Modules and Vector Spaces as Algebraic Categories

An algebraic category is defined to be any full subcategory of $\textbf{Alg}(\Omega)$, the category of all $\Omega$-algebras. We can define various algebraic structures as algebraic categories. For example, in order to define $\textbf{Groups}$, the category of Groups, we can specify $\Omega = \\{e,m,i\\}$ as the set consisting of three operators: one nullary, one unary, and one binary, such that, $(G,e,m)$ is a monoid and

$\forall x \in G$, making $\textbf{Alg}(\Omega) = \textbf{Groups}$ for $\Omega$ as specified¹. However, an issue arises when considering vector spaces over a field and modules over a ring as the operators in $\Omega$ are defined to act on the elements of the underlying set, and not on elements outside of that set, say in a field or a ring. This issue arises when trying to define scalar multiplication for these structures. Consider a vector space $V$ over a field $K$. We define a binary operator $m$ on $V$ called multiplication, then $v(m):V^2\to V$, where

for $\mathbf{N}$ the natural numbers¹. Now there is an issue since $m$ is not defined to operate on the elements of $K$. What follows is an explanation as to how we can get around this issue and define vector spaces over a field $K$ and modules over a ring $R$ as algebraic categories.

We start with $R$-modules. Let $R$ be a commutative ring with $1$ and $M$ the underlying set of the module. Define the signature $\Omega_M = \\{a_M,0_M,i_M\\}$ where $a_M$ is a commutative associative binary operator on $M$, $0_M\in M$ is a nullary operator such that

$\forall u\in M$, $i_M$ is an unary operator on $M$ such that

$\forall u\in M$. Now, in order to define scalar multiplication, we define the signature $\Omega_R = R$ where each element of $R$ is considered as a unary operator on $M$. However, we need to add some structure to $\Omega_R$ since, in a ring, we are able to add and multiply elements of the ring to get new elements in the ring. To do this, we define a signature for an $\Omega$-algebra on $\Omega_R$ by $\Omega_{R}^{\text{ring}} = \\{a,0,i,m,1\\}$ with the same properties as the operators in $\Omega_M$ above for $a,0,i$ and additionally, $m$ is an associative binary operator on $\Omega_R$ such that

$\forall r,s,t\in \Omega_R$ and $1\in \Omega_R$ is a nullary operator such that

$\forall r\in \Omega_R$. We can then make $M=(M,h_M)$ into an $\Omega_R$-set (as in Example 3.10) by defining the map $h_M:\Omega_R\times M\to M$ as $h_M(\omega,p)=\omega p$, $\forall p \in M$ ¹. However, we require additional properties for the map $h_M$:

$\forall \omega,\omega^{\prime}\in\Omega_R$ $\forall u,v\in M$. In this way, scalar multiplication of an element of M by one of R is defined using the map $h_M$. So by letting the signature of the $R$-module be $\Omega = \Omega_R \cup \Omega_ M$, we obtain $\textbf{Alg}(\Omega)=\textbf{Modules}$, showing $\textbf{Modules}$ is an algebraic category.

Similarly, for a vector space $V$ over a field $K$ we can define a signature $\Omega_V = \\{a_V,0_V,i_V\\}$ where $a_V$ is a commutative associative binary operator on $V$, $0_V\in V$ is a nullary operator such that

$\forall u\in V$, $i_V$ is an unary operator on $V$ such that

$\forall u\in V$. To define scalar multiplication of elements of $V$ by those of $K$, we define the signature $\Omega_K = K$ in which each element of $K$ is a unary operator on $V$. We add a field structure to $\Omega_K$ by defining a signature for $\Omega_K$ as $\Omega_{K}^{\text{field}}=\\{a,0,i_a,m,1,i_m\\}$ with the same properties as the operators in $\Omega_V$ above for $a,0,i$ and additionally, $m$ is a commutative associative binary operator on $\Omega_K$ such that

$\forall c,d,f\in \Omega_K$, $1\in \Omega_K$ is a nullary operator such that

$\forall c\in \Omega_K$ and $i_m$ is an unary operator on $\Omega_K$ such that

$\forall c\in \Omega_K$. Next, make $V=(V,h_V)$ into an $\Omega_K$-set by defining the map $h_V:\Omega_K\times V\to V$ as $h_V(\omega,u)=\omega u$, $\forall u \in V$ with the additional properties that

$\forall \omega,\omega^{\prime}\in\Omega_K\;\forall u,q\in V$. So, for the signature $\Omega = \Omega_ V\cup\Omega_ K$, we obtain $\textbf{Alg}(\Omega)=\textbf{Vect}$, showing that the category of vector spaces is an algebraic category.