$\providecommand{\FF}{\mathbb{F}} \providecommand{\ZZ}{\mathbb{Z}} \providecommand{\NN}{\mathbb{N}} \providecommand{\End}{\mathrm{End}} \providecommand{\Hom}{\mathrm{Hom}} \providecommand{\subgroup}{\leqslant} \providecommand{\subspace}{\leqslant}$

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 $(V,\mathbb{F},+,\cdot)$ consists of

1. A field $\mathbb{F}$, whose elements are called scalars
2. A set $V$, whose elements are called (abstract) vectors
3. A rule $(+) : V \times V \rightarrow V$ for vector addition, satisfying a. (associativity) $u+(v+w)=(u+v)+w$ b. (commutativity) $u+v = v+u$ c. (additive identity) exists $0 \in V$ with $v + 0 = v$ for all $v \in V$ d. (additive inverse) for all $v \in V$, exists $(-v) \in V$ with $v + (-v) = 0$
4. A rule $(\cdot) : \mathbb{F} \times V \rightarrow V$ for scalar multiplication, satisfying a. (scalar identity) $1_F \cdot v = v$ for all $v \in V$ b. (compatibility) $(\alpha \beta) v = \alpha(\beta(v))$ c. (distributes over vector addition) $\alpha (v + w) = \alpha v + \alpha w$ d. (distributes over field addition) $(\alpha + \beta) v = \alpha v + \beta v$

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

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

1. (scalar identity) $1_F \cdot v = v$ for all $v \in V$
2. (compatibility) $(\alpha \beta) v = \alpha(\beta(v))$
3. (distributes over addition) $\alpha (v + w) = \alpha v + \alpha w$
4. (distributes over field addition) $(\alpha + \beta) v = \alpha v + \beta v$

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

By partial application of the scaling operator $(\cdot) : \FF \times V \rightarrow V$, each scalar $\alpha \in \FF$ corresponds to a linear map $\varphi_a : v \mapsto \alpha v$ from $V$ to itself. Linear self-maps on $V$ constitute the endomorphism ring $(\End(V), +, \circ)$, whose operations are pointwise addition and function composition. The vector space axioms ensure that the map $\varphi_\boxdot : \FF \rightarrow (V \rightarrow V)$ 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 $\FF$ is a commutative group $(V,+)$ together with a ring homomorphism $\varphi : \FF \rightarrow \mathrm{End}(V)$.

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

Module over a Ring

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

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

Modules over a ring $R$ are called $R$-modules, for short. An $R$-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 $M$ be an $R$-module. An $R$-submodule of $M$ is a subgroup $N \subgroup (M,+)$ closed under the ring action, $rn \in N$ for $r \in R$, $n \in N$.

Several important examples of modules are listed below.

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

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

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

The polynomial ring $\FF[x]$ is the space of formal linear combinations of powers of an indeterminate $x$, with coefficients drawn from an underlying field $\FF$.

$p(x) = p_0 + p_1 x + p_2 x^2 + \cdots + p_d x^m \quad (m \in \NN)$

Polynomials form a ring\footnote{the polynomial ring $\FF[x]$ actually has the additional property of being an algebra, since $\FF$ embeds into the center of $\FF[x]$ via the ring homomorphism $(\alpha \in \FF) \mapsto (\alpha \cdot 1 \in \FF[x])$.} under entrywise addition and discrete convolution of coefficient sequences. The sum and product of $p, q \in \FF[x]$ have coefficients

$[p+q]_k = p_k + q_k \qquad [p\cdot q]_k = \sum_{j=0}^{\max(n,m)} p_j q_{k-j}$

Consider what it would mean for an $\FF$-vector space $V$ to be an $\FF[x]$-module. We need a ring homomorphism $\varphi : \FF[x] \rightarrow \End(V)$ describing the action of polynomials on vectors. Since $\varphi$ preserves sums and products between $\FF[x]$ and $(\End(V),+,\circ)$ as rings\footnote{We take some notational shortcuts. For instance, $\phi(x)^k$ is $\phi(x)$ composed with itself $k$ times, and $p_k$ refers to both the element of $\FF$ and to the map $(v \mapsto p_k v) \in \End(V)$.}, we find that the choice of a single linear map $\varphi(x) \in \End(V)$ determines the value of $\varphi$ on arbitrary polynomials~$p \in \FF[x]$, \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 $\phi(x) \in \End(V)$ yields a valid ring homomorphism, exposing a bijection between $\FF[x]$-modules and pairs $(V, T \in \End(V))$. $\bigg\{ \;\mathbb{F}[x]\text{-modules } V\; \bigg\} \longleftrightarrow \bigg\{ \substack{\small\text{\mathbb{F}-vector spaces V with a}\\\small \text{linear map T : V \rightarrow V}} \bigg\}$ In general, there are many different $\FF[x]$-module structures a given $\FF$-vector space $V$, each corresponding to a choice of linear $T : V \rightarrow V$.

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

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

Module Homomorphisms

An $R$-module homomorphism is a map $\phi : M \rightarrow N$ between modules which respects the $R$-module structure, by preserving addition and commuting with the ring action on $M$, \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 $\ker \phi = \phi^{-1}\{0_S\}$ as an additive group homomorphism. A bijective $R$-module homomorphism is an isomorphism. For any ring $R$, the set $\Hom_R(M,N)$ of homomorphisms between two $R$-modules forms a commutative group under pointwise addition, $(\phi + \psi)(m) = \phi(m) + \psi(m)$ for $\phi, \psi \in \Hom_R(M,N)$. Moreover,

For a commutative ring $R$, the group $\Hom_R(M,N)$ forms an $R$-module under the ring action $R \rightarrow \End(\Hom_R(M,N))$ given by \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 $R$ guarantees that $(r\cdot \phi) \in \Hom_R(M,N)$, since \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 $\Hom_R(M,M)$ form a unital ring, where \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 $\End_R(M) = \Hom_R(M,M)$ for the endomorphism ring of $M$.

Let $M$ be a module over a commutative ring $R$. The endomorphism ring $\End_R(M)$ forms an $R$-algebra, under the same ring action $r \stackrel{\varphi}{\mapsto} ( \varphi_r : m \mapsto rm)$ which defines $M$ as an $R$-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 \emph{ring actions}. There is at least one subtlety, though: When defining $M$ as an $R$-module, we required that $\varphi_\boxdot : R \rightarrow \End(M,+)$ be a ring homomorphism from $R$ to the additive group endomorphisms on $(M,+)$. Now, we are asking whether each $\varphi_r$ is also an $R$-module homomorphism.

First, the additive group homomorphism $\varphi_r \in \End(M,+)$ is also a module homomorphism, since for $r,s \in R$ and $m \in M$, \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, $\varphi_\boxdot : R \mapsto \End_R(M)$ sending $r \mapsto \varphi_r$ is a ring homomorphism. \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 $\varphi_r$ commutes with every element $\phi \in \End_R(M)$, \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 $\FF$ is a commutative ring. Therefore, the endomorphisms $\End_\FF(V)$ of any $\FF$-vector space form an $\FF$-algebra.

Quotient Modules

For groups and rings, recall that quotients are well-defined only for \emph{normal} subgroups and \emph{multiplication-absorbing} subrings (ideals), respectively. For modules $M$, it turns out that \emph{any} submodule $N \subspace M$ has a quotient $M / N$, and the natural projection map $\pi : M \rightarrow M/N$ is a ring homomorphism with kernel $\ker \pi = N$. Similarly, each $\FF$-vector subspace has a quotient $\FF$-vector space arising as the kernel of some linear map.

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

(First Isomorphism Theorem) Let $M,N$ be $R$-modules. The kernel of any module homomorphism $\phi : M \rightarrow N$ is a submodule of $M$, and $M / \ker\phi \cong \phi(M)$

Free Modules

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

Let $M$ be an $R$-module. The submodule of $M$ generated by a subset $A \subset M$ is the set of finite $R$-linear combinations $RA = \{ r_1 a_1 + \cdots + r_m a_m \mid r_k \in R, a_k \in A, m \in \NN \} \subspace M$ A submodule $N = RA \subspace M$ is finitely generated if $A \subset M$ is finite. A cyclic submodule $N = Ra$ is generated by a single element $a \in M$.

An $R$-module $F$ is free on the subset $A \subset F$ if each nonzero $x \in F$ expands uniquely as an $R$-linear combination of elements from $A$, in which case $A$ is called a basis for $F$. \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 $R$ is commutative, every basis has the same cardinality, called the module rank of $F$. 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 $T \in \Hom_\FF(V,W)$ between $\FF$-vector spaces is uniquely determined by its value on $n=\dim V$ points. $R$-linear maps between free modules enjoy the same property, which is normally stated in the following way:

(Universal Property) For any set $A$, there is a unique (up to isomorphism) free $R$-module $\mathrm{Free}(A)$ satisfying the following universal property: for any $R$-module $M$ and any function $\varphi : A \rightarrow M$, there is a unique $R$-module homomorphism $\Phi : \mathrm{Free}(A) \rightarrow M$ such that $\Phi(a) = \varphi(a)$, \begin{figure}[h!tb] \centering \begin{tikzcd} A \arrow[r, “\iota”, hook] \arrow[rd, “\varphi”’] & \mathrm{Free}(A) \arrow[d, ”{\exists!,\Phi}”, dashed] \ & M \end{tikzcd} \end{figure}