[[Module theory MOC]] 
# Free module

**Free modules** are the [[Free-forgetful adjunction|free objects]] in [[Category of left modules]]. #m/def/module 
In essence it is a module with a **basis**, hence every [[vector space]] is a free module, 
but one can form non-free modules over a ring that is not a [[Division ring]].

> [!info]+ Notation
> In these notes, we have two conventions for the free module over $R$ generated by a set $S$. The first is
> $$
> \begin{align*}
> R^{(S)} = \Span \{ \delta_{s} : s \in S \rangle \} \leq R^S
> \end{align*}
> $$
> where we think of elements as maps of finite support $S \to R$,
> and we identify $s \in S$ with $\delta_{s} : t \mapsto [s=t]$. 
> The second is
> $$
> \begin{align*}
> R \langle x_{s} \rangle_{s \in S}  = \Span \{ x_{s} : s \in S \} 
> \end{align*}
> $$
> which allows for the explicit naming of the basis to be used.

By **basis** $\mathcal{B}$ for an $R$-module $V$, we mean an $R$-spanning set $\mathcal{B}$ such that each $v \in V$ is given by a _unique_ $R$-linear combination of $\mathcal{B}$ elements. ^basis

## Universal property

Let $R$ be a ring and $S$ be a set.
The **free module** is a pair consisting of an $R$-module $R^{(S)}$ and a function $\iota : S \to R^{(S)}$ 
such that given any $R$-module $M$ and function $f : S \to M$
there exists a unique [[module homomorphism]] $\bar{f} : R^{(S)} \to M$ such that the following diagram commutes

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This has a unique extension to a [[functor]] such that 
$$
\begin{align*}
\iota : 1 \Rightarrow R^{(-)}: \Set \to \Set
\end{align*}
$$
becomes a [[natural transformation]].


## Construction as maps

Let $S$ be a set and $R$ be a ring.
The free module $R^{(S)}$ is the set of [[Support of a map|maps of finite support]] $S \to R$ with addition and scaling induced by those of $R$, #m/def/module  
i.e. for all $s \in S$
$$
\begin{align*}
(\alpha a + \beta b)(s) = \alpha a(s) + \beta b(s)
\end{align*}
$$
where we identify $s \in S$ with $\delta_{s} : t \mapsto [t=s]$ invoking an [[Iverson bracket]].

> [!check]- Proof of universal property
> Clearly $R^{(S)}$ as constructed is an $R$-module with basis $\{ \delta_{s} \}_{s \in S}$
> Now let $M$ be an $R$-module and $f : S \to M$ be a function.
> For a module homomorphism $\bar{f} : R^{(S)} \to M$ to make the diagram commute,
> we require that $\bar{f}(\delta_{s}) = f(s)$ for all $s \in S$, 
> which fully specifies $\bar{f}$ so that for $a \in R^{(S)}$
> $$
> \begin{align*}
> \bar{f}(a) = \sum_{s \in S} f(s) a(s)
> \end{align*}
> $$
> as required. <span class="QED"/>

## Properties

- $R^{(S)}$ carries the additional structure of a coring, namely the [[Free R-comonoid]]

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