How do mathematicians define free module
**This is my opinion, so please do not mind too much.
We start from the free group. Free group is a very basic free object because structures like group, ring, module are group itself. Free group only satisfies closure property and three group axioms. There is no relation among group elements. Then, the natural question is that how does free objects that only satisfy closure properties and axioms of module look like? How can we construct it?
First, let's start with a set $X (0 \in X)$ and construct the largest unitary $R$-module $F$ which is generated by $X.$ Module is a group, so we can think about elements such as $n_{i_1}x_{i_1} + n_{i_2}x_{i_2} + \cdots + n_{i_k}x_{i_k}.$ In this element, $i_k$'s may be same. Next, since module is an abelian group, this element becomes $n_1x_1 + n_2x_2 + \cdots + n_kx_k$ where $n_i, x_i$ are distinct (commutative). Now, since $F$ is a unitary module, we can define $1_Rx_i = x_i$ and $r(n_1x_1 + n_2x_2 + \cdots + n_kx_k) = n_1rx_1 + n_2rx_2 + \cdots + n_krx_k.$ By taking closure of these elements, we get $r_1x_1 + r_2x_2 + \cdots r_kx_k.$ Finally, we can define $F$ by putting all these elements and applying three module axioms.
Why $F$ is the largest? Consider a unitary $R$-module $A$ which is generated by $X.$ Since $A=\left< X \right>$ and $F=\left< X \right>$, $A$ and $F$ both have elements like $r_1x_1 + r_2x_2 + \cdots r_kx_k.$ The main difference is that in $F$, every elements are distinct but in $A$, some elements may be same. We can say this with a concept of linearly independent, $r_1x_1 + r_2x_2 + \cdots r_kx_k = 0$ if and only if $r_1, r_2, \cdots, r_k = 0.$ In $F$, $X$ is a linearly independent set but in $A$, $X$ is a linearly dependent set. Thus, $F$ is the largest and has a basis $X.$
Now, we can ask a question: Is such $F$ uniquely determined up to isomorphism?
Let's write above comment more rigorously. For above $F,$ we can prove that for any unitary $R$-module $A$ and function $f:X \rightarrow A,$ there exists a unique $R$-module homomorphism $\bar{f}:F \rightarrow A$ such that $\bar{f}\iota = f.$ Here, $\iota:X \rightarrow F$ is a natural injection. Now, suppose that there exist another $F'$ and a map $\iota':X \rightarrow F'$ which satisfy above property. Then, we can show that $F \cong F'.$
To sum up, everything we know for $F$ which we have constructed are first, it is uniquely determined up to isomorphism and second, it satisfies following properties.
Property 1: $F$ has a nonempty basis.
Property 2: there exists a nonempty set $X$ and a function $\iota:X \rightarrow F$ with the following property: given any unitary $R$-module $A$ and function $f:X \rightarrow A,$ there exists a unique $R$-module homomorphism $\bar{f}:F \rightarrow A$ such that $\bar{f}\iota = f.$
However, uniqueness came from Property 2 and we can prove that Property 1 and Property 2 are equivalent. Thus, we can define unitary free module by above properties.
I think the idea of the free module came from the vector space. So, I bet mathematicians came up with the idea of the free module first, and they defined module rigorously.
We start from the free group. Free group is a very basic free object because structures like group, ring, module are group itself. Free group only satisfies closure property and three group axioms. There is no relation among group elements. Then, the natural question is that how does free objects that only satisfy closure properties and axioms of module look like? How can we construct it?
First, let's start with a set $X (0 \in X)$ and construct the largest unitary $R$-module $F$ which is generated by $X.$ Module is a group, so we can think about elements such as $n_{i_1}x_{i_1} + n_{i_2}x_{i_2} + \cdots + n_{i_k}x_{i_k}.$ In this element, $i_k$'s may be same. Next, since module is an abelian group, this element becomes $n_1x_1 + n_2x_2 + \cdots + n_kx_k$ where $n_i, x_i$ are distinct (commutative). Now, since $F$ is a unitary module, we can define $1_Rx_i = x_i$ and $r(n_1x_1 + n_2x_2 + \cdots + n_kx_k) = n_1rx_1 + n_2rx_2 + \cdots + n_krx_k.$ By taking closure of these elements, we get $r_1x_1 + r_2x_2 + \cdots r_kx_k.$ Finally, we can define $F$ by putting all these elements and applying three module axioms.
Why $F$ is the largest? Consider a unitary $R$-module $A$ which is generated by $X.$ Since $A=\left< X \right>$ and $F=\left< X \right>$, $A$ and $F$ both have elements like $r_1x_1 + r_2x_2 + \cdots r_kx_k.$ The main difference is that in $F$, every elements are distinct but in $A$, some elements may be same. We can say this with a concept of linearly independent, $r_1x_1 + r_2x_2 + \cdots r_kx_k = 0$ if and only if $r_1, r_2, \cdots, r_k = 0.$ In $F$, $X$ is a linearly independent set but in $A$, $X$ is a linearly dependent set. Thus, $F$ is the largest and has a basis $X.$
Now, we can ask a question: Is such $F$ uniquely determined up to isomorphism?
Let's write above comment more rigorously. For above $F,$ we can prove that for any unitary $R$-module $A$ and function $f:X \rightarrow A,$ there exists a unique $R$-module homomorphism $\bar{f}:F \rightarrow A$ such that $\bar{f}\iota = f.$ Here, $\iota:X \rightarrow F$ is a natural injection. Now, suppose that there exist another $F'$ and a map $\iota':X \rightarrow F'$ which satisfy above property. Then, we can show that $F \cong F'.$
To sum up, everything we know for $F$ which we have constructed are first, it is uniquely determined up to isomorphism and second, it satisfies following properties.
Property 1: $F$ has a nonempty basis.
Property 2: there exists a nonempty set $X$ and a function $\iota:X \rightarrow F$ with the following property: given any unitary $R$-module $A$ and function $f:X \rightarrow A,$ there exists a unique $R$-module homomorphism $\bar{f}:F \rightarrow A$ such that $\bar{f}\iota = f.$
However, uniqueness came from Property 2 and we can prove that Property 1 and Property 2 are equivalent. Thus, we can define unitary free module by above properties.
I think the idea of the free module came from the vector space. So, I bet mathematicians came up with the idea of the free module first, and they defined module rigorously.
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