Linear Algebra over a Ring/Printable version

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Linear Algebra over a Ring

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https://en.wikibooks.org/wiki/Linear_Algebra_over_a_Ring

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Modules and linear functions

Definition (module):

Let be a ring. A left -module is an abelian group together with a function , denoted by juxtaposition, that satisfies the following axioms for all and :

Definition (homogenous):

Let , be left modules over a ring . A function is called homogenous if and only if for all and the identity

holds.

Definition (linear):

Let , be left modules over a ring . A function is called linear if and only if it is both homogenous and a morphism of abelian groups from to .

Theorem (first isomorphism theorem):

Let and be left modules over a ring . Let be linear. Then

.

Proof:

Exercises[edit | edit source]

  1. Prove that for a function between left -modules, the following are equivalent:
    1. is linear
    2. For all and , we have and
    3. For all and , we have
    4. For all and , we have


Free modules and matrices

Definition (free module):

Let be a ring, and let be an arbitrary set. Then the free -module over , denoted , is defined to be the -module whose elements are functions

which are zero everywhere on except on finitely many elements, together with pointwise addition and scalar multiplication.

Proposition (basis of a free module):

Let be a ring, and let be a set. Then a basis for the free -module over is given by the functions

By abuse of notation, we will write instead of . Hence, the above proposition implies that we may denote an element as a sum

,

where only finitely many are nonzero.

Proof: Let be any function that is everywhere zero except on finitely many entries, and let . Then we have

.


Direct product, direct sum and tensor product

Definition (free module over a set):

Let be any set, and let be a ring. Then the free module is defined to be the module

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together with the module operation

and the obvious addition

.

Definition (tensor product):

Let be a ring and let be -modules. The tensor product of the modules is defined as the -module

,

where is the following submodule:

.

Proposition (universal property of the tensor product):

Let be a ring and let be -modules. Then the tensor product satisfies the universal property that for each -module and each multilinear map , there exists a unique linear map such that

.

{{proposition|tensor product as multifunctor|Let be a ring. Then for each , the tensor product yields a multifunctor

.

Whenever and are -modules and for , are morphisms, the morphisms that turn into a multifunctor are given by

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} f_1 \otimes \cdots \otimes f_n: M_1 \otimes \cdots \otimes M_n \to N_1 \otimes \cdots \otimes}} {{proposition|associativity of the tensor product|Let <math>R} be a ring <math>


Homomorphism and dual modules

Proposition (multiple of module homomorphism by a ring element over a commutative ring is module homomorphism):

Let be modules over the commutative ring . Let be a homomorphism of -modules, and let . Then the function

is an -module morphism too.

Proof: Let and . Then and since is commutative, also .

{{definition|homomorphism module|Let be a commutative ring

Definition (dual):

Let be a commutative ring, and let be an -module. Then has a module structure given by addition and pointwise multipliction (since the category of modules over a ring is additive and the addition that is implied is compatible with pointwise multiplication, under which is closed), and this module, denoted more briefly by , is called the dual of .

Proposition (the category of modules is abelian):

Let be a ring. Then the category of left -modules is abelian. Also, the category of right -modules is abelian.

Proof: Existence of kernels and cokernels has been dealt with, and the same holds for existence of binary biproducts, since then product and sum coincide in this category. Also, the Noether isomorphism theorem holds. Then, the category is additive. Indeed, the summation on the homomorphism group indicated above is precisely the addition in the categorical sense.


Modules over Bézout domains

Proposition (extending a single element of a free, finite-dimensional module over a Bézout domain to a basis):

Let be a Bézout domain, and let be the -dimensional free module over . Let be such that . Then there exist elements such that is a basis of .

Proof: We proceed by induction on . If , the statement becomes trivial, so that the induction base is handled. Let now a general be given; we reduce the claim to the same statement for .

Indeed, if is given, we may set

,

and then . Hence, there are elements such that

.

We define

,

so that

and ,

where for , as usual, denotes the vector whose every component is zero except the one in the -th place, which is one. From these equations it is evident that

.

Also, and are linearly independent, because if there existed such that , then and in particular , so that either and and are linearly independent (almost) by definition, or , so that whenever (once again, if , then the vectors are automatically linearly independent). But this implies that for all , so that contrary to the assumption .

It now suffices to note that

by the first Noether isomorphism theorem applied to the homomorphism that forgets the first component, which reduces the dimension of the problem by one as we have .

Theorem (a minimum cardinality generating set of a finitely generated torsion-free module over a Bézout domain is a basis):

Let be a Bézout domain, and let be a finitely generated torsion-free module over . Then every generating set of whose cardinality is minimal is a basis of .

Proof: We proceed by induction on , where is the cardinality of a minimum cardinality generating set of . Let be a minimum cardinality generating set of . Then we have a homomorphism

such that for , where is the element of whose every entry is zero, except for the -th entry, which is one. Since is surjective, the first isomorphism theorem implies that

,

where . If , then the claim is demonstrated. We lead to a contradiction. Indeed, in this case there is a nonzero vector . We denote and set . Then , for otherwise the image of in via the canonical projection would be a torsion element. Since we have , we may extend to a basis of , and then is generated by the images of via the canonical projection. Yet this contradicts the minimality of .

Theorem (Dedekind's theorem):

Let be a Bézout domain, and let be a torsion-free module over . Then every finitely generated submodule is free.

Proof: If is torsion-free, then every submodule of is torsion-free as well. Hence, as in the theorem's statement is torsion-free and finitely generated. Hence, there exists a basis of .

Proposition (basis extension over Bézout domains):

Let be a Bézout domain, and let be a torsion-free module over . If are linearly independent vectors in such that is torsion-free and finitely generated, then we find an and vectors such that the vectors constitute a basis of .

Proof: Since is torsion-free and finitely generated, upon choosing a generating set of minimal cardinality we obtain a basis of that module. We shall denote this basis by . If moreover

is the canonical projection, we choose a for each . We claim that is a basis. First, we prove that these vectors are linearly independent. Indeed, suppose that are such that

.

Since is a homomorphism,

,

so that . Thus

,

which by virtue of the linear independence of implies .

It remains to show that is a generating set of . In order to do so, suppose that is arbitrary. We choose such that

.

We further define , so that . But by the definition of , this means that , so that there are such that

, hence ,

which shows that is a generating set of since was arbitrary.

Theorem (dimension formula):


Modules over principal ideal domains

Proposition (torsion-free modules over principal ideal domains are free):

Let be a principal ideal domain, and let be a torsion-free module over . Then is free.

(On the condition of the axiom of choice.)

Proof: We consider the set of sets such that is linearly independent and is torsion-free. This set may be equipped with the partial order that is given by inclusion. Suppose then that is a totally ordered set, and is a family such that . We claim that an upper bound for this chain is given by the union of the sets , which we shall denote by . Indeed, is linearly independent, since any linear relation within involves only finitely many elements of , and we may find a sufficiently large (w.r.t. the order of ) such that all these elements are contained within , so that by the linear independence of the given linear relation must be trivial. Moreover, has the property that is torsion-free, since if and are given such that , but (ie. the equivalence class of in is torsion), then is a linear combination of finitely many elements of , so that once more we find a sufficiently large such that , and then the equivalence class of in is torsion, a contradiction.

Thus, Zorn's lemma may be applied, and it yields a maximal linearly independent such that is torsion-free. We lead the assumption to a contradiction. Indeed, if we had , then there would be an element . The set will then be linearly independent, for if there was a linear relation

(where and ),

then we would have and the equivalence class of in would be torsion.

Theorem (Dedekind's theorem):

Let be a principal ideal domain. Whenever is a free -module and is a submodule, is free as well.

(On the condition of the axiom of choice.)

Proof: Since is a submodule of a torsion-free module, it is itself torsion-free. Thus, the theorem holds, since torsion-free modules over principal ideal domains are free.


Multilinear algebra

Definition (multilinear function):

Let be a ring, and let be -modules. Then the set of -multilinear functions from to is the set

Failed to parse (unknown function "\middle"): {\displaystyle L(M_1, \ldots, M_n, K) := \left\{ f: M_1 \times M_n \to K \middle| \forall j \in [n]: \forall m_1 \in M_1, \ldots, m_n \in M_n, l_j \in M_j, r \in R: f(m_1, \ldots, m_j + r l_j, \ldots, \right\}} .

Proposition (equivalent definition of tensor product of free modules using multilinear functions):

Let be a ring, and let be free, finitely generated -modules. Then if we alternatively define

,

and let the elementary tensors be , then the from this definition satisfies the same universal property as the usual tensor product . In particular, the two tensor products are canonically isomorphic.

Proof: For , let be a basis of , where is the respective finite index set. Given any -module and any multilinear map , we want a unique linear function such that , where is the map that sends a tuple to the respective elementary tensor.


Projective and injective modules

Theorem (Baer's criterion):

A module over a commutative ring is injective if and only if for every ideal , every homomorphism of -modules may be extended to a homomorphism of -modules , ie. extended to a homomorphism that satisfies .

(On the condition of the axiom of choice.)

Proof: The "only if" part is obvious from the definition of injectivity. Conversely, assume that satisfies the extension property described in the theorem statement. Let now be -modules, let be a homomorphism and let be an injection. By the definition of injective modules, we have to prove that there exists a homomorphism that satisfies , ie. that extends wrt. the inclusion , as the algebraists say. Now the image is a submodule of that is isomorphic to by the injectivity of via , and hence precomposition of the inverse of with yields a homomorphism from to . Now we partially order all extensions of this isomorphism by the order that if and only if

  1. the domain of is contained within the domain of , and
  2. and coincide on the domain of .

By Zorn's lemma, which applies since each chain with respect to this order has an upper bound (namely the function that assigns to an element within the union of the respective domains the value that any of the respective homomorphisms assumes on it), there exists a maximal homomorphism with respect to that order. We claim that is defined on all of . Indeed, upon assuming otherwise, we may choose an element where is not defined. Let be the domain of definition of , so that . Define the ideal

of ; it may be the zero ideal. is defined on , and yields a homomorphism on . By assumption, this homomorphism may be extended to a homomorphism . We then define ; a submodule of that is strictly larger than , since it contains . On , we define the homomorphism

for and ,

which is a proper extension of , so that was not maximal, a contradiction. Hence, was defined on all of from the beginning and yields the desired extension of .


Chain complexes of finitely generated free modules

Proposition (every chain complex of finitely generated free modules over a Bézout domain is the direct sum of some of its subcomplexes with at most two nonzero terms):

Let

be a chain complex whose objects are finitely generated free modules over a Bézout domain . This chain complex is then the countable direct sum of chain complexes of the form

,

where and .

(On the condition of the countable choice.)

Proof: We shall construct a direct sum decomposition

,

where . Once this is accomplished, we have in fact obtained a direct sum decomposition of the initial chain complex, because elements of are mapped to zero by , and elements of are mapped to due to the chain complex condition .

In order to achieve this decomposition, we invoke Dedekind's theorem for Bézout domains, which tells us that is finitely generated and free; indeed, it is finitely generated, since a generating set is given by the image (via ) of a generating set of . Let thus be a basis of . For each , we choose an arbitrary, but fixed , and then we define . This yields the desired direct sum decomposition. Indeed, , since whenever

for some elements of , applying to both sides of this equation and using its -linearity yields

,

which implies . Moreover, , since if is arbitrary, we may select such that

,

from which we may easily deduce that

.

Proposition (every chain complex of finitely generated free modules over a Bézout domain splits as the direct sum of two types of elementary chain complexes):

Let

be a chain complex whose objects are finitely generated free modules over a Bézout domain . This chain complex is the countable direct sum of copies of the following two chain complexes:

  1. for an element , ie. the arrow represents the function which is given by multiplication by
(On the condition of the countable choice.)

Proof: Using the notation of the last theorem, we have , where is finitely generated and is sent to zero by .