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I am not sure I understand the difference between free modules and finitely generated modules. I know that a free module is a module with a basis, and that a finitely generated module has a finite set of generating elements (ie any element of the ring can be expressed as a linear combination of those generators).

But then, is the difference just that the generators are not linearly independent ?

In that case, why do we specify sometimes "finitely generated free module" ? Because surely if the above is right (which I do not think it is), free would imply finitely generated so there is no need to specify finitely generated...

Thank you for your help !

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  • $\begingroup$ Please don't add "thank you" as an answer. If you are the same person as the original question poster, please visit this page for help consolidating your two user profiles. That will allow you to add comments, vote up, and accept answers for this question. $\endgroup$ Feb 15, 2013 at 15:22

4 Answers 4

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Here are very simple examples :

$$ \text{As an } \mathbb Z \text{-module, } \mathbb Z/2\mathbb Z \text{ is finitely generated but not freely generated.}$$

$$ \text{As an } \mathbb Z \text{-module, } \bigoplus_{\mathbb N} \mathbb Z \text{ is freely generated but not finitely generated.}$$

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  • $\begingroup$ $\mathbb{Z}$ is finitely generated $\endgroup$ Jun 3, 2016 at 17:09
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    $\begingroup$ Why $\mathbb{Z}/\mathbb{2Z}$ is not freely generated? It has only one basis, {1}? $\endgroup$
    – Ninja
    Jan 3, 2017 at 10:25
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    $\begingroup$ @Ninja {1} is not linearly independent. For instance, 2.1 = 0 $\endgroup$
    – HerrWarum
    Mar 15, 2018 at 5:06
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An $R$-module $M$ is called:

  • free, if $M\cong R^n=\bigoplus_{i=1}^{n}R$. $n$ is called the rank of $M$ (possibly infinite), denoted $\text{rank}(M)$ and can also be infinite. In other words, the map $$\phi:R^n\rightarrow M\tag{1}$$ is an $R$-module isomorphism.

  • finitely generated, if $M$ has a finite generating set. In other words the map $(1)$ is only surjective. This means, by the 1st isomorphism theorem, that $M\cong R^n/\ker(\phi)$.

The difference boils down to whether $\ker(\phi)=0$ or not (and whether $n$ is finite or not), and from this it is easy to construct some examples.

Examples:

  • $k[x,y]/(x^2,xy,y^2)\cong k^3$ is both finitely generated and free as a $k$-module.
  • $k[x,y]\cong \bigoplus_{i\in\mathbb{Z}}k$ is a free, but not finitely generated $k$-module.
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    $\begingroup$ Does free imply finitely generated? In the first bullet, $M$ is isomorphic to $R^n$ and so $R^n$ surjects onto $M$. By the second bullet, $M$ is also finitely generated? $\endgroup$
    – user5826
    Mar 25, 2019 at 15:32
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    $\begingroup$ @Wolfgang Only if $n<\infty$, I've edited my answer. $\endgroup$ Mar 25, 2019 at 18:09
  • $\begingroup$ @cansomeonehelpmeout, so free module with finite basis is finitely generated $\endgroup$
    – MAS
    Mar 14, 2021 at 6:32
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The basis of a free module need not be finite, so free does not imply finitely generated. You are correct that for a general finitely generated module, there may be relations between the generators, i.e. two different linear combinations of generators (with coefficients in the base ring) can represent the same element of the module.

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  • $\begingroup$ so free module with finite basis is finitely generated $\endgroup$
    – MAS
    Mar 14, 2021 at 6:32
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First, the English language

We say a basis exists if the module is finitely generated, that is there exists a finite set that spans our module. There is no such thing as 'an infinite basis' or 'an infinite rank'.

Also, we say 'a free module' if the structure we're working with is isomorphic to a free module. We do not say 'a freely generated module'.

We might argue, but allow me to say so for this specific reply.

Second, the definitions

The basis difference between a free module and a finitely generated module is this.

$$\underbrace{apples \cdot \begin{pmatrix}apples \\ apples \\ apples \\ apples \end{pmatrix}}_{\text{a free module's element}} \quad \underbrace{apples \cdot \begin{pmatrix}apples \\ lemons \\ pears \\ bananas \end{pmatrix}}_{\text{a finitely generated module's element}} $$

Here, ordered tuples and scalars of a free module all come from the same ring $R$. This might not be true for a finitely generated module in general.

Again, a module structure assumes there are elements that have the structure of tuples (=an ordered set) together with scalars that can multiply module elements.

Third, the basis

A free module has the standard basis. Assume we have some module $R^n$. This module is spanned by the standard basis $(e_1, \dots , e_n)$. Remember that a free module is comprised of elements of one single ring. A ring is defined to always have a multiplicative identity, so $e_n$ of the standard basis means we have element 1 in position $n$. The laws of composition, i.e. multiplication by a scalar is defined in such a way that we may pick an appropriate scalar to span the whole module with the standard basis.

$$ r \begin{pmatrix}a_1\\ \vdots \\ a_n \end{pmatrix} = \begin{pmatrix}ra_1\\ \vdots \\ ra_n \end{pmatrix} $$

Now, back to the vegetables. The tuple on the right is comprised of different fruit (apples, lemons, pears and bananas), analogy to the fact that elements do not come from the same ring.

Still, such module is finitely generated, as each ring has a multiplicative identity. We note existence of the zero ring 1=0, but we do not have to span it.

The finite set that may serve as a basis is this. The module is finitely generated, but it is not free.

$${ \begin{pmatrix}\text{1 apple}\\0\\0\\0\end{pmatrix}, \begin{pmatrix}0\\\text{1 lemon}\\0\\0\end{pmatrix}, \begin{pmatrix}0\\0\\\text{1 pear}\\0\end{pmatrix}, \begin{pmatrix}0\\0\\0\\ \text{1 banana}\end{pmatrix} } $$

Finally, properties of the ring

We could make a construction of a module where tuples do not have a finite length. But we do not have to. Much more interesting is what's happening in the ring when we compose ring elements.

Now we assume we deal only with a free module.

Unlike a field, a ring does not have to have the cancellation property. So the set of relations does not have to be finite, i.e. finitely generated.

Assume we have a free module $R^n$ and we try to calculate this and we get the zero element.

$$r' \begin{pmatrix}r'_1 \\ \vdots \\r'_n \end{pmatrix} + r'' \begin{pmatrix}r''_1 \\ \vdots \\r''_n \end{pmatrix} = 0 $$

Expand the composition along the top row.

$$r'r'_1+r''r''_1=0$$

For a field, given $ab=0$ we must have either $a=0$ or $b=0$, or both. But this does not apply to rings. So the module of relations need not be finitely generated in the general case.

This is cool, as a free module does have the standard basis, but the number of ways we can compose this module's to get the zero module element might be infinite, depending on the rings' properties.

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