TopicS covered: Finitely Generated Free-modules over a P.I.D. [Finitely generated free modules: examples and its Properties, properties of submodules of finitely generated free modules, Finite abelian groups and cyclic modules. ]

FACULTY OF SCIENCE

RAM JI DEPARTMENT : MATHEMATICS

PANDEY COURSE : M.A./M.Sc. (Mathematics)

Assistant Professor SEMESTER : II (Second)

PAPER : First (MAT2TH02) CONTACT ADDRESS: Department of Mathematics, References: Ewing Christian College, 1. Ramji Lal: Algebra (Volume 2), Springer Prayagraj – 211003 Publication.

2. C. Musili: Introduction to Rings and , PHONE: Narosa Publication. +91 6394787490 3. D.S. Dummit & R. M. Foote: Abstract Algebra,

John Wiley Publication. EMAIL: 4. F. W. Anderson & K. R. Fuller: Rings and [email protected] Categories of Modules. [email protected]

Finitely Generated Free Modules over a P.I.D. MODULE THEORY

OBJECTIVE In this study material you will learn the following:  Finitely generated free modules : Definition, examples and non-examples.  Properties of Finitely generated free modules.  Existence of basis of a submodule of a finitely generated free modules over a P.I.D.  Properties of submodules of finitely generated free modules over a P.I.D. and finite abelian groups.  Free abelian groups and cyclic modules.

Definition: (Finitely generated free module) A R-module M is said to be finitely generated free module if and only if it has a basis consisting of finite number of elements.

Examples: 1. The R-module , where R is any with identity, is finitely generated free module as {, ,…, } is a basis of . where = (0,0,…,1,…,0) (1 is in the place). 2. ℤ as ℤ-module is finitely generated but not free. (Justify)

[Hint: ℤ is a finite set and so it must be generated by whole ℤ, which shows ℤ is finitely generated] 3. The collection of all sequences in a ring R is a R-module, where R is a ring with identity viz free but not finitely generated. (Justify)

[Hint: Find a countable basis for above R-module as taken in Example 1 (simply extend the n-tuples to infinite sequence)]

Proposition: A finitely generated free R-module having basis consisting of n elements is isomorphic to .

Proof: Let M be a free R-module with basis ={, ,…, } consisting of n elements. As we know ={, ,…, } is a basis of , so define a map : ⟶ by, () = ;∀ = 1,2,…, then, as M is free, so it must satisfies homomorphism extension property. So, can be uniquely extended to a homomorphism :̅ ⟶ such that ̅ () = () = ; ∀ = 1,2,…, Claim: ̅ is an .

For, = ∈ ̅ then ̅ () =0

⟹ ̅ =0 ̅ ̅ ⟹ () = 0 (as f is a homomorphism)

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⟹ =0 ⟹ = 0 ;∀ = 1,2,…,

⟹ = =0 Hence, ̅ = {0} ⟹ ̅ is injective. Further, Let ∈ , then = ,∃ = ∈ such that

(̅ ) = ̅ = (̅ ) = = Showing that, ̅ is surjective. Hence, ̅ is an isomorphism. Consequently, ≅ .

Proposition: Any two R-modules having equal no. of elements in their bases are isomorphic. Proof: Proof is simple and is left for the exercise.

[Hint: Approach 1: Let M and N are two R-modules with bases = {, ,…, } and X={, ,…, } respectively and proceed in the same way as previous proposition

Approach 2: Let M and N are two R-modules with bases consisting of n elements, then by previous proposition ≅ and ≅ , showing that ≅ .]

Theorem: If N be a submodule of R-module M, such that M/N is free then ≅ ⨁ /

Sketch of the Proof: If ={ + ,…, + } is a basis of M/N (as M/N is free) and is the module generated by {,…, }, then ≅ /. (why?) Now, we claim that M is direct sum of its submodules N and L (How?), resulting in = ⨁ ≅ ⨁ / which proves the theorem.

Proof: Suppose, ={ + , + ,…, + } is basis of M/N. Claim: = {, ,… , } is linearly independent. for, consider + +⋯+ =0 ⟹ ( + +⋯+ ) + = ⟹ ( + ) + ( + ) +⋯+ ( + ) = ⟹ = =⋯= = 0 Let =< >, then L is a submodule of M with basis .

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Finitely Generated Free Modules over a P.I.D. MODULE THEORY

As, both L and M/N have basis with same no. of elements, so by defining an isomorphism : ⟶ , () = + , we have ≅ /. Claim: = ⨁ for, let ∈ ⟹ + ∈ ⟹ + = ( + ) + ( + ) +⋯+ ( + ) ⟹ + = ( + +⋯+ ) + ⟹ + = + , where = + +⋯+ ∈ ⟹ ( − ) ∈ ⟹ ( − ) = () ⟹ = + Showing that every element of M can be expressed as sum of elements of N and that of L. So, = + , for uniqueness, let ∈ , ∈ such that + =0 ⟹ ( + ) + = ⟹ + =0+ (, + = , ∈ ) ⟹ () = (0) ⟹ = 0 (, ℎ) and so we get =0. Which proves the uniqueness. Hence, = ⨁ but ≅ / and hence, ≅ ⨁ /.

Theorem: If M is a finitely generated free R-module having basis consisting of n elements, where R is a PID, then every non-zero submodule of M is also free having basis consisting of atmost n elements. Sketch of the Proof: Step – 1:

If = {, ,… , } is a basis of M then we prove it by using second principal of mathematical induction on . For = 1, M must be isomorphic to R (why?) and every submodule of R is ideal of R which must be Prinicipal Ideal generated by some element . Which proves the theorem for =1. Step – 2: For = , we first construct a module which consists a basis with (m-1) elements. If N is non-zero submodule of N then two cases arises: Step – 3: Case 1: =< > ( ), then theorem is true for all submodules with basis consists of 1 element. Case 2: ≠ < >, then ()≅ (How?) ∩ Then, () is free submodule of having basis consisting of atmost (m-1) elements. (by induction hypothesis)

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Finitely Generated Free Modules over a P.I.D. MODULE THEORY

Also, ∩< > is a submodule of N such that is free, so by previous theorem, ∩ ≅( ∩< >)⨁ , so N is free having basis consisting of atmost 1+(m-1) =m elements. ∩ Which proves the theorem.

Proof: Suppose, R is a PID and M is free R-module with basis = {, ,… , } . We apply second principal of Mathematical Induction on .

=1: Suppose =< > and R is also free R-module with basis {1} ∴ ≅ Now, let A be a non − zero submodule of R then, = , for some ∈ ∗ (as R is a PID, so every ideal (submodule) must be principal ideal) ⟹ =< >, and {} is linearly independent. ∴ has a basis {} ⟹ is free. Suppose, the theorem is true for all < . = : Let = {, ,… , } be a basis of M. Define, natural homomorphism : ⟶ by () = +< > ;∀ ∈ Claim: is free with basis = {(), (),…, ()} For, as = {, ,… , } is generating subset of M ⟹ {(), (), (),…, ()} is generating subset of . ⟹ { (), (),…, ()} is generating subset of . (as () = +< >=< > ) Again, let () + () +⋯+ () =< > ⟹ (+< >) + (+< >) +⋯+ (+< >) =< > ⟹ ( + +⋯+ )+< > = < > ⟹ ( + +⋯+ ) ∈< > ⟹ ( + +⋯+ ) = , ∈ ⟹ − + + +⋯+ =0 ⟹ = =⋯= = 0 ( ) Hence, is free with basis = {(), (),…, ()}. So, by induction hypothesis the theorem is true for . Now, Suppose N be a non-zero submodule of M, then Case I: If =< > ( ), and then < > is free with basis {} consists of one element. Hence, the theorem.

Case II: If ≠< >, then consider the restriction map of natural homomorphism, |: ⟶ () then, | = ⋂ = ∩ < > ∴ By Fundamental theorem,

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Finitely Generated Free Modules over a P.I.D. MODULE THEORY

()≅ ∩ < > but, () is free submodule of with basis consists of at most (m-1) elements (by induction hypothesis.) Also, ∩< > is a submodule of N such that is free, so by previous theorem, ∩ ≅( ∩< >)⨁ ∩ But ∩< > ⊆ < >, viz free with basis consists of at most 1 element. Hence, ( ∩< >)⨁ and consequently N is free with basis consists of at most 1+(m-1)=m. ∩ Which proves the theorem.

Remarks: 1. Any submodule of a free left R-module M, where R is a PID, is free.

[Hint: As, we saw in previous sheet, second theorem, that “If M is a finitely generated free R-module, where R is a PID, then every non-zero submodule of M is also free” and obviously zero submodule ( {0}) is also free submodule of any module with basis ]

2. Submodules of a free module need not be free.

[Hint: {0,2,4} as ℤ-module is a submodule of ℤ as ℤ-module and clearly ℤ as ℤ-module is free but its submodule {0,2,4} as ℤ-module is not free] 3. Any subgroup of finitely generated abelian is also finitely generated.

[Hint: Clearly every abelian group is a ℤ−module and ℤ is a PID. So, the remark 1 holds for abelian groups also.]

4. Every finitely generated R-module is isomorphic to quotient of a free module.

Proof: Let M be a R-module generated by the finite set = {, ,…, }. Define a relation, : ⟶ by

(, ,…, ) = + +⋯+ = . Then, obviously is an epic (surjective module homomorphism) [Prove it] Hence, by Fundamental theorem of Module Homomorphism, ≅ so, M is isomorphic to factor of free module .

5. Every submodule of finitely generated module over a PID is finitely generated.

Proof: Let M be finitely generated R-module generated by {, ,…, }. Define a relation, : ⟶ by

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Finitely Generated Free Modules over a P.I.D. MODULE THEORY

(, ,…, ) = + +⋯ = . Then, obviously is an epic (surjective module homomorphism) [Prove it] Now, let N be any submodule of R-module M, then ⟹ () is a submodule of . But we know that every submodule of is finitely generated ⟹ () is finitely generated and generated by a finite set (say S) ⟹ (()) is generated by finite set () ⟹ is generated by finite set () (as is surjective, so, () = ) Showing that, N is finitely generated and generated by finite set () Definition: (Free abelian groups) An abelian group G is said to be free abelian group if and only if G is free a ℤ- module.

Remarks: 1. A free abelian group is also known as free group. 2. Every subgroup of a finitely generated free-abelian group is free. (follows from Remark 1, as every abelian group is a ℤ−module and ℤ is a PID)

Examples: 1. ℤ is free group as it is free ℤ-module with basis {1}.

2. ℤ is not a free group as it is not free ℤ-module. Definition: (Cyclic Module) A R-module M is said to be cyclic module if and only if it is generated by a single element. If M is a cyclic R-module, generated by a single element, then =< >.

Examples: 1. {} is cyclic ℤ - module generated by {0}. 2. - module R is cyclic module generated by {1}. 3. Every submodule of a cyclic module over a PID is cyclic. (follows from Remark 1, every submodule of cyclic module must be free but the basis of any submodule of cyclic module must contain at most 1 element, hence it must be cyclic too.)

Theorem: A R-module M is cyclic if and only if it is homomorphic image of R. Proof: Suppose M is cyclic R-module generated by {x} (i.e. M=< x > ) Define : ⟶ by, () = ;∀ ∈ Then, is epic (surjective homomorphism) ) [Prove it] and hence, = () (. . ℎℎ ) Conversely, Suppose, M is equal to homomorphic image of R ⟹ ∃ a surjective homomorphism : ⟶ Now, as we know that R as R-module is generated by {1}

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⟹ () is generated by { (1) } ⟹ is generated by { (1) } (as is epic so, = () ) ⟹ is cyclic module.

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