EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 10, No. 3, 2017, 516-520
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

Module over pseudo-valuation ring and domain

Waheed Ahmad Khan

Department of Mathematics, University of Education Lahore, Pakistan

Abstract. Modules over principal ideal rings, dedekind rings and valuation rings have been dis-
cussed in the literature. In this article, we introduce and discuss the module over pseudo-valuation
ring(PVR) and its submodule. Also as a byproduct we introduce the module over pseudo-valuation
domain(PVD).

2010 Mathematics Subject Classifications: 13A05, 13A18, 12J20

Key Words and Phrases: Module, Submodule, PVR, PVD

1. Introduction and Preliminaries

Number of research articles are available upon modules over different types of rings
and domains, modules over principal ideal rings have been discussed in the literature,
modules over dedekind rings and valuation rings have been introduced by I. Kaplansky
in [6]. Similarly modules over dedekind prime rings introduced by David Eisenbud and J.
C. Robin in [3]. In this connection we introduced module over pseudovaluation ring and
pseudovaluation domain, and also their submodules.

We begin with the basics of module. R-module M over the ring R consists of an abelian
group (M, +) and an operation R × M → M (called scalar multiplication, usually just
written by juxtaposition, i.e. as rx for r ∈ R and x ∈ M) such that for all r, s in R, x, y in
M , we have r(x+y) = rx+ry , (r+s)x = rx+sx, (rs)x = r(sx), 1Rx = x if R has identity
1R. Suppose M is a left R-module and N is a subgroup of M . Then N is a submodule (or
R-submodule, to be more explicit) if for any n ∈ N and any r ∈ R, the product rn is in N
(or nr for a right module). Let R be a ring. let M be a R−module. A subset N of M is
a submodule of M ⇐⇒ (1) N 6= φ. (2) x+ ry ∈ N ∀ r ∈ R and ∀x, y ∈ N [2, proposition
1]. Similarly, it is well known that the left R-module M is said to be finitely generated if
there exist m1,m2, ...,mn ∈ M such that In this case, we say that {m1,m2, ...,mn} is a
set of generators for M . The module M is called cyclic if there exists m ∈ M such that
M = Rm. A free module is a module with a free basis, a linearly independent generating
set. For an R-module M , the set E = {e1, e2, ...en} is a free basis for M such that (1) If E
is a generating set for M ; that is to say, every element of M is a finite sum of elements of

Email addresses: sirwak2003@yahoo.com (W. A. Khan)

http://www.ejpam.com 516 c© 2017 EJPAM All rights reserved.



W. A. Khan / Eur. J. Pure Appl. Math, 10 (3) (2017), 516-520 517

E multiplied by coefficients in R; (2) E is a free set, that is, if r1e1 + r2e2 + ...+ rnen = 0,
then r1 = r2 = ... = rn = 0 (where 0 is the zero element of M and 0 is the zero element
of R).

Let R be an integral domain with quotient field K. A prime ideal P of R is called
strongly prime if x, y ∈ K and xy ∈ P imply that x ∈ P or y ∈ P (alternatively P is
strongly prime if and only if x−1P ⊂ P whenever x ∈ K\R) [5, Definition, page2]. A
domain R is called a pseudo-valuation domain if every prime ideal of R is a strongly prime
[5, Definition, page2]. An integral domain R is a pseudo-valuation domain if and only if
for every nonzero x ∈ K, either x ∈ R or ax−1 ∈ R for every nonunit a ∈ R [5, Theorem
1.5(3)]. Every valuation domain is a pseudo-valuation domain [5, Proposition. 1.1] but
converse is not true for example the valuation domain V of the form K +M , where K is
a field and M is the maximal ideal of V . If F is a proper subfield of K, then R = F +M
is a pseudo-valuation domain which is not a valuation domain. Whereas R and V have
the same quotient field L and that M is the maximal ideal of R [4, Theorem A, page
560]. A quasi-local domain (R,M) is a pseudo-valuation domain if and only if x−1M ⊂M
whenever x ∈ K\R [5, Theorem. 1.4].

Throughout we always mean by a ring R, a commutative ring having unity 1.

2. Module over pseudo-valuation ring

A commutative ring R with 1 is called a pseudo-valuation ring (PV R) if for every a,
b in R, either a divides b or b divides ac for each nonunit c in R.

We begin with the following definition.

Definition 1. If R is a pseudo-valuation ring and M be an additive abelian group then
M is said to be R-module if for all r, s, t ∈ R, where t is a nonunit of R and m,n ∈ M
satisfies:

(1) Either r/s m ∈ M or s/rt m ∈ M , for all r, s, t ∈ R, where t is a nonunit of R
and m ∈M .

(2) (r + s)m = rm+ sm
(3) r(m+ n) = rm+ rn
(4) (rs)m = r(sm)

If one writes the scalar action as fr so that fr(x) = r/sm, and f for the map which
takes each r to its corresponding map fr, then the second axiom states that every fr is
a group homomorphism of M , and the other third and fourth axioms asserts that f is a
ring homomorphism from R to the endomorphism ring End(M).

Remark 1. Following [1, Proposition 3(4)], an integral domain is a PV D ⇔ for every
a, b ∈ R either a/b or b/ac for every non unit c ∈ R. An integral domain is a PV R ⇔ it
is a PV D

Theorem 1. If R is a PV R, and M is an abelian group. then M is a module over
PV R ⇔ If ∀ r, s, t ∈ R where t is a nonunit and m ∈ M, either r/s m ∈ M or s/rt m
∈M.



W. A. Khan / Eur. J. Pure Appl. Math, 10 (3) (2017), 516-520 518

Proof. Let R is a PV R and M is a module then by definition for all r, s, t ∈ R and
m ∈ M, clearly either r/s m ∈ M or s/rt m ∈ M . Conversely suppose M is an abelian
group and R is a PV R, and either r/s m ∈ M or s/rt m ∈ M , where r, s, t ∈ R and m
∈M. We have;

(1) Either (r/s) m ∈ M or s/rt m ∈ M . If s = 1 then rm ∈ M, if s 6= 1 then
r/s ∈ R⇒ R is a PVR and let r/s = r1 ⇒ r1m ∈M .

(2) It is followed by (1) that either (r/s + a/b)m = (r/s m + a/b m) ∈ M or (r/st +
a/bc)m = (r/st m+ a/bc m) ∈M.

(3) Either r/s(m+n) = (r/s m+r/s n) ∈M or r/st (m+n) = (r/stm+r/stn) = M,
which is clear by (2).

(4) Either (r/s ·a/b)m ∈M or (r/st ·a/bc)m ∈M by definition ∀ r ∈ R and either r/s
m ∈M or s/rt m ∈M.

The above theorem reflects that M is a module. Thus we can conclude that the
condition we put in definition1 is necessary and sufficient.

Example 1. Let R = Q+XQ(y)[[X]], R is a PV R. Consider R[X] is an abelian group and
define mapping R×R[X]→ R[X] i.e., (a+ xb(y))r ∈ R[X] where a ∈ Q, b(y) ∈ Q(y)[[X]]
and r = f(x) ∈ R[X]. For all a+ xb(y), c+ xd(y) in R and r, s in R[X] we have;

(1) [(a+ xb(y))(c+ xd(y))]r = a+ xb(y)((c+ xd(y))r).
(2) [(a+ xb(y) + c+ xd(y)]r = [a+ xb(y)]r + [c+ xd(y)]r.
(3) a+ xb(y)(r + s) = [a+ xb(y)]r + [a+ xb(y)]s.

Definition 2. Let R be a pseudo-valuation ring and M be a module over PV R. A subset
N of M is a submodule over pseudo-valuation ring if:

(1) n− ń ∈ N ∀ n, ń ∈ N.
(2) r ∈ R,n ∈ N either r/s n ∈ N or s/rt n ∈ N.

Theorem 2. Let R is a PV R and M is a module. A subset N of M is a submodule over
PV R⇔ ∀ r, s, t ∈ R, n ∈ N either r/s n ∈ N or s/rt n ∈ N.

Proof. If R is a PV R and N is a subset of M , where M is a module then by definition,
we can write that ∀ r ∈ R and n, ń ∈ N, either r/s n ∈ N or s/rt n ∈ N . Conversely,
suppose that N is a submodule over PV R, since we have already define that ∀ r, s, t ∈ R
where t is a non-unit, n, ń ∈ N either r/s n ∈ N or s/rt n ∈ N.Now we can easily conclude
the properties of submodule as;

(1) n− ń ∈ N ∀ n, ń ∈ N (closure property holds).
(2) r ∈ R,n ∈ N either r/s n ∈ N or s/rt n ∈ N (by definition).
which proves that N is a submodule.

Example 2. Let R = Q+XQ(y)[[X]] is a PV R. Let R[X,Y ] is a submodule over pseudo
valuation ring if for all a + xb(y), c + xd(y) in R and p, q in R[X,Y ] where a ∈ Q,
b(y) ∈ Q(y)[[X]] and p = f(x, y), q = g(x, y) ∈ R[X,Y ], we have

(1) p− q ∈ R[X,Y ] ∀ p, q ∈ R[X,Y ].
(2) a+xb(y) ∈ R, p ∈ R[X,Y ], a+xb(y)p ∈ R[X,Y ] and either [a+xb(y)/c+xd(y)]p ∈

R[X,Y ] or [a+ xb(y)/(c+ xd(y).m+ n(y))]p ∈ R[X,Y ] (by definition).



W. A. Khan / Eur. J. Pure Appl. Math, 10 (3) (2017), 516-520 519

3. Module over pseudo-valuation domain

In this section we defined module over pseudo-valuation domain and its submodule.
By [5] let R be a domain with quotient field K. The following statements are equivalent.

(1) R is a pseudo-valuation domain.
(2) For each x ∈ K −R and for each nonunit a of R, we have (x+ a)R = xR.
(3) If module over pseudo-valuation domain R consists of an abelian group (M,+) and

an operation R×M →M scalar multiplication, usually just written by juxtaposition.

Definition 3. Let K be the quotient field of R, x ∈ K −R, a, b, r, s ∈ R and m ∈ M(an
additive abelian group) be a module over R if;

(1) For r ∈ R, m ∈ M and for each nonunit a of R, x−1am ∈M .
(2) (rs)m = r(sm) = (x−1ax−1b)m = x−1a(x−1bm)
(3) (r + s)m = rm+ sm
(4) r(m+ n) = rm+ rn

Example 3. Let R = Z+xQ[[X]] be a pseudo-valuation domain where Q[[X]] is a maximal
ideal and take Q[X] be an additive abelian group. Q[X] is a module over pseudo-valuation
domain, clearly (a+ bx)r ∈ Q[X] where a ∈ Z, b ∈ Q[[X]] and r = f(x) ∈ Q[X], ∀ a+ xb,
c+ xd is in R and r, s in Q[X] we have;

(1) [(a+ xb)(c+ xd)]r = a+ xb((c+ xd)r)
(2) (a+ xb+ c+ xd)r = a+ xbr + c+ xdr
(3) a+ xb(r + s) = a+ xbr + a+ xbs

Theorem 3. If R is pseudo-valuation domain and M is an abelian group then M is
module over R⇔ (a+ xb)m ∈M .

Proof. Assume if M is a module over pseudo-valuation domain R then for each non-unit
element a in R, x−1am ∈M where x ∈ k −R.

Conversely, let x−1am ∈ M , where x ∈ k − R, we have to prove that M is a module
over pseudo-valuation domain R. As x−1am ∈ M =⇒ (x−1ax−1b)m ∈ M , so we may
write (x−1ax−1b)m = x−1a(x−1bm).Also, if x−1am ∈ M then (x−1a + x−1b)m ∈ M =⇒
(x−1a+ x−1b)m = x−1am+ x−1bm.Finally x−1a (m+ n) = x−1am+ x−1an

for all x−1a, x−1b ∈ R and m,n ∈ M where a and b are non-unit elements of R.
As the scaler multiplication of module is given so we can write the other properties of a
module.

Suppose M is R-module and N is a subgroup of M . N is said to be a submodule (or
R-submodule, to be more explicit) if, for any n ∈ N , r ∈ R, the product rn is in N .

Definition 4. Let R be a pseudo-valuation domain and M be a R−module. If N is a
subset of R−module M then N is a submodule of M if;

(1) n− ń ∈ N for all n, ń ∈ N.
(2) For each x ∈ K − R, where K be a quotient field and for each nonunit a of R,

x−1a ∈ R, n ∈ N , x−1an ∈ N.



REFERENCES 520

Example 4. Suppose M = C[X] is a R − module where R = Z + XQ[[X]] is pseudo-
valuation domain and N = R[X] is a subgroup of M . N is a submodule (or R−submodule,
to be more explicit) since for any n = f(x) ∈ N and any a+xb in R, the product n(a+xb)
is in N where a ∈ Z, b = f(x) ∈ Q[[X]].

Theorem 4. If R is pseudo-valuation domain, M is an abelian group and N is a subgroup
of M then N is R− submodule of M ⇐⇒ x−1an ∈ N where x ∈ K −R.

Proof. Assume that N is R−submodule of M therefore N is closed under the action
of ring elements, so we have x−1an ∈ N where x ∈ K − R and a non-unit element a
in R.Conversely, assume x−1an ∈ N , where x ∈ K − R we have to prove that N is
R−submodule of M. As x−1an ∈ N we have;

(1) x−1an ∈ N ∀ x−1a ∈ R, n ∈ N.
Given that N is a subgroup of M so it is closed under subtraction then 2nd condition

of R− submodule is satisfied so;
(2) n− n′ ∈ N ∀ n, n′ ∈ N
Thus N is a R−submodule of M.

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[4] Gilmer, R. Multiplicative Ideal Theory.: Queens Papers on Pure and Applied Mathe-
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[5] Hedstorm, J. R., and Houston, E. G.: Pseudo-valuation domains. Pacific J. Math. 75
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[6] Kaplansky, I.: Modules over Dedekind rings and valuation rings. Trans. Amer. Math.
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