On the Radical of Intersection of Two Submodules

Int. J. Contemp. Math. Scences, Vol. 3, 2008, no. 31, 1535-1550 On the Radcal of Intersecton of Two Submodules J. Moghadar and R. Nekooe Department of Mathematcs Shahd Bahonar Unversty of Kerman, Kerman, Iran j.moghadar@graduate.uk.ac.r rnekooe@mal.uk.ac.r Abstract One of the most fundamental propertes of radcals of deals s that I J = I J, for every par of deals I and J of R. It s not always true that the equalty rad(n L) =radn radl holds for submodules N and L of R-module M. Lu n [4,5] and Moore and Smth n [7], gve suffcent condtons for the equalty rad(n L) =radn radl to hold. In ths paper we obtan necessary and suffcent condtons for the radcal of ntersecton of submodules of R R to be equal to the ntersecton of ther radcals,where R s a PID. Mathematcs Subject Classfcaton: 13C13; 13C99 Keywords: Prme submodules, Radcal of a submodule 1 Introducton Throughout ths paper R denotes a commutatve doman. Let M be a untary R-module. A submodule N of M s called prme f N M, and gven r R, m M, rm N mples m N or r (N : M), where (N : M) ={r R rm N}. Note that f N s prme, then (N : M) s a prme deal of R. The radcal of N s gven by rad M (N) = P, where the ntersecton s over all prme submodules of M contanng N. If there s no prme submodule contanng N, then we put rad M (N) = M. (For more nformaton about rad M (N), see [4,5,6].) Let m and n be postve ntegers, A =(a j ) M m n (R) and F be the free R-module R (n) ; We shall use the notaton A for the submodule N of F generated by rows of A. We denote a unque factorzaton doman by UFD and a prncpal deal doman by PID. Note that n a UFD, a greatest common dvsor (GCD), of

1536 J. Moghadar and R. Nekooe any collecton of elements always exsts. Recall that for any nonempty subset X of R, an element d R s sad to be a GCD of X, provded that () d dvdes all elements of X, and () f there exsts c R such that c dvdes all elements of X, then c d. Fnally, we use the notaton m n = max{m, n} and m n = mn{m, n}, for m, n N, where N = N {0}, m = max{m, 0}, for m Z and for α R, q α =0,fα =0,q α = 1, otherwse. 2 The radcal of a fnte ntersecton of cyclc submodules In ths secton we show that every full rank matrx of a fntely generated free module over a PID R, can be changed to an upper trangular matrx such that both of them generate the same submodule. Fnally we show that the radcal of a fnte ntersecton of cyclc submodules of a fntely generated free R-module, where R s a PID, s equal to ntersecton of ther radcals. Lemma 2.1. Let R be a PID and n be a postve nteger, then any A M n (R) can be changed to an upper trangular matrx such that both of them generate the same submodules of the free R-module F = R (n). Proof. The proof follows by nducton on n: Forn = 1, there s nothng to show. Let A =(a j ) M 2 (R) and let d be the GCD of a 11,a 21. There exst p 11,p 12,p 21,p 22 R such that a 11 p 11 + a 21 p 12 = d, a 11 = dp 22 and a 21 = dp 21. Let P =(p j ) M 2 (R). It s clear that PA s an upper trangular matrx n M 2 (R) and detp s a unt n R. So A = PA. Now assume that the lemma s true for any matrx n M n 1 (R) and let A M n (R). We defne A rs to be the submatrx of A obtaned by removng the r th row and the s th column. So A 1n M n 1 (R) and there exsts P 1n =(p 1n j ) M n 1 (R) such that P 1n A 1n s an upper trangular matrx n M n 1 (R) and det(p 1n ) s a unt n R. Let Q 1n = (qj 1n) M n(r), where q11 1n = 1, q1n 1 = q1 1n = 0 for all, 2 n, and qj 1n = p 1n j otherwse. So det(q 1n ) s a unt n R. Consder the matrx A = Q 1n A and A nn, the submatrx of A. There exsts P nn =(p nn j ) M n 1(R) such that P nn A nn s an upper trangular matrx n M n 1 (R) wth det(p nn ) that s a unt n R. Let Q nn =(qj nn ) M n (R), where qnn =1,qn nn = qn nn = 0, for all, 1 n 1 and qj nn = p nn j otherwse. So det(q nn ) s a unt n R. Fnally, consder the matrx A = Q nn A and ts submatrx A 11. There exsts P 11 =(p 11 j ) M n 1(R) such that P 11 A 11 s upper trangular n M n 1(R) wth det(p 11 ) a unt n R. Let Q 11 =(qj 11 ) M n (R), where q11 11 =1,q1 11 = q1 11 = 0 for all, 2 n, and qj 11 = p 11 j otherwse. So det(q 11 ) s a unt n R. NowQ 11 A = Q 11 Q nn Q 1n A s an upper trangular matrx n M n (R) such that det(q 11 Q nn Q 1n ) s a unt n R. Therefore A = Q 11 Q nn Q 1n A.

Radcal of ntersecton of two submodules 1537 In what follows we consder all full rank matrces to be upper trangular. An R-module M s called a multplcaton module, f for any submodule N of M, there exsts an deal I of R such that N=IM. By [4,Theorem 4], rad(n L) = radn radl for every par of submodules N and L of multplcaton R-module M. In partcular, every cyclc module enjoys ths property. In Theorem 2.4, where R s a UFD, we prove ths property by calculatng the ntersectons and the radcals of two cyclc submodules of a free R-module exactly. Lemma 2.2. Let R be a UFD, F = R (n), N = (a 1,...,a n ), 1 2 be submodules of F for a j R and N = N 1 N 2. Let d be the GCD of a 1,...,a n and d be GCD of d 1,d 2. If there exsts,1 n, such that (a 1 =0,a 2 0)or(a 1 0,a 2 = 0), then N=0. If d 1 = 1 then N=0 or N 2. If d 2 = 1 then N=0 or N 1.Ifd 1,d 2 1 then N=0 or N= d 1 d N 2 = d 2 d N 1. Proof. Let (x 1,..., x n ) N 1 N 2. There exst r, s R such that x = ra 1 = sa 2, for all, 1 n. If there exsts, 1 n, such that (a 1 =0,a 2 0)or(a 1 0,a 2 = 0) then t s clear that N=0. If d 1 =1 then N 2 N 1 or rs =0. SoN = N 2 or N = 0. The case d 2 = 1 s proved smlarly. If d 1,d 2 1 and d 1 d N 2 d 2 d N 1 then we have rs = 0 and so N =0. If d 1 d N 2 = d 2 d N 1 then N= d 1 d N 2 = d 2 d N 1. Theorem 2.3. Let R be a UFD and F = R (n). Suppose that 0 K = R(a 1,...,a n ) for some a R (1 n) and d s a GCD of a 1,...,a n. Then () If d s a unt, then rad F (K) =K. () If d = ud α 1 1...d αm m s a prme decomposton, then rad F (K) =Rd 1...d m ( a 1 d,...,a n d ). Proof. [1, Theorem 3.1].. Theorem 2.4. Let R be a UFD and F = R (n). Then the radcal of the ntersecton of two nonzero cyclc submodules of F s equal to the ntersecton of ther radcals. Proof. Let N = (a 1,...,a n ), 1 2 be submodules of F for a j R. Let d be the GCD of a 1,...,a n, d be the GCD of d 1,d 2 and for =1, 2 and α j N, d = l α 1 1...l α m m be the prme decompostons. By Lemma 2.2, f there exsts,1 n such that (a 1 =0,a 2 0)or(a 1 0,a 2 =0)or d 1 =1ord 2 = 1 the result s easly obtaned. If d 1,d 2 1 and N 1 N 2 = d 1 d N 2 = d 2 m m d N 1 then l u 2 radn 1 = l u 1 radn 2, where u j = α j 1 for each, 1 m and j =1, 2. So radn 1 radn 2 = Rl 1...l m ( a 21,..., a 2n ). On the d 2 d 2 other hand, rad(n 1 N 2 )=Rl 1...l m ( d 1a 21 dl,..., d 1a 2n dl ), where l = d 1d 2. So the d

1538 J. Moghadar and R. Nekooe result follows. Fnally, f d 1,d 2 1 and N 1 N 2 = 0 then radn 1 radn 2 =0 and therefore we conclude that radn 1 radn 2 = rad(n 1 N 2 ). Corollary 2.5. Let R be a PID and F = R (n). Then the radcal of a fnte ntersecton of cyclc submodules of F s equal to the ntersecton of ther radcals. Proof. The proof follows mmedately from Theorem 2.4. 3 Intersecton of two submodules of R R In [8, Example 2.3] by takng N = R(4, 0) + R(0, 1) and L = R(2, 3) t s shown that rad(n L) radn radl; how ever no necessary and suffcent condtons are gven for ths equalty to hold. In ths secton, we frst obtan the ntersecton of a cyclc and a full rank submodules of R R, where R s a PID. Then we obtan the ntersecton of two prmary full rank submodules of R R, where R s agan a PID. By [3, Proposton 1] for A M n (R) wth deta 0 and R a UFD, A s a prmary submodule of F = R (n) f and only f deta = up α for some unt u R, a prme element p R and a postve nteger α. As defned n [2], let J = {j 1,...,j α } be a subset of the nteger between 1 and n and let p R be a prme element. A matrx A M n (R), A =(a j ), s sad to be a p-prme matrx (or smply prme) f A satsfes the followng condtons: ) A s upper trangular. ) For all, 1 n, a = p f J and a =1f J. ) For all, j, 1 <j n, a j = 0 except possbly when J and j J. We call J the set of ntegers assocated wth A and often denote t by J A. By () and () t s clear that det(a) =p α. Theorem 3.1. Every full rank prme submodule of R (n) s the row space of a prme matrx and vce versa. Proof. [2, Theorem 2.5]. Lemma 3.2. Let A M n (R), deta 0 and A =(a j) be the adjont matrx of A. Then (x 1,...,x n ) A for some x R (1 n) f and only n f (deta) x a j, for every j, 1 j n. Proof. [1, Lemma 2.3]. Corollary 3.3. If A M n (R), then (deta) ( A : R (n) ). Proof. [1, Corollary 2.4]. Lemma 3.4. Let R be a PID and for prme elements p of R and α R such that GCD(p,α) = 1 for each, 1 t, m,n,k,l N, p 0,p 00 {0, 1}, N = R( p m, p r α) +R(0, p n ), L = R(p 0 p k,p 00 p l )be submodules of a free R-module F = R R. Let d =( m k + l ) ( k

Radcal of ntersecton of two submodules 1539 m + r ), w = n d. Then N L = p u L, where u s equal to () m k r l, f p 0 p 00 =1,α = 1 and for every, 1 t, m k + l = k m + r. () m k + v, f p 0 p 00 =1,α 0 ( Note that f α = 1, there must exst such that m k + l k m + r ), where v s the smallest nonnegatve nteger such that p m j k j +l j d j +v j j p k j m j +r j d j +v j j α (mod p w j j ). j=1 j=1 j=1 () (p 0 m k p 00 n l )+p 0 q α n r k m, otherwse. Proof. Let T = p u L. It s clear that T L. We show that T N and N L T. Let (x, y) N L. There exst s R, (1 3) such that x = s 1 p m 1 1...p mt t = s 3 p 0 p k 1 1...p kt t, y = s 1 p r 1 1...prt t α + s 2p n 1 1...pnt t = s 3 p 00 p l 1 1...p lt t. So there exsts s 3 R such that s 3 = s 3 Assume that α =0,p 0 p 00 = 1. Therefore s 2 p n = s 3 p m k, s 1 = s 3 and p u s 3, where u = m k n l. Hence (x, y) T. On the other hand, p u ( p k, p l ) = p u +k m ( p l p m, 0) + p u +l n (0, p k m. p n ) N. The proofs n other cases are smlar. Lemma 3.5. Let R be a PIDand, for a prme element p of R, m, l, n, t, k, s N wth m t and α, β R such that GCD(p, α) =GCD(p, β) =1, N 1 = R(p m,p l α)+r(0,p n ), N 2 = R(p t,p k β)+(0,p s ) be prmary submodules of R R. Then N 1 N 2 = N, where N = R(p t,γ)+r(0,p n ),n = n s, t = t + u and () If q α q β 0 then u = { 0, f t m + l = k and α = β (s n) (k (t m + l)) Au, otherwse

1540 J. Moghadar and R. Nekooe and γ = p k+u β( s n +1 1) + p t m+l+u α( n s 1), where u s the greatest nteger such that p u (α β) and A =1,f(k = t m+l < (n s) and α β) else A=0. () If q α q β = 0 then u = (q α + q β )(n s)+q α (m l t) q β k γ = p s (t m+l) αq α ( n s 1) + p n k βq β ( s n 1). Proof. For the proof, we need to dscuss four man cases: (αβ 0), (α 0,β = 0), (α =0,β 0) and (α = β = 0). As the proofs n all cases are smlar, we only consder the case αβ 0. Frst assume that m 0ort 0. Let (x, y) N 1 N 2. There exst r 1,r 2,s 1,s 2 R such that x = r 1 p m = r 2 p t and y = r 1 p l α + s 1 p n = r 2 p k β + s 2 p s.asm t, we have r 1 = r 2 p t m and y = r 2 p t m+l α + s 1 p n = r 2 p k β + s 2 p s. ( ) We prove the statement when n s. In ths case N = R(p t+u,p k+u β)+r(0,p s ). It s clear that N N 2. Consder (x, y) =(r 2 p t,r 2 p k β + s 2 p s )=r 2 (p t,p k β)+s 2 (0,p s ). ) If t m + l = k and α = β, we have u = 0 and (x, y) N. On the other hand, (p t,p k β)=p t m (p m,p l α) and so N N 1. Therefore N = N 1 N 2. ) If k<nand k<t m + l, we have u = n k and by ( ), p u r 2. Now there exsts r 2 R such that r 2 = p u r 2. Therefore x = r 2 pt+u, y = r 2 pk+u β + s 2 p s and hence (x, y) N. On the other hand, (p t+u,p k+u β)=p t m+u (p m,p l α)+(β p t m+l k α)(0,p n ), mples that N N 1. ) If k = t m + l<nand α β, we have u = n k u and by ( ), r 2 p k (α β) =p n (s 2 p s n s 1 ). Therefore p u r 2 and there exsts r 2 R such that r 2 = p u r 2.Sox = r 2p t+u, y = r 2p k+u β + s 2 p s and so (x, y) N. On the other hand, (p t+u,p k+u β)=p t m+u (p m,p l α) p k+u n z(0,p n ), where there exsts z R such that α β = p u z and GCD(p, z) = 1, hence N N 1. v) If t m + l<kand t m + l<nthen u = n t + m l and by ( ), we have p u r 2. By an argument smlar to (), (x, y) N. On the other hand, (p t+u,p k+u β)=p t m+u (p m,p l α)+(p k t+m l β α)(0,p n ), hence N N 1. v) If n t m + l and n k, we have u =0. Thus(x, y) N. On the other hand, (p t+u,p k+u β) = (p t,p k β) = p t m (p m,p l α)+(p k n β p t m+l n α) (0,p n ), hence N N 1. Therefore when n s, we have N 1 N 2 = N. By a smlar argument we can show that when s n, N 1 N 2 = N. Smlarly, f m = t = 0, we can obtan that N 1 N 2 = N. Corollary 3.6. Let R be a PID and N = R(p m,e )+R(0,p n )be prme submodules of a free R-module R R, for a prme element p of R and

Radcal of ntersecton of two submodules 1541 m,n {0, 1} wth m 1 m 2, e R, {1, 2}. Then N 1 N 2 = N = R(p m,e)+r(0,p n ), where n = n 1 n 2, { { m1 m m = 2, f e 1 = e 2 e1, f e and e = 1 = e 2 1 Au, f e 1 e 2 p 1 Au e 1, f e 1 e 2 u s the greatest nteger such that p u (e 1 e 2 ) and f (m 1 = m 2,n 1 = n 2 = 1,e 1 e 2 ), A = 1, else A =0. Proof. The proof follows from Lemma 3.5. Proposton 3.7. Let R be a PID and N = R(x,α )+R(0,y ), 1 t be submodules of R R, where x,y,α R such that for all, j, 1, j t, j, GCD(x,x j )=GCD(y,y j ) = 1 and x,y 0. Then for all, 1 t, there exst l R such that β t = l y + x 1...ˆx...x t α and N 1...N t = N = R(x 1...x t,β t )+R(0,y 1...y t ). Proof. Snce GCD(y 1,y 2 ) = 1, there exst r 1,r 2 R such that r 1 y 1 + r 2 y 2 =1. (1) So r 1 (x 1 α 2 x 2 α 1 )y 1 + x 2 α 1 = r 2 (x 2 α 1 x 1 α 2 )y 2 + x 1 α 2. Set k 11 = r 1 (x 1 α 2 x 2 α 1 ) and k 12 = r 2 (x 2 α 1 x 1 α 2 ) (2) Now let β 2 = k 11 y 1 + x 2 α 1 = k 12 y 2 + x 1 α 2. Then R(x 1 x 2,β 2 )+R(0,y 1 y 2 ) N 1 N 2. On the other hand, f (a, b) N 1 N 2, then there exst u, v 1,v 2 R such that a = ux 1 x 2, b = ux 2 α 1 + v 1 y 1 = ux 1 α 2 + v 2 y 2 (3) Now by (1), (2) and (3),we have b = uβ 2 +(v 1 r 2 + v 2 r 1 )y 1 y 2. Therefore (a, b) R(x 1 x 2,β 2 )+R(0,y 1 y 2 ) and hence N 1 N 2 = R(x 1 x 2,β 2 )+ R(0,y 1 y 2 ). Smlarly, there exst k 21,k 22 R such that β 3 = k 21 y 1 y 2 + x 3 β 2 = k 22 y 3 + x 1 x 2 α 3 and N 1 N 2 N 3 = R(x 1 x 2 x 3,β 3 )+R(0,y 1 y 2 y 3 ). By substtutng the value for β 2, we have k 21 y 1 y 2 + x 3 k 11 y 1 + x 2 x 3 α 1 = k 21 y 1 y 2 + x 3 k 12 y 2 + x 3 x 1 α 2 = k 22 y 3 +x 1 x 2 α 3. Let k 21 = k 21y 2 +x 3 k 11, k 21 = k 21y 1 +x 3 k 12, k 21 = k 22. Fnally, β 3 = k 21 y 1 + x 2 x 3 α 1 = k 21 y 2 + x 1 x 3 α 2 = k 21 y 3 + x 1 x 2 α 3. Repeatng ths step t tmes we can fnd an l R such that β t = l y + x 1...ˆx...x t α, for all, 1 t and N 1... N t = N. Proposton 3.8. Let R be a PID and for prme elements p of R and α R, m,n j N, N = R( p m,α)+r(0, p n ) be a submodule of the free R- module F = R R. Then there exst α R such that α p m 1 1... pm ˆ...p mt t α (mod p n ) and N has a prmary decomposton N = N 1...N t, where N = R(p m,α )+R(0,p n ). Proof. The proof follows easly from Theorem 2, n [3].

1542 J. Moghadar and R. Nekooe 4 Necessary and suffcent condtons for rad(n L) =radn radl In ths secton, we frst obtan the necessary and suffcent condtons for the radcal of the ntersecton of a cyclc and a full rank submodule of a free R- module R R to be equal to the ntersecton of ther radcals, where R s a PID. Fnally, we fnd the necessary and suffcent condtons for the radcal of the ntersecton of two full rank submodules of free R-module R R to be equal to the ntersecton of ther radcals, where R s a PID. After each theorem we gve examples to show that the condtons of the theorem are essental. Proposton 4.1. Let R be a PID and F = R (n). Let N = A, for A M n (R) where deta 0, for all, 1 t. IfGCD(detA, deta j )=1, for all, j, 1, j t, j. Then rad( N )= radn. Proof. It s clear that rad( N ) radn. Let P be a prme submodule of F contanng N. Suppose that for all, 1 t, N P ; as (N : F )=( N : F ) (P : F ) and (P : F ) s a prme deal and there exsts j, 1 j n, such that (N j : F ) (P : F ). On the other hand, by Corollary 3.3, (deta ) (N : F ), for all. Hence deta j (P : F ) and deta N j N P. Snce P s a prme submodule of F and N j P, j we have deta (P : F ). Therefore there exst, 1 n, j, such j that deta (P : F ). Now snce R s a PID, there exsts a prme element q n R such that (P : F )= q. Soq deta, q deta j whch s a contradcton. Hence there exsts l, 1 l n, such that N l P. So radn l P and therefore radn P. We conclude that radn rad( N ). Lemma 4.2. Let R be a PID and N = R(p m,α)+r(0,p n ) be a prmary submodule of F = R R, for a prme element p of R, where α R, m, n N. Then radn = R(p u,a)+r(0,p v ), where u = m 1, v = 0 f (mn 0,p α) and v = n 1; otherwse, a p m α(modp). Proof. There are fve dfferent cases: ) If m 0,n = 0 then R(p, 0) + R(0, 1) s the only prme submodule of F

Radcal of ntersecton of two submodules 1543 whch contans N; So radn = R(p, 0) + R(0, 1). ) If mn 0,p α then radn = R(p, 0) + R(0,p). ) If mn 0,p α then R(p, 0) + R(0, 1) s the only prme submodule of F whch contans N; So radn = R(p, 0) + R(0, 1). v) If m =0,n 0,α = 0 then R(1, 0) + R(0,p) s the only prme submodule of F whch contans N; So radn = R(1, 0) + R(0,p). v) If m =0,n 0,α 0 then R(1,a)+R(0,p) s the only prme submodule of F whch contans N, where a α(modp); So radn = R(1,a)+R(0,p). Theorem 4.3. Let R be a PIDand F = R R. Let N = R( p m, p r α)+ R(0, p n ) and L = R(0, p l ), where p s are prme elements of R, α R such that GCD(α, p ) = 1, for all, 1 t, m,r,n,l N. Then rad(n L) =radn radl f and only f α = 0 or, for each, (m n =0or l 0orr 0). Proof. By Lemma 3.4, N L = By Theorem 2.3, rad(n L) = p u L, where u = n l. p ( n l +l ) 1 R(0, 1). On the other hand, by Proposton 4.1, Lemma 4.2 and Proposton 3.7, radn radl =[R( R(0, p n )] p l 1 R(0, 1) = p n (l 1) +(l 1) R(0, 1). p m,β)+ If α =0orn =0orn l 0 or (n 0,l = m = 0) t easly follows that rad(n L) =radn radl. Ifn m 0,l = 0 then for the equalty to hold t s necessary that r 1. Example 4.4. Let R = Z, F = R R, N = (4, 3), (0, 4) and L = (0, 1).Then radn radl = (0, 1) and rad(n L) = (0, 2). Sorad(N L) radn radl. In fact mn 0,l = r = 0 and α 0 and the condtons of Theorem 4.3, are not satsfed. Theorem 4.5. Let R be a PIDand F = R R. Let N = R( p m, p r α)+ R(0, p n ) and L = R( p k, 0), where the p s are prme elements of R, α R such that GCD(α, p ) = 1, for all, 1 t, m,r,n,k N. Then rad(n L) =radn radl f and only f α = 0 or, for each, (k m > 0or n r or r = 0). Proof. If α = 0 then by Lemma 3.4, N L = p u L, where u = m

1544 J. Moghadar and R. Nekooe k. By Theorem 2.3, rad(n L) = p ( m k +k ) 1 (1, 0). On the other hand, by Proposton 4.1, Lemma 4.2 and Proposton 3.7, radn radl = [R( p m, 0) + R(0, p n )] p k 1 R(1, 0) = p m (k 1) +(k 1) R(1, 0). It easly follows that rad(n L) =radn radl. If α 0 then N L = p u L, where u = m k + n r k m and rad(n L) = p (u +k ) 1 R(1, 0). On the other hand, by Proposton 4.1, Lemma 4.2 and Proposton 3.7, radn = R( p m,β)+r(0, p n ). Now, f β = 0 then radn radl = p m (k 1) +(k 1) R(1, 0). For each,1 t, fk 0 or (k =0,m 0) t easly follows that rad(n L) =radn radl. Ifk = m = 0 then we should have n r =0 or n r. The condton for case α 0 and β 0 s smlarly obtaned. Example 4.6. Let R = Z, F = R R ) If N = (1, 2), (0, 4) and L = (1, 0), then radn radl = (1, 0) and rad(n L) = (2, 0). So rad(n L) radn radl. In fact α 0, k = m =0,n>r,r 0 and the condtons of Theorem 4.5, are not satsfed. ) If N = (1, 10), (0, 12) and L = (1, 0), then radn radl = (3, 0) and rad(n L) = (6, 0). So rad(n L) radn radl. In fact for p 1 =2, α 0 and k 1 = m 1 =0,n 1 >r 1,r 1 0 and the condtons of Theorem 4.5, are not satsfed. Theorem 4.7. Let R be a PID and F = R R. Let N = R( p m, 0) + R(0, p n ) and L = R( p k, p l ), where p s are prme elements of R, m,n,k, l N. Then rad(n L) =radn radl f and only f for each, one of the followng condtons hold: ) If k = m =0,n >l then l =0. ) If k m 0,l = n = 0 then m k. Proof. The proof s smlar to that of Theorem 4.5. Example 4.8. Let R = Z, F = R R. ) If N = (1, 0), (0, 4) and L = (1, 2), then radn radl = (1, 2) and rad(n L) = (2, 4). Sorad(N L) radn radl. In fact k = m =0,n>l but l 0 and the condtons of Theorem 4.7, are not satsfed. ) If N = (4, 0), (0, 1) and L = (2, 1),then radn radl = (2, 1) and rad(n L) = (4, 2). Sorad(N L) radn radl. In fact km 0,l = n =0

Radcal of ntersecton of two submodules 1545 but m>kand the condtons of Theorem 4.7, are not satsfed. Let R be a PID, F = R R, N = R( p m, p r α)+r(0, p n ) and L = R( p k, p l ), where for all, 1 t, p s are prme elements of R, α R such that α 0 and GCD(α, p ) = 1 and m,r,n,k,l N. Then by Proposton 4.1, Lemma 4.2 and Proposton 3.7, radn = R( p m,β)+ R(0, p n ). Set x = k l 1 and β = p r β, where GCD(p,β ) = 1, for all, 1 t and for all j, 1 j t, m j k j + l j = k j m j + r j. ( ) (m j 1) (x j +k j )+(k j l j ) +x j +l j (k j l j )= x j +k j (k j l j ) (m j 1) +r j. ( ) In the followng Theorem we use these notatons and replace v wth v n Lemma 3.4, for radn radl. Theorem 4.9. Let R be a PID, F = R R, N = R( p m, p r α)+ R(0, p n ) and L = R( p k, p l ), where p s are prme elements of R, α R such that α 0 and GCD(α, p ) = 1, for all, 1 t, m,r,n,k,l N. Then rad(n L) =radn radl f and only f, for each, one of the followng condtons hold: I) f α 0,β =0,k = 0, then m 0orn =0or 1 l v 1. II) f α 0,β =0,k 0, then l 0orm =0orn ( m k +v ) 1. III) f α 0,β =1,( ) holds, k = 0, then m 0orv =0. IV)f α 0,β =1,( ) holds, k 0, then l 0 or (m = 1 and v = 0). V) f α 0,β 0,k = 0, then m 0orv (v 1). VI) f α 0,β 0,k 0, then m =0orl 0orv ( m k +v ) 1. VII) f α =1,( ) holds, β =1,( ) holds, then k =0orl 0orm =1. VIII) f α =1,( ) holds, β = 0, then k =0orl 0orn m k 1. IX) f α =1,( ) holds, β 0, then k =0orl 0orv 0orm = k. Proof. By Lemma 3.4, N L = p u L, where u s are as n Lemma 3.4. By Theorem 2.3, radl = p x R( p k (k l ), p l (k l ), where x = k l 1, and rad(n L) = p y R( p k (k l ), p l (k l ) ), where y = (u +(k l )) 1. Now obtan necessary and suffcent condtons for the

1546 J. Moghadar and R. Nekooe equalty radn radl = rad(n L). By Lemma 3.4, there are sx cases. Assume that α 0 and β = 0. The equalty holds f and only f for each, x +[ m (x + k )+(k l ) n (x + l )+(k l ) ] [ m k + v +(k l )] 1. Now f (k =0,m 0)or(k = m = n =0)or (k l 0)or(k 0,m = l = 0) t easly follows that the above nequalty holds. If k = m =0,n 0 then for the nequalty to hold t s necessary that 1 l v 1. Fnally, f k m 0,l = 0 then for the nequalty to hold t s necessary that n ( m k + v ) 1. The proofs n the other cases are smlar. Example 4.10. In the followng examples R = Z, F = R R and rad(n L) radn radl. ) Let N = (1, 6), (0, 8) and L = (1, 2).Then radn radl = (1, 2) and rad(n L) = (2, 4). In fact α 0,β =0,k = m =0,n 0, 1 l = 0 <v 1 = 1 and the condtons of Theorem 4.9, are not satsfed. ) Let N = (12, 21), (0, 27) and L = (6, 9).Then radn radl = (6, 9) and rad(n L) = (12, 18). In fact for p 1 =2,α 0,β =0,k 1 m 1 0,l 1 =0, n 1 =0< ( m 1 k 1 + v 1 ) 1 = 1 and the condtons of Theorem 4.9, are not satsfed. ) Let N = (1, 3), (0, 8) and L = (1, 1).Then radn radl = (1, 1) and rad(n L) = (2, 2). In fact α 0,β =1,( ) holds and k = m =0, v = 2 and the condtons of Theorem 4.9, are not satsfed. v) Let N = (2, 7), (0, 12) and L = (6, 3).Then radn radl = (6, 3) and rad(n L) = (12, 6). In fact for p 1 =2,α 0,β =1,( ) holds and k 1 0,l 1 =0,m 1 =1,v 1 = 1 and the condtons of Theorem 4.9, are not satsfed. v) Let N = (1, 3), (0, 40) and L = (1, 1).Then radn radl = (5, 5) and rad(n L) = (10, 10). In fact for p 1 =2,α 0,β 0,k 1 = m 1 =0, v 1 =0<v 1 1 = 1 and the condtons of Theorem 4.9, are not satsfed. v) Let N = (4, 6), (0, 25) and L = (2, 1).Then radn radl = (10, 5) and rad(n L) = (20, 10). In fact for p 1 =2,α 0,β 0,k 1 m 1 0,l 1 =0, v 1 =0< ( m 1 k 1 + v 1 ) 1 = 1 and the condtons of Theorem 4.9, are not satsfed. v) Let N = (4, 2), (0, 3) and L = (6, 3).Then radn radl = (6, 3) and rad(n L) = (12, 6). In fact for p 1 =2,α =1,( ) holds, β =1, ( ) holds, k 1 0,l 1 =0,m 1 1 and the condtons of Theorem 4.9, are not satsfed. v) Let N = (4, 6), (0, 9) and L = (2, 3).Then radn radl = (2, 3) and rad(n L) = (4, 6). In fact for p 1 =2,α =1,( ) holds, β =0,k 1 0,l 1 =0,n 1 =0< m 1 k 1 1 = 1 and the condtons of Theorem 4.9, are not satsfed. x) Let N = (8, 2), (0, 3) and L = (12, 3).Then radn radl = (12, 3) and rad(n L) = (24, 6). In fact for p 1 =2,α =1,( ) holds, β 0,k 1

Radcal of ntersecton of two submodules 1547 0,l 1 = v 1 =0,m 1 =3 k 1 = 2 and the condtons of Theorem 4.9, are not satsfed. Theorem 4.11. Let R be a PIDand F = R R. Let N = [R(p m 1,p l α 1 )+ R(0,p m 2 )] and L = [R(p m 3,p k α 3 )+R(0,p m 4 )], where p are dstnct prme elements of R, m j,l,k N and α j R such that GCD(α j,p ) = 1 for each, 1 t. Let a 1 p l α 1 (modp ), a 3 p k greatest nteger such that p u A = (α 1 α 3 ), p u { 1, f k = l < (m 2 m 4 ),α 1 α 3 0, otherwse α 3 (modp ), u, u be the (a 1 a 3 ) and Then rad(n L) =radn radl f and only f for each, one of the followng condtons hold: I) If m 2 =0,m 3 m 4 0,k = 0, then α 1 0orm 3 m 1. II) If m 4 =0,m 1 m 2 0,l = 0, then α 3 0orm 1 m 3. III) If m 1 m 2 m 3 m 4 0,l = k = 0, then m 1 = m 3 ; moreover, f α 1 α 3, then m 2 m 4 u. IV) If m 1 = m 3 =0,a 1 = a 3,q α1 q α3 0,q α1 q α3 = 0, then m 2 m 4 k q α3 + l q α1. V) If m 1 = m 3 =0,a 1 = a 3,q α1 q α3 0, (l k or α 1 α 3 ), then m 2 m 4 (k l )+A u. VI) If m 1 = m 3 =0,m 2 m 4 0,a 1 a 3,u 0,α 1 α 3,q α1 q α3 0, then m 2 m 4 u. Proof. Let N = N 1... N t, where N = R(p m 1,p l α 1 )+R(0,p m 2 ) and L = L 1... L t, where L = R(p m 3,p k α 3 )+R(0,p m 4 ), for all, 1 t. Now, by Proposton 4.1, radn = radn 1... radn t and by Lemma 4.2, radn = R(p u 1,a 1 )+R(0,p v 2 { a1 p l α 1 (modp ), f m 1 =0 a 1 =0, f m 1 0 ) where u 1 = m 1 1 and { 0, f m1 m, v 2 = 2 0,l =0 m 2 1, otherwse Smlarly, radl = radl 1... radl t, where radl = R(p u 3,a 3 )+R(0,p v 4 ) and we agan obtan u 3,a 3,v 4. By Corollary 3.6, radn radl = [R(p n,β )+R(0,p v 2 v 4 )], where n = and { u1 u 3, f a 1 = a 3 1 A u, f a 1 a 3. { a1, f a 1 = a 3, β = p 1 A u a 1, f a 1 a 3. { 1, f A u1 = u = 3,v 2 = v 4 =1,a 1 a 3 0, otherwse

1548 J. Moghadar and R. Nekooe By Proposton 3.7, radn radl = R( p n,β)+r(0,. there exst b R such that β = p n 1 n 1... ˆp...p nt t β + b p v 2 v 4 On the other hand, by Lemma 3.5, p v 2 v 4 ), where t N L = [R(p s,γ )+R(0,p m 2 m 4 )] [R(p s,γ )+R(0,p m 2 m 4 )], =t +1 where there exsts t N such that for all, 1 t, m 1 m 3 and for all, t +1 t, m 3 m 1 and s,γ are obtaned by Lemma 3.5. By Proposton 4.1 and Lemma 4.2, we have rad(n L) = [R(p l,f )+ R(0,p k )], where l = s 1, f γ (modp )fs =0,f = 0, otherwse and { 0, f s (m k = 2 m 4 ) 0,p γ (m 2 m 4 ) 1, otherwse Now to have radn radl = rad(n L); t s enough that for all,1 t, R( p n,β)+r(0, p v 2 v 4 ) R(p l,f )+R(0,p k ). So t s necessary that for all, 1 t, I) k v 2 v 4. Ths means that, f v 2 v 4 = 0 then k = 0. But v 2 v 4 = 0 f and only f [m 2 = 0 or (m 1 m 2 0,l = 0)] and [m 4 =0or (m 3 m 4 0,k = 0)]. If m 2 =0,m 3 m 4 0,k = 0 then k = 0 f and only f p γ and p γ f and only f α 1 0orm 3 m 1. The proofs n the other cases are smlar. II) ( ) n l. p n,β) R(p l,f )+R(0,p k ). Ths means that, for all, 1 t ) there exsts r R such that β = f j=1 j p n j j p n l + r p k. In part (), assume that m 1 = m 3 =0,a 1 = a 3 and q α1 =0,q α3 =1. By Lemma 3.5, s = (m 2 m 4 ) k. So for the nequalty to hold t s necessary that m 2 m 4 k. The proofs of the other cases are smlar. Part (), s easly obtaned. Example 4.12. In the followng examples R = Z, F = R R and rad(n L) radn radl. ) Let N = (4, 0), (0, 1) and L = (2, 1), (0, 2).Then radn radl = (2, 0), (0, 1) and rad(n L) = (2, 0), (0, 2). In fact m 21 =0,m 31 m 41 0,k 1 =0,α 11 =0,m 31 <m 11 and the condtons of Theorem 4.11, are not satsfed.

Radcal of ntersecton of two submodules 1549 ) Let N = (2, 1), (0, 2) and L = (4, 0), (0, 1).Then radn radl = (2, 0), (0, 1) and rad(n L) = (2, 0), (0, 2). In fact m 41 =0,m 11 m 21 0,l 1 =0,α 31 =0,m 11 <m 31 and the condtons of Theorem 4.11, are not satsfed. ) Let N = (2, 1), (0, 2) and L = (4, 1), (0, 2).Then radn radl = (2, 0), (0, 1) and rad(n L) = (2, 0), (0, 2). In fact m 11 m 21 m 31 m 41 0,l 1 = k 1 = 0, but m 11 m 31 and the condtons of Theorem 4.11, are not satsfed. v) Let N = (1, 0), (0, 4) and L = (1, 2), (0, 4).Then radn radl = (1, 0), (0, 2) and rad(n L) = (2, 0), (0, 2). In fact m 11 = m 31 =0,a 11 = a 31,q α11 q α31 0,q α11 q α31 = 0, but m 21 m 41 > k 1 q α 31 + l 1 q α 11 and the condtons of Theorem 4.11, are not satsfed. v) Let N = (1, 1), (0, 4) and L = (1, 3), (0, 8).Then radn radl = (1, 1), (0, 2) and rad(n L) = (2, 0), (0, 2). In fact m 11 = m 31 =0,a 11 = a 31,q α11 q α31 0,l 1 = k 1 = 0 and α 11 α 31, but m 21 m 4 > (k 1 l 1)+A 1 u 1 and the condtons of Theorem 4.11, are not satsfed. v) Let R = Z[], F = R R. Consder N = (1, ), (0, (1 + ) 2 ) and L = (1, 1), (0, (1 + ) 2 ).Then radn radl = (1,),(0, 1+) and rad(n L) = (1 +, 0), (0, 1+). In fact m 11 = m 31 =0,m 21 m 41 0,a 11 a 31,u 1 0,α 11 α 31,q α11 q α31 0, but m 21 m 41 >u 1 and the condtons of Theorem 4.11, are not satsfed. Acknowledgements. Ths research has been supported by Mahan mathematcal research center. REFERENCES [1] S. Hedayat and R. Nekooe, Characterzaton of prme submodules of a fntely generated free module over a PID, Houston Journal of mathematcs, 1 (31), (2005), 75-85. [2] S. Hedayat and R. Nekooe, Prme and radcal submodules of free modules over a PID, Houston Journal of mathematcs, 2 (32), (2006), 355-367. [3] S. Hedayat and R. Nekooe, Prmary Decomposton of submodules of a fntely generated module over a PID, Houston Journal of mathematcs, 2 (32), (2006), 369-377. [4] C.P. Lu, M-radcals of submodules n modules, Math. Japonca 34 (2), (1989), 211-219. [5] C.P. Lu, M-radcals of submodules n modules II, Math. Japonca 35 (5), (1990), 991-1001. [6] R.L. McCasland and M.E. Moore, On radcals of submodules of fntely generated spectra, Houston J. of Math., 22 (1996), 457-471.

1550 J. Moghadar and R. Nekooe [7] M.E. Moore and S.J. Smth, Prme and radcal submodules of modules over commutatve rngs, Communcatons n Algebra, 10 (30), (2002), 5037-5064. Receved: Aprl 13, 2008