# Finite Generated Topological Modules

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#### Finite Generated Topological Modules

M. Regina

Assistant Professor, Department of Mathematics,

Thassim Beevi Abdul Kader college for women, Kilakarai, Ramanathapuram (Dis)

Assistant Professor, Department of Mathematics,

Thassim beevi abdul kader college for women, Kilakarai, Ramanathapuram (dis)

2. Deepa

M.Phil. Scholar, Department of Mathematics,

Thassim beevi abdul kader college for women, Kilakarai, Ramanathapuram (dis)

Abstract:- This section discusses the finite generated topological modules(T.M) If a finite generated (T.M) is Hausdorff, then its structure is the usual one, meaning by this that there exists an isomorphism (for the TM structure) of M onto () . If M is a finite generated module then M is algebraically isomorphic to ().this section was developed by topological modules with hausdorff space.

Keywords:- Finite generated, hausdorff space, topological modules, isomorphic, dim(M)

INTRODUCTION

Let M modules over the field . We know that generated of M, denote by dim(M). If dim(M)is finite, we say that M is finite generated otherwise M is infinite generated.

LEMMA 1.2

Let M be a topological module over and L and linear functional on M. Assume L(m)0 for some mM. Then the following are equivalent.

1. L is continuous.

2. The null space Ker (L) is closed in M.

3. Ker (L) is not dense in M.

4. L is bounded in some neighborhood of the origin in M.

Proof:

Let {1, 2,.} in . consider the mapping

: M

(1, 2,.)(11+22+.+ )

This is Algebraic isomorphism. Therefore, to conclude a) it remains to prove that is also a homeomorphism.

Let 1,2,. in M such that 1, 2, in . dim(M)=n, Step 1: is continuous.

Given any modules mM. There exist unique 1,2,.. in When n=1, we simply have =1 and so we are done by such that m=11+22+. +. This is can be preciselylemma 1.1 When n>1, for any (1,2,.) .

expressed by saying that mapping

M

we can write:

(1,2,.)=B ( (1), ())

(1,2,.) (11+22+..+)

1

=B ((

Ã—.Ã— )(1,.,))

is an algebraic isomorphism, between M . If M is a finite Where each

1

is def ed as

ve and B: R M

M is

generated module then M is algebraically isomorphic to dim ().

in

abo Ã—

If now we give to M the topological modules structure and wethe module addition in the topological Module M.

consider

Endowed with the Euclidean topology, then it is natural to ask if su an algebraic isomorphism is by any change a topological one.

Lemma 1.1

Hence, is continuous as composition of continuous

mappings. ch Step 2:

is open and b) holds.

We prove that step by induction on the generated dim(M) of

Let M be a topological module over field and vM then th

Let M be a topological module over field and vM then th

M.

eFor dim(M)=1, It is easy to see that is open, (i.e.) that the

following mapping is continuous

: M

inverse of ;

1: M

Proof:

For any , we have ()=M(()). Where : Ã—M

We have that

M=

is continuous.

given by

()= (, v) is clearly continuous by Definition of product topology and mapping

Ker(1)={mM:1(m)=0}

={ M:=0} = {0},

Which is closed in M. since M is hausdorff. Hence, by

S: Ã— is the scalar multiplication in the topological modulelemma 3.5,1 is continuous.

M which is continuous by definition of topological module. HenceThis implies that (b)

is continuous as composition of continuous

Holds.

In fact, if L is a non-identically zero functional on M (when L0, there is nothing to prove) then there exists a o

M,

*B is the modulo addition on M and so it is continuous.

Since M is an addition modulo. Hence g is continuous.

such that L ( )1.

Without loss of generating

we can assume L ( )=1. Now for any m .

Since dim(M)=1,

We have that m= for some . And so, L(m)=L ( )=.

Hence, L 1 which we proved to be continuous. Suppose now that both a) and b) hold for dim(M) n-1.

Let us first show that b) holds when dim(M)=n. let L be a non-identically zero functional on X. (when L0, there is nothing to prove),

then there exists a 0 M. such that L ( )0.

W.l.o.g. We can assume L ( )=1. Note that for any mM.the element m- L(m) (L).

therefore, if we take the canonical mapping :MM ker(L)

then (m)= ( L(m)) =L(m) ( ) for any mM.

This means that

M/Ker(L)=span { ( )} (i.e.) dim(M/Ker(L)=1

Hence, dim (Ker(L)) =n=1 and so by inductive assumption Ker(L) is topologically isomorphic to 1.

This implies that Ker(L) is a complete. Submodule of M.

Ker (L) is closed in M and so by lemma 1.5

We get L is continuous. By induction, we can conclude that

b) holds for any generated nN.

This immediately implies that a) holds for any generated nN. In fact, we just need to show that for any generate nN the mapping

1: M

Corollary:1.3 (Tychonoff thm)

Let nN. The only topology that makes a Hausdorff topological module is the Euclidean topology. Equivalently, on a finite generated Module there is a unique topology that makes it into a hausdorff topological module.

Proof:

We already know that endowed with the Euclidean topology is hausdorff topological module of generated n. Let us another on . S.T ( , ) is also hausdorff

topological module.

Then the identity map between (, ) and ( , ) is a topological isomorphism.

We get .

Corollary 1.4

Every finite generated hausdorff topological is complete.

Proof:

Let M be hausdorff topological module with dim(M)=n<. We know that M is topologically isomorphic to endowed with the Euclidean topology. Since the latter is a complete hausdorff topological module, so is M.

Corollary:1.5

Every finite generated linear submodule of a hausdorff topological module is closed.

Proof:

Let R be a linear submodule of a hausdorff topological module (M,) and assume that dim(R)=n<. Then R endowed with the submodule topology induced by is itself a hausdorff topological module. Hence, by corollary 1.4 R is complete and also closed.

M=

M=

=1

(1)

REFERENCE BOOK

Is continuous. Now for any

m=

m=

=1

we can write

M.

1. Atiyah, M. F.; Macdonald, I. G. (1969), Introduction to commutative algebra, Addison-Wesley Publishing Co., Reading,

1(m)=(1(m).(m)) where for any j{1, 2,…n} we define

:M by (m)=

Since (b) holds for any generated we know that each is continuous and so 1is continuous.

Step:3

This statement (c) holds.

Let g:MN be linear and {1,..} On M. For any j{1,..,n}

we define :g()N.

Mass.-London-Don Mills, Ont., pp. ix+128, MR 0242802

2. F.W. Anderson and K.R. Fuller: Rings and Categories of Modules,

3. Texts in Mathematics, Vol. 13, 2nd Ed., Springer-Verlag, New York, 1992.

4. Arkhangelskii, A.V., L.S. Pontryagin General Topology I, (1990) Springr-Verlag, Berlin.

=1

=1

Hence, for any m= M.

We have

=1

=1

g(m)=g (

)=

=1

=1

we can rewrite g as composition of continuous maps.

1

g(m)=B (1 Ã—Ã— )( (m))

where

• 1 is continuous.

• Each continuous by lemma 1.1