Abstract Algebra: Modules

Posted in math on May 13, 2020
PDF

Table of Contents

Vector Space over a Field

Vector spaces over a field are a special case of the more general notion of modules over a ring. Normally, textbooks define a vector space as a set equipped with two operations which obey a long list of axioms:

An (abstract) vector space (V,F,+,⋅)(V,\mathbb{F},+,\cdot) consists of

  1. A field F\mathbb{F}, whose elements are called scalars
  2. A set VV, whose elements are called (abstract) vectors
  3. A rule (+):V×V→V(+) : V \times V \rightarrow V for vector addition, satisfying
    1. (associativity) u+(v+w)=(u+v)+wu+(v+w)=(u+v)+w
    2. (commutativity) u+v=v+uu+v = v+u
    3. (additive identity) exists 0∈V0 \in V with v+0=vv + 0 = v for all v∈Vv \in V
    4. (additive inverse) for all v∈Vv \in V, exists (−v)∈V(-v) \in V with v+(−v)=0v + (-v) = 0
  4. A rule (⋅):F×V→V(\cdot) : \mathbb{F} \times V \rightarrow V for scalar multiplication, satisfying
    1. (scalar identity) 1F⋅v=v1_F \cdot v = v for all v∈Vv \in V
    2. (compatibility) (αβ)v=α(β(v))(\alpha \beta) v = \alpha(\beta(v))
    3. (distributes over vector addition) α(v+w)=αv+αw\alpha (v + w) = \alpha v + \alpha w
    4. (distributes over field addition) (α+β)v=αv+βv(\alpha + \beta) v = \alpha v + \beta v

We can state these properties more concisely by noticing that Property III is equivalent to the requirement that (V,+)(V,+) forms a commutative group.

An (abstract) vector space over the field F\FF is a commutative group (V,+)(V,+) together with a rule (⋅):F×V→V(\cdot) : \FF \times V \rightarrow V satisfying

  1. (scalar identity) 1F⋅v=v1_F \cdot v = v for all v∈Vv \in V
  2. (compatibility) (αβ)v=α(β(v))(\alpha \beta) v = \alpha(\beta(v))
  3. (distributes over addition) α(v+w)=αv+αw\alpha (v + w) = \alpha v + \alpha w
  4. (distributes over field addition) (α+β)v=αv+βv(\alpha + \beta) v = \alpha v + \beta v

Definitions 1a and 1b seem to present the set VV as the primary object of interest, relegating the scalars F\FF to the sidelines. The key to understanding modules is to turn this presumption on its head by treating F\FF as the distinguished object instead.

By partial application of the scaling operator (⋅):F×V→V(\cdot) : \FF \times V \rightarrow V, each scalar α∈F\alpha \in \FF corresponds to a linear map φa:v↦αv\varphi_a : v \mapsto \alpha v from VV to itself. Linear self-maps on VV constitute the endomorphism ring (End(V),+,∘)(\End(V), +, \circ), whose operations are pointwise addition and function composition. The vector space axioms ensure that the map φ⊡:F→(V→V)\varphi_\boxdot : \FF \rightarrow (V \rightarrow V) from field elements to linear self-maps is a ring homomorphism. We arrive at our third and final definition,

An (abstract) vector space over the field F\FF is a commutative group (V,+)(V,+) together with a ring homomorphism φ:F→End(V)\varphi : \FF \rightarrow \mathrm{End}(V).

The ring homomorphism defines the additive and multiplicative group actions on VV by scalars from the field F\FF.

Module over a Ring

For modules, we require only that the set acting on VV be a ring, rather than a field.

A module over the ring RR is a commutative group (M,+)(M,+) together with a ring homomorphism φ:R→End(M)\varphi : R \rightarrow \End(M) defining an action of RR on MM, where End(M)\End(M) is the set of group homomorphisms M→MM \rightarrow M.

Modules over a ring RR are called RR-modules, for short. An RR-module is called left if it arises from a left action, and right otherwise. As for vector spaces, we could unfold this definition into a list of axioms, but this would obfuscate the real purpose of modules: Many mathematical objects happen to be rings, and modules allow us to study rings by their action on a set (much like we can study groups via their representations).

Let MM be an RR-module. An RR-submodule of MM is a subgroup N⩽(M,+)N \subgroup (M,+) closed under the ring action, rn∈Nrn \in N for r∈Rr \in R, n∈Nn \in N.

Several important examples of modules are listed below.

  • If F\FF is a field, then F\FF-modules and F\FF-vector spaces are identical.
  • Every ring RR is an RR-module over itself. In particular, every field F\FF is an F\FF-vector space. Submodules of RR as a field over itself are ideals.
  • If SS is a subring of RR with 1S=1R1_S = 1_R, every RR-module is an SS-module.
  • If GG is a commutative group of finite order mm, then m⋅g=0m \cdot g=0 for all g∈Gg \in G, and GG is a (Z/mZ)(\ZZ / m \ZZ)-module. In particular, if GG has prime order pp, then GG is a vector space over the field (Z/pZ)(\ZZ/ p \ZZ).
  • The smooth real-valued functions C∞(M)\mathscr{C}^\infty(\mathcal{M}) on a smooth manifold form a ring. The smooth vector fields on M\mathcal{M} form a C∞(M)\mathscr{C}^\infty(\mathcal{M})-module.
  • For a ring RR, every RR-algebra has natural (left/right) RR-module structure given by the (left/right) ring action of RR on AA.

(Z\ZZ-modules) By definition, every Z\ZZ-module is a commutative group. Likewise, every commutative group (G,+)(G,+) becomes a Z\ZZ-module under the ring action defined for n∈Zn \in \ZZ, g∈Gg \in G by n⋅g={−a+a+⋯+a−(n times)if n>0−0if n=0−a−a−⋯−a(−n times)if n<0 n \cdot g = \begin{cases} \hphantom{-}a + a + \cdots + a \quad \hphantom{-}\text{(} n \text{ times)} & \text{if } n > 0 \\ \hphantom{-}0 & \text{if } n = 0 \\ -a - a - \cdots - a \quad \text{(} {-}n \text{ times)} & \text{if } n < 0 \end{cases} We conclude that Z\ZZ-modules and commutative groups are one in the same.

Modules over a Polynomial Ring F[x]\FF[x]

The polynomial ring F[x]\FF[x] is the space of formal linear combinations of powers of an indeterminate xx, with coefficients drawn from an underlying field F\FF. p(x)=p0+p1x+p2x2+⋯+pdxm(m∈N) p(x) = p_0 + p_1 x + p_2 x^2 + \cdots + p_d x^m \quad (m \in \NN)

Polynomials form a ring under entrywise addition and discrete convolution of coefficient sequences. The sum and product of p,q∈F[x]p, q \in \FF[x] have coefficients [p+q]k=pk+qk[p⋅q]k=∑j=0max⁡(n,m)pjqk−j [p+q]_k = p_k + q_k \qquad [p\cdot q]_k = \sum_{j=0}^{\max(n,m)} p_j q_{k-j}

Consider what it would mean for an F\FF-vector space VV to be an F[x]\FF[x]-module. We need a ring homomorphism φ:F[x]→End(V)\varphi : \FF[x] \rightarrow \End(V) describing the action of polynomials on vectors. Since φ\varphi preserves sums and products between F[x]\FF[x] and (End(V),+,∘)(\End(V),+,\circ) as rings, we find that the choice of a single linear map φ(x)∈End(V)\varphi(x) \in \End(V) determines the value of φ\varphi on arbitrary polynomials~p∈F[x]p \in \FF[x], φ(p)v=φ(∑k=1mpkxk)v=∑k=1mpkφ(x)kv \begin{aligned} \varphi(p) v = \varphi\left( \sum_{k=1}^m p_k x^k \right) v = \sum_{k=1}^m p_k \varphi(x)^k v \end{aligned} Similarly, any choice of ϕ(x)∈End(V)\phi(x) \in \End(V) yields a valid ring homomorphism, exposing a bijection between F[x]\FF[x]-modules and pairs (V,T∈End(V))(V, T \in \End(V)). {  F[x]-modules V  }⟷{F-vector spaces V with alinear map T:V→V} \bigg\{ \;\mathbb{F}[x]\text{-modules } V\; \bigg\} \longleftrightarrow \bigg\{ \substack{\small\text{$\mathbb{F}$-vector spaces $V$ with a}\\\small \text{linear map $T : V \rightarrow V$}} \bigg\} In general, there are many different F[x]\FF[x]-module structures a given F\FF-vector space VV, each corresponding to a choice of linear T:V→VT : V \rightarrow V.

The F[x]\FF[x]-submodules of an F[x]\FF[x]-module~VV are precisely the TT-invariant subspaces of VV, where T∈End(V)T \in \End(V) denotes the action of xx.

Each F[x]\FF[x]-submodule of VV is closed under actions by ring elements, including TT. Likewise, every TT-invariant subspace is closed under ring actions, which are all polynomials in TT.

Module Homomorphisms

An RR-module homomorphism is a map ϕ:M→N\phi : M \rightarrow N between modules which respects the RR-module structure, by preserving addition and commuting with the ring action on MM, ϕ(x+y)=ϕ(x)+ϕ(y)∀ x,y∈Mϕ(r⋅x)=r⋅ϕ(x)∀ x∈M,r∈R \begin{aligned} \phi(x + y) &= \phi(x) + \phi(y) & \forall\, x,y \in M \\ \phi(r \cdot x) &= r \cdot \phi(x) & \forall\, x \in M, r \in R \end{aligned}

The kernel of a module homomorphism is its kernel ker⁡ϕ=ϕ−1{0S}\ker \phi = \phi^{-1}\{0_S\} as an additive group homomorphism. A bijective RR-module homomorphism is an isomorphism. For any ring RR, the set HomR(M,N)\Hom_R(M,N) of homomorphisms between two RR-modules forms a commutative group under pointwise addition, (ϕ+ψ)(m)=ϕ(m)+ψ(m)(\phi + \psi)(m) = \phi(m) + \psi(m) for ϕ,ψ∈HomR(M,N)\phi, \psi \in \Hom_R(M,N). Moreover,

For a commutative ring RR, the group HomR(M,N)\Hom_R(M,N) forms an RR-module under the ring action R→End(HomR(M,N))R \rightarrow \End(\Hom_R(M,N)) given by (r⋅ϕ)(m)≡r⋅ϕ(m)∀ r∈R,m∈M,ϕ∈HomR(M,N) \begin{aligned} (r \cdot \phi)(m) &\equiv r \cdot \phi(m) &\forall\, r \in R, m \in M, \phi \in \Hom_R(M,N) \end{aligned}

Commutativity of RR guarantees that (r⋅ϕ)∈HomR(M,N)(r\cdot \phi) \in \Hom_R(M,N), since (r⋅ϕ)(s⋅m)=r⋅ϕ(s⋅m)(by definition)=rs⋅ϕ(m)(ϕ is a homomorphism)=sr⋅ϕ(m)(commutativity)=s⋅(r⋅ϕ(m))(by definition) \begin{aligned} (r \cdot \phi)(s \cdot m) &= r \cdot \phi(s \cdot m) & \text{(by definition)} \\ &= rs \cdot \phi(m) & \text{(} \phi \text{ is a homomorphism)} \\ &= sr \cdot \phi(m) & \text{(commutativity)} \\ &= s \cdot (r \cdot \phi(m)) & \text{(by definition)} \end{aligned}

Ring of Module Endomorphisms

Endomorphisms HomR(M,M)\Hom_R(M,M) form a unital ring, where (ϕ+ψ)(m)=ϕ(m)+ψ(m)(pointwise addition)(ϕψ)(m)=(ϕ∘ψ)(m)(composition)1HomR(M,M)=IdM(multiplicative identity) \begin{aligned} (\phi + \psi)(m) &= \phi(m) + \psi(m) & \text{(pointwise addition)} \\ (\phi \psi)(m) &= (\phi \circ \psi)(m) & \text{(composition)} \\ 1_{\Hom_R(M,M)} &= \mathrm{Id}_M & \text{(multiplicative identity)} \end{aligned} We write EndR(M)=HomR(M,M)\End_R(M) = \Hom_R(M,M) for the endomorphism ring of MM.

Let MM be a module over a commutative ring RR. The endomorphism ring EndR(M)\End_R(M) forms an RR-algebra, under the same ring action r↦φ(φr:m↦rm)r \stackrel{\varphi}{\mapsto} ( \varphi_r : m \mapsto rm) which defines MM as an RR-module.

This property is normally stated without reference to ring homomorphisms, but in these notes we wish to emphasize that the study of modules is really the study of . There is at least one subtlety, though: When defining MM as an RR-module, we required that φ⊡:R→End(M,+)\varphi_\boxdot : R \rightarrow \End(M,+) be a ring homomorphism from RR to the additive group endomorphisms on (M,+)(M,+). Now, we are asking whether each φr\varphi_r is also an RR-module homomorphism.

First, the additive group homomorphism φr∈End(M,+)\varphi_r \in \End(M,+) is also a module homomorphism, since for r,s∈Rr,s \in R and m∈Mm \in M, φr(s⋅m)=r⋅(s⋅m)(by definition)=(rs)⋅m1(associativity of scalars)=s⋅(r⋅m)(associativity of scalars)=s⋅φr(m)(by definition) \begin{aligned} \varphi_r(s \cdot m) &= r \cdot (s \cdot m) &\text{(by definition)} \\ &= (rs) \cdot m_1 &\hspace{4em}\text{(associativity of scalars)}\\ &= s \cdot (r \cdot m) &\text{(associativity of scalars)}\\ &= s \cdot \varphi_r(m) &\text{(by definition)} \end{aligned} Futher, φ⊡:R↦EndR(M)\varphi_\boxdot : R \mapsto \End_R(M) sending r↦φrr \mapsto \varphi_r is a ring homomorphism. φr1+r2(m)=(r1+r2)⋅m(by definition)=r1⋅m+r2⋅m(distributivity of scalars)=φr1(m)+φr2(m)(by definition)φr1r2(m)=(r1r2)⋅m(by definition)=r2⋅(r1⋅m)(R commutative)=(φr2∘φr1)(m)(by definition) \begin{aligned} \varphi_{r_1 + r_2}(m) &= (r_1 + r_2) \cdot m &\text{(by definition)} \\ &= r_1 \cdot m + r_2 \cdot m &\text{(distributivity of scalars)} \\ &= \varphi_{r_1}(m) + \varphi_{r_2}(m) &\text{(by definition)} \\ \varphi_{r_1 r_2}(m) &= (r_1 r_2) \cdot m & \text{(by definition)} \\ &= r_2 \cdot (r_1 \cdot m) &\text{($R$ commutative)} \\ &= (\varphi_{r_2} \circ \varphi_{r_1})(m) &\text{(by definition)} \end{aligned} Finally, each φr\varphi_r commutes with every element ϕ∈EndR(M)\phi \in \End_R(M), (φr∘ϕ)(m)=φr(ϕ(m))(composition)=r⋅ϕ(m)(by definition)=ϕ(r⋅m)(module homomorphism)=ϕ(φr(m))(by definition) \begin{aligned} (\varphi_r \circ \phi)(m) &= \varphi_r (\phi(m)) &\text{(composition)} \\ &= r \cdot \phi(m) &\text{(by definition)}\\ &= \phi(r \cdot m) &\text{(module homomorphism)}\\ &= \phi( \varphi_r(m)) &\text{(by definition)} \end{aligned}

By definition, every field F\FF is a commutative ring. Therefore, the endomorphisms EndF(V)\End_\FF(V) of any F\FF-vector space form an F\FF-algebra.

Quotient Modules

For groups and rings, recall that quotients are well-defined only for subgroups and subrings (ideals), respectively. For modules MM, it turns out that submodule N⩽MN \subspace M has a quotient M/NM / N, and the natural projection map π:M→M/N\pi : M \rightarrow M/N is a ring homomorphism with kernel ker⁡π=N\ker \pi = N. Similarly, each F\FF-vector subspace has a quotient F\FF-vector space arising as the kernel of some linear map.

Let RR be a ring. Let N⩽MN \subspace M be a submodule of the RR-module MM. The (additive, commutative) quotient group M/NM / N can be made into an RR-module under the ring action R→End(M/N)R \rightarrow \End(M/N) given by r⋅(x+N)=(r⋅x)+N∀ r∈R,x+N∈M/N \begin{aligned} r \cdot (x + N) &= (r \cdot x) + N &\forall\, r \in R, x + N \in M/N \end{aligned} The natural projection π:M→M/N\pi : M \rightarrow M/N mapping x↦x+Nx \mapsto x+N is an RR-module homomorphism with kernel ker⁡π=N\ker\pi = N.

(First Isomorphism Theorem) Let M,NM,N be RR-modules. The kernel of any module homomorphism ϕ:M→N\phi : M \rightarrow N is a submodule of MM, and M/ker⁡ϕ≅ϕ(M) M / \ker\phi \cong \phi(M)

Free Modules

The vector space concepts of linear combinations, bases, and span all have analogues in RR-module theory. We normally assume RR is a ring with identity.

Let MM be an RR-module. The submodule of MM generated by a subset A⊂MA \subset M is the set of finite RR-linear combinations RA={r1a1+⋯+rmam∣rk∈R,ak∈A,m∈N}⩽M RA = \{ r_1 a_1 + \cdots + r_m a_m \mid r_k \in R, a_k \in A, m \in \NN \} \subspace M A submodule N=RA⩽MN = RA \subspace M is finitely generated if A⊂MA \subset M is finite. A cyclic submodule N=RaN = Ra is generated by a single element a∈Ma \in M.

An RR-module FF is free on the subset A⊂FA \subset F if each nonzero x∈Fx \in F expands uniquely as an RR-linear combination of elements from AA, in which case AA is called a basis for FF. x=r1a1+⋯+rmam∃! rk∈R,ak∈A,∀ x∈F \begin{aligned} x &= r_1 a_1 + \cdots + r_m a_m & \exists!\, r_k \in R, a_k \in A, \forall\, x \in F \end{aligned}

In general, more than one basis may exist. If RR is commutative, every basis has the same cardinality, called the module rank of FF. Unlike for vector spaces, not every module has a basis (not every module is free).

Universal Property of Free Modules

Recall that every linear map T∈HomF(V,W)T \in \Hom_\FF(V,W) between F\FF-vector spaces is uniquely determined by its value on n=dim⁡Vn=\dim V points. RR-linear maps between free modules enjoy the same property, which is normally stated in the following way:

(Universal Property) For any set AA, there is a unique (up to isomorphism) free RR-module Free(A)\mathrm{Free}(A) satisfying the following universal property: for any RR-module MM and any function φ:A→M\varphi : A \rightarrow M, there is a unique RR-module homomorphism Φ:Free(A)→M\Phi : \mathrm{Free}(A) \rightarrow M such that Φ(a)=φ(a)\Phi(a) = \varphi(a),