J. Comp. & Math. Sci. Vol.2 (5), 719-724 (2011)
Cyclic P- Fuzzy Modules over Dedekind domains SOURIAR SEBASTIAN1 and GEORGE MATHEW2 1
Department of Mathematics, St. Albert’s College Ernakulam, Kochi- 682018, Kerala, India 2 Department of Mathematics, B.C.M. College, Kottayam-686001, Kerala, India ABSTRACT
We have introduced the notion of P-fuzzy ideals of a commutative ring R where P is a prime ideal of R and discussed it’s various properties in15. Extending this notion to modules, we have introduced and studied P-fuzzy modules in16. In this paper we confine our study to P- fuzzy modules of a cyclic module M = Rz over a Dedekind domain R. We prove that a fuzzy submodule µ of M is a P- fuzzy R- submodule ⟺ every non-zero level set of µ is of the form Prz for some non-negative integer r whether or not M is torsion free. We also prove that if and µ and ν are P- fuzzy R- submodules of M, then µ + ν is a P- fuzzy R- submodule of M. Keywords: Prime ideals, localization, prime factorization, fuzzy ideals, P- fuzzy ideals, fuzzy modules, P-fuzzy modules.
1. INTRODUCTION Zadeh17 defined a fuzzy subset µ of a given universal set X as a function µ: X ⟶[0,1]. Rosenfeld8 used this idea to introduce the notion of fuzzy groupoids and fuzzy groups and derived their elementary properties. Liu5-6 defined and studied fuzzy ideals in groups and rings. The concept of fuzzy modules was introduced by Negoita and Ralescue7 and has been further developed by Golan4 and Bambri and Kumar1-2. Some works done in this area by the authors are reported in9-16.
We have introduced the notion of Pfuzzy ideals of a commutative ring R where P is a prime ideal of R and discussed it’s various properties in15. Extending this notion to modules, we have introduced and studied P-fuzzy modules in16. In this paper we confine our study to P- fuzzy modules of a cyclic module M = Rz over a Dedekind domain R. We prove that a fuzzy submodule µ of M is a P- fuzzy R- submodule ⟺ every non-zero level set of µ is of the form Prz for some non-negative integer r, whether or not M is torsion free. We also prove that if and µ and ν are P- fuzzy R- submodules of M, then
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Souriar Sebastian, et al., J. Comp. & Math. Sci. Vol.2 (5), 719-724 (2011)
µ + ν is a P- fuzzy R- submodule of M. 2. PRELIMINARIES Throughout this paper, R denotes a commutative ring with unit element, unless otherwise stated. Algebraic terms and notations used here are either standard or as in3 and18. Given a universal set X, a function μ :X ⟶ [0, 1] is called a fuzzy subset of X. For a fuzzy subset µ of R and t ∈ [0,1] the set µt = {x ∈ R : µ(x) ≥ t} is called the level set of µ corresponding to t. We denote the set { x ∈ R : µ(x) = µ(0) } by µ∗. It may be recalled that a fuzzy subset µ of R is called a fuzzy ideal2 of R if (i) µ (x + y) ≥ µ(x) ∧ μ(y), (ii) µ(-x) = µ(x) and (iii) µ (xy) ≥ µ(x) ∨ µ(y) for all x, y ∈ R. If M is an R- module, then a fuzzy subset µ : M ⟶ [0, 1] is called a fuzzy R- submodule4 of M if (i) µ(x + y) ≥ µ(x) ∧ µ(y) ∀ x, y ∈ M (ii) µ(rx) ≥ µ(x) ∀ x ∈ M and r ∈ R and (iii) µ(0) = 1. Here ∧ and ∨ denote the min (or inf) and max (or sup) operators respectively. Also recall that if P is a prime ideal of R, then the localization RP of R at P is the ring of quotients S-1R where S = R – P. 2.1. Definition15. Let R be a ring and P be a prime ideal of R. Then a fuzzy ideal µ of R is said to be a P – fuzzy ideal if it satisfies any of the following two equivalent conditions. (i) r1/s1, r2/s2 ∈ RP and r1/s1 = r2/s2 ⟹ µ(r1) = µ(r2). (ii) r ∈ R and s ∈ R – P ⟹ µ(rs) = µ(r). 3. CYCLIC P-FUZZY MODULES OVER DEDEKIND DOMAINS Recall that an integral domain R is a Dedekind domain if R satisfies any of the following equivalent conditions.
(i)
Every non-zero proper ideal is a product of prime ideals (ii) If A and B are any two fractional ideals of R, then B ⊆ A ⟹ B = CA for some integral ideal C of R (iii) R is Noetherian, integrally closed and every prime ideal of R is maximal. (iv) R is Noetherian and RP is a discrete valuation ring for all prime ideals P of R. 3.1. Proposition. Let R be an integral domain and M = Rz be a cyclic R- module. If P is a prime ideal of R and µ is a P- fuzzy R- submodule of M, then every proper level set of µ is a submodule of PM. Proof. Since µ is a fuzzy submodule, every level set of µ is a submodule of M. Let µt be a proper level set of µ. Then z ∉ µt , for, otherwise Rz = M ⊆ µt so that M = µt which is not possible. Since µ is a P- fuzzy submodule, µ(sx) = µ(x) for all s ∈ S and x ∈ M. Let x ∈ µt. Then x = rz for some r ∈ R. If r ∉ P, then r ∈ S. But then µ(rz) = µ(z) < t so that x = rz ∉ µt which is not possible. Therefore r ∈ P. Hence x = rz ∈ PM. Hence µt ⊆ PM. 3.2. Theorem. Let R be a Dedekind domain and M = Rz be a cyclic R- Module. Let P be a prime ideal of R. Then a fuzzy submodule µ of M is a P- fuzzy R- submodule ⟺ every non zero level set of µ is of the form Prz for some non-negative integer r. Proof. Let µ be a P- fuzzy submodule of M. Then by proposition 3.1, every proper level set of µ is a submodule of PM = Pz. But a submodule of Pz is of the form Qz where Q is an ideal of R contained in P. Let µt = Qz where Q ⊆ P. If Q ⊊ P, since R is a
Journal of Computer and Mathematical Sciences Vol. 2, Issue 5, 31 October, 2011 Pages (693-779)
Souriar Sebastian, et al., J. Comp. & Math. Sci. Vol.2 (5), 719-724 (2011)
Dedekind domain, Q = CP for some ideal C of R. ∴ µt = CPz. We prove that C ⊆ P. If C ⊈ P, then there exists r ∈ C such that r ∉ P. But then r ∈ S. Let y ∈ Pz. Then y = pz for some p ∈ P. ∴ ry = rpz ∈ CPz = µt. Hence µ(ry) ≥ t. But µ(ry) = µ(y), since r ∈ S. ∴ µ(y) ≥ t, hence y ∈ µt = Qz. ∴ Pz ⊆ Qz so that Pz = Qz. But this is not possible as Qz is a proper submodule of Pz. ∴ C ⊆ P. Since R is Dedekind domain, C = C1P. ∴ µt = C1P2z. If C1 ⊈ P, there exists r1 ∈ C1, but r1 ∉ P. But then r1 ∈ S. Let y ∈ P2z. Then r1y ∈ C1P2z = µt so that µ(r1y) ≥ t. Since µ(y) = µ(r1y) we get y ∈ µt. ∴ P2z ⊆ µt, hence µt = P2z. On the other hand, if C1 ⊆ P, then C1 = C2P. Continuing the process, we finally get µt = Prz for some non negative integer r. Conversely, suppose that every non zero level set of µ is of the form Prz. If s ∈ S and x ∈ M, we prove that sx and x belong to the same level set of µ. Suppose sx ∈ Pk+1z. Assume the contrary that x ∉ Pk+1z, but x ∈ Pkz. Since R is a Dedekind domain, R = P + (s). ∴ 1 = α – βs, α ∈ P and β ∈ R, hence α -1 = βs. ∴ αx – x = βsx. Since x ∈ Pkz and α ∈ P, αx ∈ Pk+1z. Now αx – x ∈ Pk+1z – Pkz, for, if αx – x = y ∈ Pk+1z, then x = αx – y ∈ Pk+1z which is not possible. ∴ αx – x = βsx ∈ Pk+1z – Pkz. But since sx ∈ Pk+1z, βsx ∈ Pk+1z. This is a contradiction. Hence x ∈ Pk+1z. On the other hand, if x ∈ Pk+1z, then sx ∈ Pk+1z. Thus x ∈ Pk+1z ⟺ sx ∈ Pk+1z. Now to prove that µ(sx) = µ(x) for all s ∈ S and x ∈ M, we have to consider two separate cases. Case I : M is torsion free If Az and Bz are any two submodules of M, then Az ∩ Bz = (A ∩ B)z, for, if y ∈ Az ∩ Bz, then y = rz = tz, where r
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∈ A and t ∈ B. Since M is torsion free, this implies r = t so that y ∈ (A ∩B)z. On the other hand, if y ∈ (A ∩ B)z, then y = rz where r ∈ A ∩ B. But then r ∈ A and r ∈ B, hence y = rz ∈ Az ∩ Bz. n
Consider ∩ Piz. Let x ∈ M. If x ∉ i=1
n
∩ P z, then x ∉ Piz for some i. Let j be the i
i=1
lowest power such that x ∉ Pjz. Then x ∈ Pjz – Pjz. But then sx ∈ Pj-1z – Pjz for all s ∈ S. Since the level sets of µ are of the form Prz, it follows that µ(sx) = µ(x). On the other
1
n
hand, if
x ∈ ∩ Piz, then by the above i=1
n
paragraph, x ∈ ( ∩ Pi)z.
Since R is a
i=1
n
Noetherian domain, ∩ Pi = (0), hence x = 0. i=1
Therefore sx = 0 and hence µ(sx) = µ(x). Case II : M is a torsion module In this case M ≅ R/I where I is the torsion ideal defined by I = { r ∈ R : rz = 0 }. The correspondence Qz ↔ Q/I is a one to one correspondence between the set of all sobmodules of M and the set of all ideals of R/I. Since R is a Dedekind domain, R/I has finite number of ideals only. Therefore M has finite number of submodules only. Hence µ has only finite number of level sets. Since Thus x ∈ Pk+1z ⟺ sx ∈ Pk+1z and each level set of µ is of the form Prz, it follows that µ(sx) = µ(x). Thus in either case µ(sx) = µ(x) for all s ∈ S and x ∈ M.. ∴ µ is a P- fuzzy Rsubmodule of M. This proves the result. 3.3. Examples (i) In the ring of integers Z, the ideal M = (2) is an Z-module and is a cyclic module
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generated by 2. The ideal P = (3) is a prime ideal of Z. Let µ be the fuzzy sobmodule of M defined by (54) ⟼ 1, (18) – (54) ⟼ ½, (6) – (18) ⟼ 1/3, (2) – (6) ⟼ ¼. Since µ(sx) = µ (x) for all s ∈ S and x ∈ M, µ is a P- fuzzy Z- submodule of M. Note that the level sets of µ are P3z = (54), P2z = (18), Pz = (6) and M = (2). That is every level is of the form Prz, r = 0, 1, 2, 3. (ii) Consider the abelian group A = { 0, 1, 2, ..........., 23 } under addition modulo 24. Then A is an Z- module. Let M = (3) = { 0, 3, 6, 9,12, 15, 18, 21 } be the cyclic Zsubmodule generated by 3. Let P = (2). Then P is a prime ideal of Z. The fuzzy module µ : M ⟼ [0, 1] defined by 0, 12 ⟼1; 6, 18 ⟼ 1/5; 3, 9, 15, 21 ⟼ 1/7 is a P- fuzzy Z- submodule of M whose level sets are P2z, Pz and M. (iii) The ring R = Z[√2] is a Dedekind domain and P = (2, √2 ) is a prime ideal of R. Let M be the cyclic R- module generated by z = √2. Then M = { (a + b√2 ) √2 : a, b ∈ Z } = { 2b + a√2 : a, b ∈ Z } = (2, √2 ) = P. The powers of P are P2 = (4, √2 ), P3 = (4, 2√2 ), P4 = (4, 4√2 ) and so on .We can prove that if s ∈ S and x ∈ M, then x ∈ Prz ⇔ sx ∈ Prz. We prove a particular case and the general case is similar. Let s ∈ S. Then s = a + b√2 where a, b ∈ Z and a is odd. Let x = 2c + d√2 ∈ M where c, d ∈ Z. Then sx = (a + b√2 )(2c + d√2) = (2ac + 2bd) + (ad + 2bc) √2 . Suppose that sx ∈ P3z = (4, 4√2 ). Then (2ac + 2bd) + (ad + 2bc) √2 = 4u + 4v√2 where u, v ∈ Z. ∴ ac + bd = 2u (1)
ad + 2bc = 4v
(2)
From (2), ad = M(2). Since a is odd, d = M(2). Therefore from (1), ac = M(2). Since a is odd, c = M(2). But then from (2), ad = M(4). Since a is odd, we get d = M(4). Thus c = M(2) and d = M(4). ∴ x = 2c + d√2 = 4α + 4β√2 ∈ P3z. Thus x ∈ P3z ⟺ x ∈ P3z. Hence if µ is a fuzzy submodule of M all whose level sets are in the form Prz, then µ (sx) = µ (x) for all s ∈ S and x ∈ M. That is µ is a P- fuzzy R- submodule of M. 3.4. Remark. The cyclic R- modules M discussed in examples 3.3(i) and 3.3(iii) are torsion free modules where as the cyclic Rmodule M discussed in 3.3(ii) is a torsion module. 3.5. Theorem. Let R be a Dedekind domain and M = Rz be a cyclc R- module. Let P be a prime ideal of R and µ and ν be P- fuzzy Rsubmodules of M. Then µ + ν is a P- fuzzy R- submodule of M. Proof. Since µ and ν are P- fuzzy Rsubmodules of M and R is a Dedekind domain, every level set of µ and ν is of the form Prz. We prove that if s ∈ S and x ∈ M, then (µ + ν)(sx) = (µ + ν)(x). Suppose that sx ∈ Pkz – Pk+1z. We have (µ + ν)(sx) = ∨ { µ(u) ∧ ν(v) : sx = u + v } = ∨ A where A ={ µ(u) ∧ ν(v) : sx = u + v }. Since sx = sx + 0 and sx = 0 + x and µ(0) = ν(0) = 1, µ(sx), ν(sx) ∈ A ∴ ∨A ≥ ∨ { µ (sx), ν(sx) }. We prove that ∨A = ∨ { µ(sx), ν(sx) }. In order to do so, assume the contrary that ∨A > ∨ { µ(sx), ν(sx) }. Then there exist u0, v0 ∈ M such that sx = u0 + v0 and that
Journal of Computer and Mathematical Sciences Vol. 2, Issue 5, 31 October, 2011 Pages (693-779)
Souriar Sebastian, et al., J. Comp. & Math. Sci. Vol.2 (5), 719-724 (2011)
µ(u0) ∧ ν(v0) > ∨ { µ(sx), ν(sx) }. ∴ µ(u0) ∧ ν(v0)> µ(sx) and µ(u0) ∧ ν(v0) > ν(sx). Hence µ(u0) > µ(sx), ν(v0) > µ(sx), µ(u0) > ν(sx), ν(v0) > ν(sx). Since µ(u0) > µ(sx), u0 ∈ Pk+iz where i ≥ 1. Since ν(v0) > ν(sx), v0 ∈ Pk+jz where j ≥ 1. Now suppose i ≥ j. Then u0, v0 ∈ Pk+jz. ∴ u0 + v0 ∈ Pk+jz. i.e., sx ∈ Pk+jz. But this is a contradiction since sx ∈ Pkz – Pk+1z. ∴ (µ + ν)(sx) = ∨ { µ(sx), ν(sx) }. Taking s = 1, we get (µ + ν)(x) = ∨{ µ(x), ν(x) }. Now since µ and ν are P- fuzzy Rsubmodules, (µ + ν)(sx) = ∨ { µ(sx), ν(sx) } = ∨ { µ(x), ν(x) } = (µ + ν)(x). ∴ µ + ν is a P- fuzzy R- submodule of M. 3.6. Theorem. Let R be a Dedekind domain and P be a prime ideal of R. If µ and ν are P- fuzzy ideals of R, and µ(0) = ν(0), then µ + ν is a P- fuzzy ideal of R. Proof. Since µ and ν are fuzzy ideals µ + ν is a fuzzy ideal. If we follow similar steps as in Theorem we can prove that (µ + ν)(rs) = (µ + ν)(r) for all r ∈ R and s ∈ S. Hence µ + ν is a P- fuzzy ideal. 3.7. Proposition16. Let M = Rz be a cyclic R- module and µ be a fuzzy R- submodule of M. Then µ' : R → [0, 1] defined by µ'(r) = µ(rz) is a fuzzy ideal of R. If µ is a P- fuzzy R- submodule of M for some prime ideal P of R, then µ' is a P- fuzzy ideal of R. 3.8. Theorem. Let R be a Dedekind domain and M = Rz be a cyclic R- module. If µ and ν are P- fuzzy R- submodules of M, then µ + ν is a P- fuzzy R-submodule of M and that (µ + ν)' = µ' + ν'. Since µ and ν are P- fuzzy Rsubmodules of M= Rz, by proposition 3.7, µ' and ν' are P-fuzzy ideals of R. Again since
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R is a Dedekind domain, by theorem 3.5, µ + ν is a P- fuzzy R- submodule of M. Correspondingly by proposition 3.7 , (µ + ν)' is a P- fuzzy ideal of R. Also since µ' and ν' are P- fuzzy ideals of R with µ'(0) = µ(0) = 1 = ν(0) = ν'(0) and R is a Dedekind domain, by theorem3.6, µ' + ν' is a P- fuzzy ideal of R. Now to prove that (µ + ν)' = µ' + ν', we have (µ + ν)' (r) = (µ + ν)(rz) = ∨ { µ(uz) ∧ ν(vz) : rz = uz + vz } = ∨ A where A = { µ(uz) ∧ ν(vz) : rz = uz + vz } and (µ' + ν')(r) = ∨ { µ'(x) ∧ ν'(y) : r = x + y } = ∨ { µ(xz) ∧ ν(xz) : r = x + y } = ∨ B where B = { µ(xz) ∧ ν(yz) : r = x + y } Consider any element of B say, µ(xz) ∧ ν(yz). Then r = x + y. ∴ rz = xz + yz. Hence µ(xz) ∧ ν(yz) ∈ A. ∴ B ⊆ A. On the other hand, if we take any element of A, say, µ(uz) ∧ ν(vz), then rz = uz + vz. If r = u + v, then µ(uz) ∧ ν(vz) ∈ B. If r ≠ u + v, then there exist w ∈ R such that r = u + w. But then rz = uz + wz. It follows that uz + vz = uz = wz which gives vz = wz. ∴ ν(vz) = ν(wz). Hence µ(uz) ∧ ν(vz) = µ(uz) ∧ ν(wz). Since r = u + w, µ(uz) ∧ ν(wz) ∈ B, hence µ(uz) ∧ ν(vz) ∈ B. ∴ B ⊆ A. Thus A = B, hence ∨ A = ∨ B. Therefore (µ + ν)'(r) = (µ' + ν')(r), hence (µ + ν)' = µ' + ν'. REFERENCES 1. Bhambri S. K., Kumar R. and Kumar P., On fuzzy primary submodules, Bull. Cal. Math. Soc. 86, 445-452 (1994). 2. Bhambri S. K., Kumar R. and Kumar P., Fuzzy prime submodules and radical of a fuzzy submodule, Bull. Cal. Math.Soc. 87, 163-168 (1995).
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Journal of Computer and Mathematical Sciences Vol. 2, Issue 5, 31 October, 2011 Pages (693-779)