Finitely generated modules as free modules

[Kreizhn]

I'm reading up on the classification of finitely generated modules over principal ideal domains. In doing so, I continuously come up on the statement "Let M be a finitely generated, free R-module."

My question is, is this statement redundant? It seems to me that all finitely generated R-modules are necessarily free as R-modules. In particular, if M is an finitely generated R-module with minimal generating set $A \subseteq M$, then isn't the free R-module on A also M? Or am I missing a technical point?

 

[micromass]

Hi Kreizhn!

What you say is certainly true if R is a field (thus if we're working with vector spaces), as every module is free there. But R doesn't need to be a field.

For example, let $R=\mathbb{Z}$ here. Then the free modules are all of the form $\mathbb{Z}^n$. However, there are much more finitely generated modules. For example $\mathbb{Z}_2$ (the integers modulo 2) is certainly finitely generated, but it is not free.

 

[Kreizhn]

Ah yes, because in this instance, 1 generates $\mathbb Z_2$ but the free module on the singleton would be $\mathbb Z$?

 

[micromass]

Ah yes, because in this instance, 1 generates $\mathbb Z_2$ but the free module on the singleton would be $\mathbb Z$?

Indeed!

 

[Kreizhn]

I think I know where I made the mistake in my logic. M is finitely generated over R if there is a surjective homomorphism $R^{\oplus A} \to M$ for some finite subset $A \subseteq M$. On the other hand, M is free if there exists a set B such that $R^{\oplus B} \to M$ is an isomorphism.

So for this example, certainly $\mathbb Z \to \mathbb Z /2\mathbb Z$ is surjective, but there's no way this is could be an isomorphism.

 

[Kreizhn]

On this note, I would be interested in making something clear.

When we think of free modules, does the word "free" essentially characterize the existence of a basis? Or is it possible for some modules to have a basis but not be free.

In the case of our $\mathbb Z_2$ example, I'm trying to think of whether {1} is a basis. Clearly {1} is a minimal generating set, but is it linearly independent? I feel that it isn't, since viewing $\mathbb Z_2$ as a $\mathbb Z$ module, we have $2 \cdot 1 = 0$. Is this correct?

 

[micromass]

On this note, I would be interested in making something clear.

When we think of free modules, does the word "free" essentially characterize the existence of a basis? Or is it possible for some modules to have a basis but not be free.

In the case of our $\mathbb Z_2$ example, I'm trying to think of whether {1} is a basis. Clearly {1} is a minimal generating set, but is it linearly independent? I feel that it isn't, since viewing $\mathbb Z_2$ as a $\mathbb Z$ module, we have $2 \cdot 1 = 0$. Is this correct?

Hmm, I don't think we like using the word basis when not working in vector spaces. All free means is that there is an isomorphism $\varphi:R^n\rightarrow M$. I guess you could see $\varphi(1,0,...,0),...,\varphi(0,0,...,1)$ as a basis of M. But we don't use that terminology (not sure why actually).

Anyway, what you do have is that a module is finitely generated and free if there exists a finite set $\{x_1,...,x_n\}$ that generates the set and such that

$$\sum{r_i x_i=0}~\Rightarrow~r_i=0$$

I guess we can call this linearly independent. But it's not standard terminology.

 

[Kreizhn]

I guess maybe the bases thing was a slight abuse of terminology, though I do have a definition of linear independence for general R-mods.

Let $i: I \to M$ be a non-empty mapping from an index set I to an R-module M, and consider the free R-module on I denoted $F^R(I)$. We know there is a canonical inclusion $\iota: I \to F^R(I)$, and so by the universal property of free R-modules, it follows that there exists an R-module homomorphism $\phi: F^R(I) \to M$, such that $i = \phi \circ \iota$. If $\phi$ is injective, then $i: I \to M$ is linearly independent.

I guess this answers my question though, since again there's no way that $\mathbb Z \to \mathbb Z_2$ is injective. And I guess in particular, the kernel is the ideal $2 \mathbb Z$.

Okay, maybe a better question then. We know that all finite dimensional k-vector spaces (for k a field) are isomorphic. Does this hold in general for any given cardinals? Namely, let V and W be vector spaces whose module rank is both the cardinal $\omega$, is it necessary that $V \cong_k W$ as k-vector spaces?

I use module rank because I'm not sure if "dimension" is appropriate in this context. I want to say that this is true, since if V and W are free k-modules of rank $\omega$ then there are isomorphisms $k^{\oplus A} \to V, k^{\oplus B} \to W$ where $|A|=|B| = \omega$. But then there is a set bijection $A \to B$, so I figure this must make $k^{\oplus A} \cong k^{\oplus B}$ making V and W isomorphic as k-vector spaces.

I've never seen this stated though, so I question whether or not I've done something wrong.

 

[micromass]

Yes, this is true! Two vector spaces are equal if and only if they have the same dimension (=module rank). This holds for infinite cardinalities as well!

So if V and W both have dimension $\aleph_0$, then they are isomorphic!

 

[Kreizhn]

Excellent. Thanks so much.

 

[lavinia]

I'm reading up on the classification of finitely generated modules over principal ideal domains. In doing so, I continuously come up on the statement "Let M be a finitely generated, free R-module."

My question is, is this statement redundant? It seems to me that all finitely generated R-modules are necessarily free as R-modules. In particular, if M is an finitely generated R-module with minimal generating set $A \subseteq M$, then isn't the free R-module on A also M? Or am I missing a technical point?

Here is an important example that you might like to think about.

A linear transformation of a finite dimensional vector space makes the vector space into a module over the ring of polynomials with coefficients in the base field. Since the ring of polynomials over a field is a principal ideal domain, the vector space is now a finitely generated module over a PID.

The module structure is x.v = L(v) then extend by linearity to all of the polynomials.

This module is not free. In fact it is a torsion module.