On the Radical of Intersection of Two Submodules

Transcription

1 Int. J. Contemp. Math. Scences, Vol. 3, 2008, no. 31, On the Radcal of Intersecton of Two Submodules J. Moghadar and R. Nekooe Department of Mathematcs Shahd Bahonar Unversty of Kerman, Kerman, Iran 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

2 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.

3 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 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 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

4 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 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

5 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 p mt t = s 3 p 0 p k p kt t, y = s 1 p r prt t α + s 2p n pnt t = s 3 p 00 p l 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

6 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

7 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 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].

8 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

9 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

10 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

11 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

12 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 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

13 Radcal of ntersecton of two submodules ,l 1 = v 1 =0,m 1 =3 k 1 = 2 and the condtons of Theorem 4.9, are not satsfed. Theorem 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

14 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 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.

15 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), [2] S. Hedayat and R. Nekooe, Prme and radcal submodules of free modules over a PID, Houston Journal of mathematcs, 2 (32), (2006), [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), [4] C.P. Lu, M-radcals of submodules n modules, Math. Japonca 34 (2), (1989), [5] C.P. Lu, M-radcals of submodules n modules II, Math. Japonca 35 (5), (1990), [6] R.L. McCasland and M.E. Moore, On radcals of submodules of fntely generated spectra, Houston J. of Math., 22 (1996),

16 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), Receved: Aprl 13, 2008