If $F_1$ and $F_2$ are free modules on the same set $A$, then $F_1 \cong F_2$

Question

I am trying to prove the following:

If $F_1$ and $F_2$ are free modules on the same set $A$, there is a unique isomorphism between $F_1$ and $F_2$ which is the identity map on $A$.

The definition I am using is

An $R$-module $F$ is said to be free on a subset $A$ of $F$, if for every nonzero element $x\in F$, there exist unique nonzero elements $r_1,\cdots,r_n \in R$ and unique $a_1,a_2,\cdots,a_n \in A$ such that $x = r_1 a_1 + \cdots + r_n a_n$ for some $n \in \mathbb{Z}^+$.

Then regarding the universal mapping property:

For any set $A$, there is a free $R$-module $F(A)$ satisfying the following universal property:...

I actually tried the following:

Since $F_1$ is a free module on set $A$, then for every $x\in F_1$, it can be written as $x = r_1 a_1 + \cdots + r_n a_n$ for unique nonzero $r_i\in R$ and unique $a_i \in A$. Similarly, for every $y\in F_2$, $y = s_1 a_1 + \cdots + s_m a_m$.

I wanted to say that let's just consider

$$\phi: r_1 a_1 + \cdots + r_n a_n \longrightarrow r_1 a_1 + \cdots + r_n a_n$$

But then I am not sure if $r_1 a_1 + \cdots + r_n a_n$ is in $F_2$ or not? $F_2$ being a free module does not imply that all possible combinations $\sum r_i a_i$ is in it, but only implies that if something is in it, it can be uniquely written in that form, right? How should I show surjectivity?

Answer 1

$\renewcommand{\phi}{\varphi}$ Addendum

OP has given the definition of free modue he assumes. Let us prove that it satisfies the universal property I use below.

Let $f : A \to M$ a map, where $M$ is a module. Since every element of $F(A)$ can be uniquely written as $\sum_{i=1}^{n} r_{i} a_{i}$, for $r_{i} \in R$ and $a_{i} \in A$, a morphism $\phi: F(A) \to M$ such that $\phi(a) = f(a)$ for $a \in A$ (provided it exists) is well defined and uniquely defined as $$ \phi\left(\sum_{i=1}^{n} r_{i} a_{i}\right) = \sum_{i=1}^{n} r_{i} \phi(a_{i}). $$ Now it is not difficult to verify that this map is indeed a morphism $F(A) \to M$, $$ \phi\left(\sum_{i=1}^{n} r_{i} a_{i} + \sum_{i=1}^{n} s_{i} a_{i}\right) = \phi\left(\sum_{i=1}^{n} (r_{i}+s_{i}) a_{i}\right) = \sum_{i=1}^{n} (r_{i}+s_{i}) \phi(a_{i}) =\\= \sum_{i=1}^{n} r_{i} \phi(a_{i}) + \sum_{i=1}^{n} s_{i} \phi(a_{i}) = \phi\left(\sum_{i=1}^{n} r_{i} a_{i}\right) + \phi\left(\sum_{i=1}^{n} s_{i} a_{i}\right). $$

---

You should have defined a free module over $A$ as a module $F$ containing $A$ such that if $M$ is any module, and $f: A \to M$ is a map, then there is a unique module morphism $\phi : F \to M$ such that $\phi(a) = f(a)$ for each $a \in A$.

So if $F_{1}, F_{2}$ are two free modules over $A$, consider $f_{2} : A \to F_{2}$ to be the identity (or inclusion) map, i.e. $f_{2}(a) = a$ for $a \in A$. Since $F_{1}$ is free on $A$, there is a unique morphism $\phi_{2}: F_{1} \to F_{2}$ such that $\phi_{2}(a) = a$ for $a \in A$. Similarly, there is a unique morphism $\phi_{1}: F_{2} \to F_{1}$ such that $\phi_{1}(a) = a$ for $a \in A$.

Now the composition $\phi = \phi_{1} \circ \phi_{2}$ is a morphism $F_{1} \to F_{1}$ such that $\phi(a) = a$ for $a \in A$. Since $F_{1}$ is free, this is unique, and thus is the identity, as the identity also maps all elements $a \in A$ to $a$. Similary, $\phi_{2} \circ \phi_{1}$ is the identity on $F_{2}$, and thus $\phi_{1}, \phi_{2}$ are isomorphisms, one the inverse of the other.

Answer 2

All linear combination $\sum_{i=1}^n r_ia_i$ are in $F_2$ because is a module and as such closed under addition and multiplication by ring elements and $F_2$ contains $A$.

Answer 3

You have given the right definition of $\phi$. Let's be clear about the notation: Define $\phi(r_1\cdot a_1 + \cdots + r_n a_n) = r_1\odot a_1 \oplus \cdots \oplus r_n \odot a_n$, where I'm using $\cdot$ and $+$ for multiplication and addition in $F_1$, and $\odot$ and $\oplus$ for the multiplication in $F_2$. After all, they may be different.

You ask whether the right-hand side is "in $F_2$". Yes, certainly it is; it's an $R$-linear combination of elements of $F_2$, and $F_2$ is an $R$-module.

Note that $\phi$ is well-defined precisely because any element of $F_1$ can be written uniquely in the form $r_1\cdot a_1 + \cdots + r_n a_n$.

We should check that $\phi$ is really an $R$-module homomorphism. This is not completely trivial, but it should be intuitive: if $x=\sum_i r_i a_i$ and $y=\sum_i s_i a_i$, then $x+y=\sum_i (r_i + s_i) a_i$ and $r\cdot x = \sum_i (r r_i)\cdot a_i$. There is a little more work to be done here, involving the operations $\odot$ and $\oplus$, but I leave it to you.

Finally, why is $\phi$ a bijection? Well, this comes down to the fact that $F_2$ is a free module over $A$. In fact, constructing $\phi$ only required that $F_2$ is some $R$-module containing $A$, but to prove injectivity, we need to know that representations in $F_2$ are unique, and to prove surjectivity, we need to know that these representations exist for all elements of $F_2$.