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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?

user26857
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3x89g2
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    It would help to give the definition of "free module on a set" you are using, since there are many different possible meanings. For example, I would probably not define a free module on $A$ to necessarily have $A$ as a subset, but you seem to have assumed this. – Andrew Dudzik Nov 15 '16 at 17:58
  • To answer the question about being in $F_2$: if you are assuming that $A\subset F_2$, then all $R$-linear combinations of elements of $A$ are also in $F_2$, since $F_2$ is an $R$-module. – Andrew Dudzik Nov 15 '16 at 18:00
  • @Slade Edited. Basically I always see that there is a homomorphism between $F_1$ and $F_2$, but somehow I need to use universal property (perhaps) to show that it's actually an isomorphism – 3x89g2 Nov 15 '16 at 18:05
  • @Misakov: Notice that, according to your definition, if $a \in A$, then $Ra$ is a free module on $A$ over $R$, which is clearly not what you want. Something is missing from it, maybe a maximality condition? In particular, with this definition the statement that you are trying to prove is not true. – Alex M. Nov 15 '16 at 18:08
  • @AlexM. I am copying the definition from Dummit&Foote literally. Why $Ra$ being a free module is bad? – 3x89g2 Nov 15 '16 at 18:13
  • @Misakov: According to your definition, if $a, b \in A$ then both $\langle a \rangle$ and $\langle a, b \rangle$ are free modules on $A$ over $R$. Does it seem to you that they are isomorphic? In fact, one definition of the free module generated by $A$ over $R$ is the one found in my answer (that somebody mindlessly downvoted). Please notice the definite article: it's the free module, not a free module. In the example shown by me above, which one is the free module? – Alex M. Nov 15 '16 at 18:16
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    @AlexM No, according to his definition a free module on $A$ must contain $A$, which the examples you're giving are not guaranteed to. – Andrew Dudzik Nov 15 '16 at 18:21
  • @Slade: His definition (the second yellow paragraph) clearly does not enforce $A$ being contained in $F$. – Alex M. Nov 15 '16 at 18:22
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    @AlexM. It literally says that $A$ is a subset of $F$. – Andrew Dudzik Nov 15 '16 at 18:23
  • @AlexM. If that's the case, assuming the book is right about the theorem on universal property, then $\langle a\rangle$ and $\langle a,b\rangle$ must be the same up to unique isomorphism... – 3x89g2 Nov 15 '16 at 18:27
  • @Slade: You are right, I had missed that. It is a very weird definition, though, precisely because it requires $A \subseteq F$, when the whole point of this construction is to be performed starting with only an abstract set of generators and a ring. Of course, in the end they prove to be equivalent, but I find it a strange choice of definition. – Alex M. Nov 15 '16 at 18:28
  • @Misakov If $\langle a\rangle$ and $\langle a,b\rangle$ are free on the same subset, then they are isomorphic. But neither of these needs to be free on anything. – Andrew Dudzik Nov 15 '16 at 18:30
  • @AlexM. The way that this construction is always done is to make an abstract definition of "freeness", prove that it characterizes free things up to some kind of isomorphism, then give an explicit construction of an object that satisfies the property, at which point we can start saying "the". (I don't know why you keep trying to skip to the last step, which has nothing to do with this question.) Hagen von Eitzen gave the most standard universal property, but the one used here, from Dummit and Foote, is basically the same, though less categorical and more concrete. – Andrew Dudzik Nov 15 '16 at 18:36

3 Answers3

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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 $F_2$.

Andrew Dudzik
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$\renewcommand{\phi}{\varphi}$ 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.

Andreas Caranti
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  • My book actually first defined free module in a different way, then showed that free module satisfies the universal property that you mentioned. Need some times digesting the last part. Thanks! – 3x89g2 Nov 15 '16 at 18:18
  • So in general, if composite of two homomorphisms is identity, then each of those is isomorphism? – 3x89g2 Nov 15 '16 at 18:30
  • $\renewcommand{\phi}{\varphi}$If both $\phi_{1} \circ \phi_{2}$ and $\phi_{2} \circ \phi_{1}$ are the identity on their respective domains, then they are bijective maps (a general fact in set theory), and thus isomorphisms. – Andreas Caranti Nov 15 '16 at 18:33
  • OK, now I realize why it's good to use universal property as definition. I am having trouble justifying "Since $F_1$ is free on $A_1$, there is a unique..." part because according to my definition, we do not know if that's true or not. We only know there is some $F(A)$ satisfying the property, but that might not be $F_1$... – 3x89g2 Nov 15 '16 at 19:00
  • You want to prove uniqueness of the free module up to isomorphism. So you take two free modules, and show they are isomorphic, that's it. – Andreas Caranti Nov 15 '16 at 19:26
  • Yes, but I cannot assume that $F_1$ and $F_2$ have the universal property, right? (using my definition) – 3x89g2 Nov 15 '16 at 19:27
  • If you start with your definition, you may prove that the universal property holds, I will add this to my answer - just a few minutes. – Andreas Caranti Nov 15 '16 at 19:41
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    I already proved that there is some free module $F(A)$ such that $F(A)$ satisfies the universal property. Now I need to show that (1) if $F_1$ and $F_2$ are free modules, they are isomorphic and thus I can finally show that every free module on $A$ is essentially $F(A)$. At least that's the approach my book took... – 3x89g2 Nov 15 '16 at 19:42
  • And this is exactly what I did above! – Andreas Caranti Nov 15 '16 at 19:51
  • Oops, my bad, did not see your update – 3x89g2 Nov 15 '16 at 19:57
  • @Misakov, thanks! – Andreas Caranti Nov 15 '16 at 20:02
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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 $A$.

Hagen von Eitzen
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