E Q+.
(iii) (r¡ + r2) 8x = lr1 + r21x S lr1lx + hlx = ri 8 x + r2 8 x, Vr1,r2 E íZ., Vx E Q+.
(iv) 18 X= X, Vx E Q+.
(v) O 8 x =O, Vx E Q+ and r 8 O= O, Vr E íZ..
A4 : [Q+]o := { ~ E Q+ : ~ $ ~'V~ E Q+" {~}} = {O}. Again, 1 0 ~ + (-1) 0 ~ = O <*
E + E = Ü <* 2E = Ü <* E = Ü. q q q q
A5 : For each ~ E Q+ we ha ve O S ~.
Thus (Q+, +, 0, s) is a qmod over íZ..
Example 2. 7. Let {p1 , p2, ... } be a complete enumeration of all primes in order i.e. 2 =
p1 < p2 < · · ·. Now any integer m ?'. 1 can be expressed uniquely as m = pf1 p~2 .. . pf' .. .,
where a; E z+ := {n E íZ.: n ?'. O}, Vi and al! but finitely many a;'s are zero. Thus we
can identify the integer m with the sequence ( a 1, a 2, ... ) in z+. In other words, we can
say that m can be identified with an element of (z+r whose all but finitely many terms
are zero and for convenience Jet us denote this set as (íZ.+)~0 i.e.
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(z+)~o := { ( ai , 0'.2, . . . ) : O'.; E z+, O'.; = O far ali but finitely many i's}
Let us first introduce sorne notations : if a := ( ai , a 2, ... ) E (Z+)~0 and p := (Pi, p2 , . .. )
be the sequence of all primes in strictly increasing order, as stated above, we denote
pª := pf1 p~2 . . . which is valid since ali but finitely many factors in this infinite product are
l. Also if a, (3 E (Z+ )~0 then pªrJ3 = po:+f3, where a+(3 is the sequence in (z+)~0 obtained by
term by term addition of a and (3 . Again if r is any non-negative integer then (pª)" = pro:.
Thus the usual product of two integers (2: 1) can be viewed as the sum of two elements in
(Z+)~0 ; also far any integer m (2: 1) and r E z+ the exponent operation mr can be viewed
as the operation ra := (rai , ra2, ... ), where m := pª and a := (ai, a2, . . . ) E (Z+ )~0 .
These facts now culminate into the fallowing example of quasi module.
(z+)~0 is a commutative semigroup with respect to the usual term by term addition of
two sequences; it contains an identity element, namely zero sequence 'O' (i.e. the sequence
ali of whose terms are zero). With the help of the unitary ring Z, we define a ring multipli­cation
'-': Z x (Z+ )~0 ~ (z+ ) ~0 as (r, a) f---t lr la. We now define an order '~' by a ~ (3
iff pª :::; r}3. It is obvious that ~ is a partial order in (z+)~0 . We show below that (Z+ )~0
is a qmod over Z with respect to the afaresaid operations and partial order.
A 2 : Let a , (3, ¡ E (Z+ )~0 with a ~ (3. Then pª :::; rJ3 =? p1 pª :::; p1 pf3 =? p-r+o: :::; p-r+f3
=?¡+a~ ¡+(3. Again far any r E Z we have (p°')lrl:::; (r}3)1rl =? pirla::::; plrl(3 =? r·O'. ~ r ·f3.
A 3 : Let a , (3 E (Z+)~0 and n, ni, n2 E Z. Also !et a:= (a;) i.E N , (3 := (!3;).iE N . Then
(i) n ·(a+ (3) = (lnl(a; + (3;))iEN= ( ln la;) iEN + (lnlf3;) iEN = n ·a + n · (3.
(ii) ni· (n2 ·a)= ni · (ln2 la;) i.E N = ( lnilln2la;) i.E N = ( lnin2la;) i.E N = (nin2) · a .
(iii) (ni+ n2) · a = (lni + n2la;) iEN. Now lni + n2 la; :::; lnila; + ln2 la;, Vi E N. So
p\n1+n2ia,:::; p\n1lo:•p¡n210:', Vi EN=? pln1+n2lo::::; pln1lo:+ln2lo: =?(ni+ n2). a~ ni. a + n2 . a.
(iv) 1 · a = (ll la;)iEN = O'..
(v) O· a= (IOla;)iEN = O and n ·O = O.
A4 : Since any a E (Z+)~0 corresponds to an integer ::::: 1 it fallows that, zero sequence 'O'
is the only one order element of (Z+)~0 . Again 1 · a+ (-1) ·a = O <* (a;+ a;).iE N = O
<* 2a; = O, Vi <* a = O. So axiom A4 fallows.
A5 : For each a E (Z+ )~0 , since pª ::::: 1 = pº we have O ~a.
Thus it fallows that ( (Z+)~0 , + ,., ~ ) is a quasi module over Z.
3 Order morphism
In this section we introduce a morphism-like structure between two quasi modules over a
common unitary ring and study sorne of its properties.
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Definition 3.1. A mapping f : X ----+ Y (X, Y being two quasi modules overa unitary
ring R) is called an order-morphism if
(i) f (X + y) = f (X) + f (y)' V X' y E X
(ii) f(rx) = rf(x), 't:/r E R, 't:/x E X
(iii) x S y (x, y E X)=? f(x) S f(y)
(iv) p s q (p, q E f(X)) =? ¡- 1(p) <;;;.J,. ¡- 1(q) and ¡- 1(q) <;;;t ¡- 1(p) , where
t A := {x E X : x ~ a for sorne a E A} and -i A := {x E X : x ::; a for sorne a E A} for
any A<;;; X.
A surjective (injective, bijective) order-morphism is called an order-epimorphism ( order­monomorphism,
order-isomorphism).
Note 3.2. If f : X ----+ Y be an order-morphism and B, ()' be the identity elements of
X, Y respectively then f(B) = f(O.B) = O.f(B) = ()'. Again if x0 be an one order element
of X then so is f (xo) of Y. In fact, Xo E Xo =? xo - xo = () =? f( xo) - f(xo) = ()'
=? f(x0 ) E Y0 . Also if Yo E Yo n f(X) then :lx E ¡-1(yo) =? :lxo E Xo such that
xo S x =? f(xo) S f(x) = Yo =? f(xo) = Yo [·: Yo is an one order element of Y]. Thus
¡ - 1 (yo) n Xo =f. 0.
Before proceeding further let us first introduce the following concept which will be useful
in the seque!.
Definition 3.3. A subset Y of a qmod X is said to be a sub quasi module (subqmod in
short) if Y itself be a quasi module with al! the compositions of X being restricted to Y.
Note 3.4. A subset Y of a qmod X (overa unitary ring R) is a sub quasi module iff Y
satisfies the following conditions :
(i) rx + sy E Y , 't:/r,s E R, 't:/x,y E Y .
( ii) Yo <;;; X o n Y , w her e Yo : = { z E Y : y '/:. z, V y E Y "- { z} }
(iii) 't:/y E Y, :3y0 E Yo such that Yo s y
If Y be a subqmod of X then actually Y0 = X0 n Y , since for any Y <;;; X we have
Xo n Y <;;; Yo.
Proposition 3.5. If f : X ----+ Y (X, Y being two quasi modules over a unitary ring R)
be an order-morphism then f(M) := {f(m): m E M} is a subqmod of Y , for any subqmod
M of X.
Proof. For x, y E M and r, s E R we have r f(x) + sf(y) = f(rx + sy) E J(M), since
rx+sy E M for, Mis a subqmod of X. Clearly, f(M) nY0 <;;; [f(M)]0 . Now !et y E [f(M)]o
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=} :3m E M such that y= f(m). So :3p E Ma such that p :::'.: m =} f(p) :::'.: f (m) =} f(p) =
f(m) =y [·:y is an one order elernent of f (M) and f(p) E f(M)]. Since Ma = M n Xa
so p E Xa =} f(p) is an one order elernent of Y and hence y = f(p) E Yo n f(M). Thus
[f(M)]a <;;; f (M) n Ya. Therefore [f(M)]a = f(M) n Ya . Again for any m E M, :3 ma E Ma
such that ma :::'.: m =} f(ma) :::'.: f(m). Here f(ma) is an one order elernent of Y and hence
f(ma) E [f(M)]a. Thus it follows that f(M) is a subqrnod of Y. D
Proposition 3.6. Let X , Y, Z be three qmods over an unitary ring R and f : X -----+ Y ,
g : Y -----+ Z be two order-morphisms. Then their composition g o f : X -----+ Z is an
order-morphism, provided f is onto.
Proof. (g o f)(rx1 + x2) = g(f(rx1 + x2)) = g(r f(x1) + f (x2)) = r.(g o f)(x1 ) + (g o f)(x2) ,
Vx1 , x2 E X and Vr E R. Moreover, x1 :::'.: x2 (x1,x2 E X)=} f( x1) :::'.: f(x2) =} g(f(x1)) :::'.:
g(f(x2 )) =;. (g o !)(xi):::'.: (g o f)(x2).
Now let Z¡, Z2 E (g o !)(X) such that Z¡ ::::: Z2 . Let X E (g o n-1(z¡). Then (g o f)(x) =
z1 =} g(f(x)) = z1 =} f(x) E g- 1(z1) <;;;_J.. g- 1(z2) =} f (x) :::'.: y, for sorne y E g- 1(z2)
=} g(y) = z2 . Now f being onto, y E f(X) and hence x E-!. ¡ - 1(y) =} x :::; x', where
f(x') =y. Therefore (g o f)( x') = g(f(x')) = g(y) = Z2 =}X E_j,. (g o f) - 1 (z2) .
... (g o n - 1(z1) <;;;-!. (g o n - 1(z2).
Again Xa E (gof)- 1(z2) =} (gof)(xa) = z2 =} g(f(xa)) = z2 =} f(xa) E g- 1(z2) <;;;t g- 1(z1)
=} f( xa) 2 y', for sorne y' E g- 1 (z1). So g(y') = z1 . Now f being onto, y' E f(X ) and hence
xa Et ¡ -1 (y1) =} xa 2 x", where f( x") = y'. Therefore (g o f)(x") = g(f(x")) = g(y') = z1
=} Xa Et (g o f) - 1(z1) .
... (g o n - l(z2) <;;;t (g o n - l(z¡).
Thus it follows that (g o!) is an order-rnorphisrn. D
Proposition 3.7. Jf f: X -----+ Y, g : Y -----+ Z (X, Y, Z being qmods over the same unitary
ring R) be two order-morphisms such that g o f : X -----+ Z is also an order-morphism then
(i) g o f is onto iff both f , g are onto;
(ii) g o f is injective iff both f , g are injective.
Proof. First of all, g o f is an order-rnorphisrn provided f is onto. So (i) is irnrnediate. For
(ii), we are only to show that g is injective whenever g o f is injective. Actually g o f is
injective irnplies gis injective on f(X); in fact, if g(f(x 1)) = g(f(x2)) for f(x 1) 1- f(x2 )
(and hence for x1 1- x2 ) then injectivity of g o f would be contradicted. Since for g o f to
be an order-rnorphisrn f needs to be onto, (ii) follows. D
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Remark 3.8. The above proposition readily implies that two order-isomorphisms, after
composition, ge11erates again an order-isomorphism; i11verse of a11 order-isomorphism is
a11 order-isomorphism a11d the ide11tity map 011 any qmod is an order-isomorphism. Thus
'order-isomorphism' induces an equivalence relation 011 the collection of all qmods over the
same unitary ring and we can identify two such qmods related by this equivalence relation.
Definition 3.9. Let f : X ----+ Y (X, Y being two qmods over the same unitary ring R)
be an order-morphism. We define ker f := { (x, y) E X x X: f(x) = J(y)} and cal! it the
'kernel off'.
It is immediate from definitio11 that (x,x) E ker f, Vx E X and thus if we write
6 := { (x, x) : x E X} the11 6 <;;; ker f, equality holds iff f is injective.
We 110w show that ker J is a subqmod of X x X, but for doing so we have to first discuss
the Cartesian product of qmods.
4 Arbitrary product of quasi modules
In this section we shall discuss arbitrary product of quasi modules and show that the
product is also a quasi module.
Definition 4.1. Let {Xμ : μ E J\} be an arbitrary family of quasi modules over the
unitary ring R. Let X := IJ Xμ be the Cartesian product of these quasi modules defined
μ E/\.
as : x E X if and only if x : J\ ----+ LJ Xμ is a map such that x(μ) E Xμ , Vμ E J\. Then
μE/\.
by the axiom of choice we know that X is nonempty, since J\ is nonempty and each Xμ
contains at least the additive ide11tity Bμ (say) .
Let us denote Xμ := x(μ) , Vμ E J\. Also we write each x E X as x = (xμ), where
Xμ= Pμ(x), Pμ : X ----+Xμ being the projection map, Vμ E J\. Now we define addition,
ring multiplication and partial order as follows : for x = (xμ), y= (yμ) E X and r E R
(i) X+ y= (x¡.i +Yμ); (ii) r.x = (rxμ); (iii) X::; y if Xμ::; Yμ, Vμ E J\.
We now show that (X,+,., ::;) is a quasi module over R.
A1 : Clearly X is a commutative semigroup with identity fJ, where fJ = (Bμ)·
A2 : x ::; y => Xμ ::; Yμ , Vμ E J\ => Xμ + zμ ::; Yμ + zμ and rxμ ::; ryμ, Vμ E J\ and V r E R
=> x + z::; y+ z and rx::; ry, where z = (zμ) E X.
A3 : For x = (x1,), y = (yμ) E X and for r, s E R we have
(i) r(x +y) = (r(xμ +Yμ)) = (rxμ + ryμ) = (rxμ) + (ryμ) = rx + ry.
(ii) r(sx) = r(sxμ) = rs(xμ) = rsx.
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(iii) each Xμ being a quasi module, (r + s)xμ:::; rxμ + sxμ, \::/μE A => (r + s)x:::; rx + sx.
(iv) l.x = (l.xμ) = (xμ) = x, 1 being the multiplicative identity of R.
(v) O.x = (0.xμ) = (Bμ) =B. Again r.B = (r.Bμ) = (Bμ) =B.
A4 : x - x = () {o} Xμ - Xμ = Bμ, \::/μ E A {o} Xμ E [Xμ]o, \::/μ E A, where [Xμ]o is the set
of ali one arder elements of Xμ- We claim that X0 = { (xμ) E X : Xμ E [Xμ]o, \::/μE A} =
II [Xμ]o. In fact, x = (xμ) ~ II [Xμ]o =>X.>-~ [X.>-]o far sorne A E A. Then :3i.>- E [X.>-]o
μEA μEA
such that i.>- :::; X.>,, i.>- f= X.>,. Let y = (yμ) where Yμ = Xμ, μ f= .\ and Y>- = i.>,. Then
y :::; x, y f= x => x ~ X0 . Conversely if Xμ E [Xμ]o , \::/μ E A then x = (xμ) E X0 . Thus
Xo = rr [Xμ]o. So X - X = () {o} X E Xo.
μEA
A5 : Let x = (xμ) E X. Then Xμ E Xμ, \::/μ E A => :3 tμ E [Xμ]o, \::/μ E A such that tμ :::; xμ,
\::/μE A=> t = (tμ) :<::: (xμ) = x, where t E X0.
:. (X,+,.,:<:::) is a quasi module over R.
Proposition 4.2. Let {Xi: i E A} be an arbitrary family of quasi modules over an unitary
ring R and X := Il Xi be the product qmod oj these qmods. Then each projection map
iEA
Pj : X --+ Xj is an order-epimorphism.
Proof. Let x =(xi), y= (yi) E X and r E R. Then Pj(rx +y)= rxj + yj = rpj(x) + Pj(y) .
Again if x :<:::y then xi:<::: Yi, Vi E A=> Pj(x) :<::: Pj(y).
Now !et a,b E pj(X) = Xj (since every projection map is onto) with a :::; b. Let
x =(xi) E p~¡1(a). Then pj(x) =a i.e. Xj =a. We choose y= (yi) where Yi =Xi far i f= j
and yj = b. Then x :<:::y and pj(y) = yj = b i.e. y E pj1(b). Thus we have pj1(a) <;;;;_J.. pj1(b).
Similarly we can show that pj1(b) <;;;;t pj1(a). So pj being onto the proposition fallows. O
Proposition 4.3. Let f : X --+Y be an order-morphism (X, Y being two qmods over an
unitary ring R). Then ker f is a subqmod of X x X.
Proof. Far (x1 , y1) , (x2, y2) E ker f and r, s E R we have f(rx1 + sx2) = r f(x1) + sf(x2) =
r f(yi)+sf(y2) = f(ry1 +sy2) => (rx1 +sx2, ry1 +sy2) E ker f i.e. r(x1, Y1)+s(x2, Y2) E ker f.
Now !et (x, y) E ker f but (x, y) ~ X0 x X0 . Without loss of generality assume that
x ~ X0 . Then ::Ja E X0 such that a :<::: x. So f(a) :<::: f(x) = f(y). Now f being an
order-morphism, :3 z E ¡-1(J(a)) such that z:::; y. Then :3t E X0 such that t :<::: z => f(t) :<:::
f( z) = j(a). Now a being one arder, f(a) is so and hence J(t) = f(a) => (a, t) E ker f.
Also (a, t) :::; (x , y) and (a, t) f= (x, y) . This ensures that (x, y) ~ [ker f] 0 . Thus we have
[ker f]o <;;;; ker f n (X0 x X0). Also from this argument we find that far any (x, y) E ker f ,
:3 (a, t) E ker f n (X0 x X0) = [ker f]o such that (a, t) :::; (x, y) (if (x, y) itself be of arder
one we need not find (a, t)). The proposition then fallows from the note 3.4. O
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5 Order isomorphism theorem
In this section we present an isomorphism theorem between quasi modules.
Lemma 5.1. Let X, Y, Z be three quasi modules over the unitary ring R, a : X --t Y
be an order-epimorphism and (3 : X --t Z be an order-morphism such that ker a i;::: ker (3.
Then 3 a unique order-morphism ¡ : Y --t Z such that ¡ o a = (3.
y
Proof. We first show that if an order-morphism ¡:Y --t Z exists satisfying ¡ oa = (3 then
that must be unique. In fact, if ¡ ' be another such order-morphism then ¡ o a = (3 = ¡'o a.
This shows that ¡ , ¡ ' coincide on a(X). a being onto, ¡, ¡' really coincide on Y.
To prove the existence !et y E Y. a being order-epimorphism, a - 1 (y) =!= 0. Now
kera i;::: ker (J ==? (3 is constant on a - 1(y). So it is reasonable to define ¡ (y):= (3(a - 1(y)),
Vy E Y. Clearly then ¡ o a= (3. Now let y, y' E Y, r E R and x E a - 1(y), x' E a - 1(y').
Then ¡ (y) = (J(x) and ¡ (y')= (J (x') . Now a being an order-morphism, we have y+ ry' =
a(x) + ra(x') = a(x + rx') ==? x + rx' E a - 1(y + ry'). Then (3 being an order-morphism we
have ¡ (y)+ r¡(y') = (J (x ) + r (J (x') = (J(x + rx') =¡(y + ry').
Next let y :::; y' (y, y' E Y). Then a-1(y) i;:::..¡. a-1 (y'). So for x E a - 1(y) , 3 x' E a-1(y')
such that x:::; x'. Thus ¡(y) = (J(x) :::; (J(x') =¡(y').
For declaring ¡ to be an order-morphism it now remains to show that ,-1 (z) i;:::..¡. ¡ - 1 (z')
and , - 1 (z') i;:::t , - 1 (z), whenever z :::; z' [z,z' E ¡ (Y)]. To prove the first inclusion !et
y E , - 1 (z). Then a - 1(y) i;::: (3 - 1 (z ) i;:::..¡. (3 - 1 (z'). So for x E a - 1(y) , 3 x' E (3- 1(z') such
that x :::; x'. Then y = a(x) :::; a(x') = y' (say). Now ¡(y') = ¡ (a(x')) = (J(x') = z'
==?y' E , - 1 (z'), where y:::; y'. The second inclusion can be similarly disposed of. O
Lemma 5.2. Let X , Y , Z be three quasi modules over the unitary ring R , a: Y --t X be
an order-monomorphism and (3 : Z --t X be an order-morphism such that a(Y) = (J(Z).
Then 3 a unique order-epimorphism ¡ : Z --t Y such that the following diagram commutes.
z
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Proof. o: being an order-monomorphism it follows that o: is an order-isomorphism from Y
onto the sub quasi module o:(Y) ( = f3(Z)) of X. Thus we may define ¡ := 0:-1 o (3. It
then follows from the remark 3.8 and proposition 3.6 that ¡ is an order-morphism, since f3
is 'onto' the domain of 0:- 1 . Since o:- 1 (/3(Z)) = Y , so¡ is surjective. Also o: o¡= f3 i.e.
the above diagram is commutative.
If, together with ¡, any other ¡ ' makes the above diagram commutative then, o: o¡ =
f3 =o: o¡' ==? ¡ = ¡' (since o: is an order-isomorphism). D
Before going to prove the isomorphism theorem we need to construct a quotient struc­ture
which again necessitates the introduction of the concept of 'congruence'. So let us
define this concept first.
Definition 5.3. An equivalence relation E , defined on a quasi module X over an unitary
ring R, is said to be a congruence on X if,
(i) (x, y) E E=} (a+ x, a+ y) E E, V a E X
(ii) (x, y) E E=} (rx, ry) E E, Vr E R
(iii) x :S: y :S: z and (x, z) E E=} (x, y) E E (and hence (y, z) E E)
(iv) a :S: x :S: b and (x,y) E E=} :lc,d E X with e :S: y :S: d such that (a,c) , (b ,d) E E.
Proposition 5.4. Jf <P : X ---+ Y (X, Y being two qmods over an unitary ring R) be an
order-morphism then ker <P is a congruence on X.
Proof. Clearly ker <P is an equivalence relation on X. Let (x, y) E ker </J, a E X and r E R.
Then <P(a+x) = <P(a) +<P(x) = <P(a) +</J(y) = <P(a+y) and <P(rx) = r<P(x) = r</J(y) = <P(ry).
Thus (a+x,a+y),(rx,ry) E ker</J. Again x :S: y :S: z =} <P(x) :S: <P(y) :S: <P(z). So
(x,z) E ker </J =} <P(x) = <P(z) = <P(y) =} (x,y) E ker</J.
Next let b 2': x. Then <P(b) 2': <P(x) = </J(y) =}y E </J-1 (</J(y)) <;;;;_J.. <P- 1 (</J(b)) =} :ld E
<P-1 (</J(b)) such that y :S: d. Now <P(d) = <P(b) =} (b,d) E ker </J. Similarly for a :S: x we have
<P(a) :S: <P(x) = <jJ(y). So y E </J-1 (</J(y)) <;;;;t <P-1 (</J(a)) =}:le E <P-1 (</J(a)) such that y 2': c.
Now <P(c) = <P(a) =} (a , c) E ker</J. Thus ker</J is a congruence on X. D
We now give a quotient structure on X using the above congruence. For this !et us
construct the quotient set X/ ker <P := { [x] : x E X}, where [x] is the equivalence class
containing x obtained by the congruence ker </J. We define addition, ring multiplication and
partial order on X/ ker <P as follows : For x, y E X and r E R,
(i) [x] +[y] := [x +y]; (ii) r[x] := [rx]; (iii) [x] :S: [y] if and only if <P(x) :S: <P(y).
Theorem 5.5. Jf <P : X ---+ Y (X, Y being two qmods over an unitary ring R) be an
order-morphism then X/ ker <P is a quasi module over R.
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Prooj. A1 : Clearly (X/ ker ef¡, +) is a commutative semigroup with identity [B], where B is
the identity of X.
A2: Let [x], [y], [z] E X / ker ef¡ and [x] :S [y]. Now, efi(x + z) = efi(x) + efi(z) :S efi(y) + efi(z) =
efi(y + z ) =? [x + z] :S [y+ z] =? [x] + [z] :S [y]+ [z]. Also, efi(rx ) = refi(x) :S refi(y) = efi(ry) ,
'fr E R =? [rx] :S [ry] =? r[x] :S r[y], V r E R.
A3 : (i) r([x] + [y]) = r[x + y] = [rx + ry] = [rx] + [ry] = r[x] + r[y]
(ii) r(s[x]) = r [sx] = [rsx] = rs[x], where r, s E R
(iii) (r + s)x :S rx + sx =? efi((r + s)x) :S efi(rx) + efi(sx) =? (r + s)[x] = [(r + s)x] :S
[rx] + [sx] = r[x] + s[x]
(iv) lR[x] = [x]
(v) O[x] = [Ox] = [B] and r[B] = [rB] = [B], Vr E R
A4 : [x] + (-l)[x] = [B] <* [x] + [-x] = [B] <* [x - x] = [B] <* efi(x - x) = ef¡(B) <*
efi(x) - efi(x) = B' (where B' is the identity in Y) <=? efi(x) E Y0 . Now the set of ali one order
elements of X/ ker ef¡ is given by [X/ ker efi] 0 := {[x] E X/ keref¡: [y] i [x], V[y] =/= [xJ} =
{[x] : efi(y) i efi(x),Vefi(y) =/= efi(x)} = {[x]: efi(x) E [efi(X)]o = ef¡(X)n Yo} [·: efi(X) is a
subqmod of Y]. Thus we have [x] + (- l) [x] = [B] if and only if [x] E [X/ kerefiJo.
A5 : Let [x] E X/ ker ef¡. Then :3p E X0 such that p :S x =? efi(p) :S efi(x) =? [p] :S [x]. Here
p being an one order element of X , efi(p) is so in Y and hence [p] E [X/ ker ef¡ Jo. D
Proposition 5.6. Let ef¡ : X -----+ Y (X, Y being two qmods over an unitary ring R ) be an
order-morphism. Then the canonical map 7r : X -----+ X/ ker ef¡ defined by 7r(x) := [x],
V x E X is an order-epimorphism.
Proof. Since ef¡ is an order-morphism it follows immediately that 7r satisfies the first three
axioms of an order-morphism. Also 7r is an onto map. So we are only to show that
7r- 1 (¡xJ) ~.j.. 7r- 1 (¡y]) and 7r- 1 (¡y]) ~t 7r- 1 (¡x] ) , whenever [x] :S [y] in X/ ker ef¡. For this
let a E 7r- 1 ([xJ). Then [a] = 7r(a) = [x] =? efi(a) = efi(x) :S efi(y). Now a E ef¡-1 (ef¡(a)) ~.j..
ef¡-1 (ef¡(y)) =? :3 b E ef¡- 1(ef¡(y)) such that a :S b. Again efi(b) = efi(y) =? 7r(b) = [b] =[y] . Thus
we have a E.j.. 7r- 1 (¡y]) i.e. 7r - 1 (¡x]) ~.j.. 7r - 1 (¡y]) , whenever [x] :S [y] in X/ ker ef¡. Similarly
7r - 1 (¡yJ) ~t 7r - 1 (¡x] ) . D
We now have the following isomorphism theorem.
Theorem 5.7. Jf ef¡ : X -----+ Y (X, Y being two qmods over an unitary ring R) be an
order-morphism then X / ker ef¡ is order-isomorphic to efi(X) .
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X/ ker </J
Proof. ken := {(x,y): 7r(x) = 7r(y)} = {(x, y): [x] = [y]}= {(x,y): </J(x) = </J(y)} =
ker </J. Since </J : X ---+ </J(X) <:;::: Y is an order-morphism and 7r : X ---+ X / ker </J is
an order-epimorphism (by proposition 5.6) , by lemma 5.1 we can fiad a unique order­morphism
'!/; : X / ker </J---+ </J(X) such that '!/;o 7r =</J. Now </J : X ---+ </J(X) is onto implies
'!/; is onto. Again 1/;[x] = 1/;[y] =? 1/;(7r(x )) = 1/;(7r(y)) =? </J(x) = </J(y) =? (x, y) E ker</J
=? [x] = [y], where [x], [y] E X / ker </J. Therefore '!/; is injective and hence bijective. D
Acknowledgement : The second author is thankful to UGC, INDIA for financia!
assistance.
References
[1] T. S. Blyth; Module theory : an approach to linear algebra; Oxford University Press,
USA (1977)
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