Advances in Pure Mathematics, 2012, 2, 344-348 http://dx.doi.org/10.4236/apm.2012.25049 Published Online September 2012 (http://www.SciRP.org/journal/apm)
The Primary Radical of a Submodule Lamis J. M. Abulebda University College, Abu Dhabi Universityn, Abu Dhabi, UAE Email:
[email protected] Received April 5, 2012; revised April 29, 2012; accepted May 7, 2012
ABSTRACT In this paper we introduced a definition for the primary radical of a submodule with some of its basic properties. We also define the P-radical submodule and review some results about it. We find a method to characterize the primary radical of a finitely generated submodule of a free module. Keywords: Primary Submodule; Prime Radical of a Submodule; Radical Submodule; Free Module; Noetherian Module; Finitely Generated Submodule
1. Introduction
2.2. Proposition
The prime radical of a submodule N of an R-module M , denoted by rad M N is defined as the intersection of all prime submodules of M which contain N , if there exists no prime submodule of M containing N , we put rad M N M [1]. We naturally seek a counterpart in the primary radical of a submodule of module. Firstly we introduced a definition for the primary radical of a submodule with some of its basic properties. We also define the P-radical submodule and review some results about it. Finally, we find a method to characterize the primary radical of a finitely generated submodule of a free module.
Let N and L be submodules of an R-module M . Then 1) N prad M N
2. Some Basic Properties of the Primary Radical In this section we introduce the concept of the primary radical and give some useful properties about it.
2.1. Definition The primary radical of a submodule N of an R-module M , denoted by prad M N is defined as the intersection of all primary submodules of M which contain N . If there exists no primary submodule of M containing N , we put prad M N M . If M R , since the primary submodules and the primary ideals are the same, so if I is an ideal of R , prad R I is the intersection of all primary ideals of R , which contain I . Now, we give useful properties of the primary radical of a submodule. Copyright © 2012 SciRes.
2) prad M N L prad M N prad M L 3) prad M prad M N prad M N Proof. 1) It is clear. 2) Let H be primary submodule of M containing L, since N L L H so prad M N L H . Thus prad M N L prad M L . By the same way prad M N L prad M N . It follows prad M N L prad M N prad M L . 3) By 1) we have prad M N prad M prad M N . Now prad M N L where the intersection is over all primary submodules L of M with L N . prad M prad M N prad M L prad M L L
In the following two propositions, we give a condition under which the other inclusion of 2) holds, that is; prad M N L prad M N prad M L provided that every primary submodule of M which contains N L is completely irreducible submodule. Where a submodule K of an R -module M is called Completely Irreducible if whenever N L K , then either N K or L K where N and L are submodules of M .
2.3. Proposition Let N and L be submodules of an R -module M . If every primary submodule of M which contains N L APM
L. J. M. ABULEBDA
is completely irreducible submodule, then:
Proof. By proposition (2.2, (2)) prad M N L prad M N prad M L . If prad M N L M , clearly prad M N prad M L M . If prad M N L M , there exists a primary submodule K of M such that, N L K by hypothesis either N K or L K so that either prad M N K or prad M L K , because every primary submodule containing N L , so either prad M N prad M N L or prad M L prad M N L therefore prad M N prad M L prad M N L .
2.4. Proposition
N : M L : M R , then prad M N L prad M N prad M L .
such that
N L : M K : M . So K : M
N : M K : M . If
or
N : M K : M ,
thus rl K N , P
for
L : M K : M .
K N , P n m M cm N P n M for some c P
K N, P M . n
P K N , P : M , that is P K N , P : M . Let r K N , P : M , r M K N , P for some t Z , thus c r M N P M for some c P . If r P then cr P this implies M K N , P , P K N , P n : M , we have proved above that n
n
n
t
t
n
n
t
n
which is a contradiction. Therefore r P thus
Let N be a submodule of a module M over a Noetherian ring R . Then prad M N
K N , Pn
P is a prime ideal of R, n Z
To show K N , P
Copyright © 2012 SciRes.
n
is P -primary,
Proof. By proposition (2.2), for each positive integer n and any prime ideal P we have K N , P n is a P -primary submodule containing N . Hence
prad M N
K N , P n P is a prime ideal of R, n Z
For every primary submodule H containing N with P
H : M
there exists a positive integer r such
H . So K N , P P is a prime ideal of R, n Z K N, P H. r
n
Let N be a submodules of an R-module M and P be a prime ideal of R. For each positive integer n : K N , P n M or K N , P n is a P -primary submodule of M . Proof. Let n be any positive integer, it is clear that K N , P n is a submodule of M . Assume
c P such that
that K N , P
2.5. Proposition
, there exists
r P , therefore K N , P n : L P . So K N , P n is a primary submodule
n
c rl N P M . If r P , then cr P and this implies l K N , P n , which is a contradiction. It follows
Let N be a proper submodule of an R-module M . Let P be a prime ideal of R. For each positive integer n , we shall denote by K N , P n the following subset of
n
which
is a contradiction. Therefore N K . Now, applying proposition (2.3), we can conclude that prad M N L prad M N prad M L . We conclude
2.6. Theorem
L : M K : M
then
N : M L : M K : M
the same result if
is a prime ideal, either
L : M K : M otherwise R
K N , Pn : M P . The following theorem gives a description of the primary radical of a submodule.
N : M L : M N L : M K : M . Since
Proof. If K is a primary submodule containing N L ,
that is P n K N , P n : M . Now, let L be a submodule of M properly containing K N , P n , let r K N , P n : L , rL K N , P n . Since K N , P n L , let l L , but l K N , P n
n
Let N and L be submodules of an R -module M
Pn M K N , Pn
prad M N L prad M N prad M L .
then
345
r
Thus
K N , P n P is a prime ideal of R, n Z
prad M N .
We will give the following definition. APM
L. J. M. ABULEBDA
346
2.7. Definition A proper submodule N of an R-module M with prad M N N will be called P-Radical Submodule. Now, we are ready to consider the relationships among the following three statements for any R-module M . 1) M satisfies the ascending chain condition for pradical submodules. 2) Each p-radical submodule is an intersection of a finite number of primary submodules 3) Every p-radical submodule is the p-radical of a finitely generated submodule of it.
2.8. Proposition Let M be an R -module. If M satisfies the ascending chain condition for p-radical submodule of M is an intersection of a finite number of primary submoules. Proof. Let N be a p-radical submodule of M and put N iI N i , where Ni is a primary submodule for each i I , and the expression is reduced. Assume that I is an infinite index set. Without loss of generality we may assume that I is countable, then
i 1
i 2
i 3
N N i N i N i is an ascending chain of
p-radical submodules, since by proposition (2.2),
N i prad M N i prad M N i N i i i i i i
By hypothesis this ascending chain must terminate, so there exists j I such that i j N i i j 1 N i , whence i j 1 N i N j which contradicts that the expression N i N i is a reduced. Therefore I must be finite.
2.9. Proposition Let M be an R-module. If M satisfies the ascending chain condition for p-radical submodules, then every p-radical submodule is the p-radical of finitely generated submodule of it. Proof. Assume that there exists a p-radical submodule N of M which is not the p-radical of a finitely generated submodule of it. Let m1 N and let N1 prad M Rm1 so N1 N , hence there exists
m2 N N1 . Let N 2 prad M Rm1 Rm2 , then N1 N 2 N , thus there exists m3 N N 2 , etc. This
implies an ascending chain of p-radical submodules, N1 N 2 N 3 which does not terminate and this
generated submodule of it, then M satisfies the ascending chain condition for primary submodules. Proof. Let N1 N 2 N 3 be an ascending chain of primary submodules of M . Since M is finitely generated then, N i N i is a primary submodule of M . Thus by hypothesis, N is the p-radical for some finitely generated submodule L R m1 , m2 , , mn , hence L prad M L N i N i , then there exists j I such that L N j hence N prad M L prad M N j N j . Thus i Ni N j for some j I . Therefore the chain of primary submodules N i terminates
3. The Primary Radical of Submodules of Free Modules In this section we describe the elements of prad F N , where N is a finitely generated submodule of the free module F . Let n be a positive integer and let F be n the free R -module R . Let xi F 1 i m for some m Z , then xi xi1 , xi 2 , , xin , 1 i m , for some xij R , 1 i m , 1 j n . We set x11 x x1 x2 xm 21 xm1
x12 x1n x22 x2n M m n R xm 2 xmn
Thus the jth row of the matrix x1 x2 xm consists of the components of the element x j in F . Let A aij M mn R . By a t t minor of A we mean the determinant of a t t submatrix of A , that is a determinant of the form: ai 1 j 1 ai 1 j t
ai t j 1 ai t j t
where 1 i 1 i t m , 1 j 1 j t n . For each 1 t min m, n . We denote by At the ideal of R generated by the t t minors of A . Note that A1 in1 mj 1 Raij A2 A3 Ak , where k min m, n . The key to the desired result is the following two propositions.
contradicts the hypothesis.
3.1. Proposition 2.10. Proposition Let M be a finitely generated R-module. If every primary submodule of M is the p-radical of a finitely
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Let R be a ring and F be the free R -module R , m for some positive integer n . Let N i 1 Rxi be a finitely generated submodule of F where m n . If n
APM
L. J. M. ABULEBDA
rx1 x2 xm t
r prad F N , then
in
P l for every prime ideal P such that
P for every maximal ideal P such that
x1 xn t P for some l Z , 1 t m 1 Proof. Suppose r r1 , r2 , , rn prad F N where
l
ri R , 1 i n . Let P be any maximal ideal of R and Z such that 0 x1 xm t P . By proposition (2), there exists c R \ P , si R , 1 i m and pi P such that cr s1 x1 s2 x2 sm xm p , where p p1 , p2 , , pn , that is, if xi xi1 , xi 2 , , xin where xij R 1 i m,1 j n , then 3.1) cri s1 x1i s2 x2i sm xmi pi ; 1 i n Suppose that 1 t m 1, let 1 i 1 i 2 i t 1 m , 1 j 1 j t n. Let
rj (1) Xt
xi (1) j (1)
xi (t 1) j (1)
r
which is a t t minor of (3.1)
cX t
xi (1) j (1)
xi (t 1) j (1)
xi (1) j (1) xi (t 1) j (1)
xi (1) j (t )
xi (t 1) j ( t ) x1 x2 xm . Then by
r x1 xm t Pl for every prime ideal P such that 0 x1 xm t P l for some l Z and 1 t m 1 . Let P be any prime ideal of R and k any positive integer. It is enough to show that r K N , P k for all k Z+ . If 0 x1 x2 xm 1 P k , then
ri r x1 x2 ··· xm 1 P k , hence
0 x1
x2 xm 1 P . Note that 0 x1 x2 xm m 1 0 P k for all k . Thus there exists 1 t m such that 0 x1 kx2 xm t P, but 0 x1 x2 xm t 1 is a subset of P , there exists 1 i 1 i t m, 1 j 1 j t n , such that xi (1) j (1) xi (t ) j (1)
rj
rj (1)
xi (t 1) j (t )
xi (1) j
xi (1) j (1)
xh j (t )
xi (1) j (t )
xi (t ) j
xi (t ) j (1)
xi (t 1) j ( t )
xi (1) j (t ) xi (t ) j (t )
P
rj ( t ) xi (1) j (t )
Pk
xi (t ) j (t )
Expanding this determinant by first column we find that drj ai (1) xi (1) j ai (t ) xi (t ) j P k where
p j (t ) xi (1) j (t ) P l .
rj (1) xi (1) j (1)
xi (t 1) j ( t )
ai ( h ) 1 xi ( h -1) j (1) xi ( h 1) j (1) h
t
P with 0 x1 x2 xm t P for some Z and
1 t m 1 .
xi (t ) j (1)
rj ( t ) xi (1) j (t )
xi ( h -1) j (t )
xi ( h 1) j (t )
xi (t ) j (t )
For each 1 h t. Note that d and ai ( h ) 1 h t are independent of j . Thus drj ai(1) xi(1) j ai (t ) xi (t ) j P k 1 j n.
3.2. Proposition Let R be a ring and F be the free R -module R , n
m
for some positive integer n . Let N Rxi be a finii 1
tely generated submodule of F r x1 x2 xm t in
r r1 , r2 , , rn P k F K N , P k . Suppose SSS
By hypothesis, for each 1 j n
xi (1) j (t )
which is primary with c P (note that, here p j (1) , , p j (t ) P ) hence X t P P or X t P . It follows r x1 x2 ··· xm P for every maximal ideal
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Proof. Suppose
crj (t )
xh j (1) m sh xi (1) j (1) h 1 xi (t 1) j (1) p j (1)
rj ( t )
0 x1 xm t Pl for some l Z 1 t m 1 , then r prad F N .
d
crj (1)
347
where m n . If
i.e. dr Rx1 Rx2 Rxm P k F N P k F with d P , hence r K N , P k .Thus r prad F N .
3.3. Proposition Let M 1 and M 2 be R -modules and APM
L. J. M. ABULEBDA
348
M M1 M 2
m , m m M , i 1, 2 1
2
i
i
Let N be a proper submodule of M 1 , then x prad M1 N if and only if x, 0 prad M N 0 . Proof: Suppose first that x prad M1 N . Let K be any primary submodule of M such that N 0 K . Let K ' m M 1 m, 0 K . K is a submodule of M 1 and if K M 1 then K is a primary submodule of M 1 since, if rm K where r R and m M 1 , then rm, 0 K , so r m, 0 K , which is primary submodule of M , hence either m, 0 K thus m K or r n M K for some n Z that is, r n M 1 M 2 K , so r n M 1 0 r n M 1 M 2 K , therefore r n M 1 0 ) r n M 1 0 K , thus r n M 1 K for some n Z , that is r K : M 1 . Hence K is a
primary submodule of M 1 containing N . Thus x K , so x, 0 K . It follows x, 0 prad M N 0 . Conversely, suppose that x, 0 prad M N 0 . Let Q be a primary submodule of M 1 such that N Q . Then Q M 2 is a primary submodule of M containing N 0 . Hence x, 0 Q M 2 , that is x Q so x prad M1 N . Now, we have the main result of this section.
1 t min m 1, n . Proof. Let k min m 1, n . Suppose first k m 1 , that is m n , by proposition (3.1), if r prad F N , then r x1 x2 xm t in P for every maximal ideal P such that
0 x1 xn t Pl for some l Z ;
1 t min(m 1, n) . Now suppose k n, i.e. n m 1 . Let G R ( m 1) , r r1 , r2 , , rn , xi xi1 , xi 2 , , xin for some rj R and xij R , 1 i m, 1 j n . By proposition (3.3), r prad F N if and only if r1 , r2 , , rn , 0, 0, , 0 in pradG N . Where
N i 1 R xi1 , xi 2 , , xin , 0, 0, , 0 . m
Now apply proposition (3.1) to obtain the result. The following example will illustrate application of the proposition (3.2).
3.5. Example Let R Z , F Z 3 and N be the submodule Z 1,3,5 Z 2 1,1,1 of F . Then r1 , r2 , r3 prad F N if 3r1 r2 , 5r1 r3 , 5r2 r3 in 2Z and r1 2r2 r3 0 .
3.4. Theorem Let R be a ring and F be the free R -module R , m for some positive integer n . Let N i 1 Rxi be a finitely generated submodule of F where m n . If r prad F N , then r x1 x2 xm t in
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n
[1]
R. L. McCasland and M. E. Moore, “On Radicals of Submodules of Finitely Generated Modules,” Canadian Mathematical Bulletin, Vol. 29, 1986, pp. 37-39.
P for every maximal ideal P such that
0 x1 xn t Pl for some l Z ;
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