5
$\begingroup$

What is the difference between free groups and free modules/vector/algebras spaces, or in other words what does free mean in algebra?

I have seen two uses of the term free:

  • something of infinite order that is generate by a linear independent set called a basis. I have seen this applied to (Groups/abelian groups/vector spaces/modules/algebras)

  • an algebraic structure in which there is no defined relation between elements. I have seen this apllied to vector spaces/modules-used in tensor product definition.

Thanks

$\endgroup$
  • 4
    $\begingroup$ The Category theory answer is that anything that is "free" is an adjoint of a forgetful functor. But then, that gets to the problem that "forgetful functor" is not actually defined. That said, the most common "free" objects are the adjoints of the usual "forgetful" functors from your category ($R$-modules, groups, etc.) to the category of sets. $\endgroup$ – Thomas Andrews Aug 16 '14 at 21:55
  • $\begingroup$ So really, if I am undetstanding correctly, the two definitions I gave are the same. So all algebraic structures that have a basis and are infinite with no relation between elements are free. But ones with the first two properties, but do have relations between elements I.e. vector space are free objects too? $\endgroup$ – dylan7 Aug 16 '14 at 22:11
  • 1
    $\begingroup$ No. For example, there is no such thing as a "free field," or even a "free field of characteristic zero." It's more complicated than "no relationship." $\endgroup$ – Thomas Andrews Aug 16 '14 at 22:13
  • 2
    $\begingroup$ Yeah the point is that a free function imposes no relations except those stipulated in the target category. So why free groups look so much wierder and wilder is simply because the category of groups stipulates a hell of a lot fewer relations than the other examples. A free abelian group on $S$ satisfies no relations other than that pesky commutativity thing that was stipulated in the category of abelian groups. $\endgroup$ – JHance Aug 16 '14 at 22:56
  • $\begingroup$ Though there is such a thing as a free algebraically closed field of a given characteristic. In fact, every such field is free (just like every vector space is free), courtesy of exchange property. $\endgroup$ – tomasz Aug 16 '14 at 23:04
6
$\begingroup$

The idea of 'free' is that you have some set $S$ (finite or infinite), and you want to create some object (group, module, algebra, etc.) which is both generated by the elements in S, and there are also no restrictions/relations between the elements in S.

So take $S = \{x,y\}$ for example. The free group generated by S would contain $x$ and $y$, but it would also have to contain $xy$, $x^{-1}$, $y^{-1}$, $xyyxy^{-1}x^{-1}y$, etc. And these are all distinct elements. The free abelian group would also contain $x$ and $y$, but since we have the 'abelian' restriction the elements $xy$ and $yx$ are the same. (Abelian group operations are typically written as $+$, so instead we would say $x + y$ and $y + x$ are the same).

One consequence of this is that to describe any homomorphism from the free object generated by $S$ to some other object, it's enough to say where the elements of $S$ go AND there are no restrictions on where the elements of $S$ can go. So for instance the complex numbers are not a free real algebra: $\mathbb{C}$ is generated (as an $\mathbb{R}-$algebra) by $1$ and $i$, but any homomorphism $\mathbb{C} \rightarrow A$ must send $i$ to some element that squares to $-1$. And it's important to know that the notion of 'free' depends on the category being considered. The polynomial ring $\mathbb{R}[x,y]$ is a free abelian $\mathbb{R}-$algebra, but not a free $\mathbb{R}-$algebra.

This last property (that any homomorphism is uniquely described by an arbitrary choice of where the elements of $S$ get sent) is exactly what Thomas Andrews means when he says that 'free' is an adjoint of a forgetful functor.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ What does $ A $ represent? Is that just another $\mathbb {R} -algebra $? $\endgroup$ – dylan7 Aug 16 '14 at 23:17
7
$\begingroup$

(This is essentially a down-to-earth translation of Thomas Andrews comment)

If a foo is a set with certain additional structure, then a free foo over a certain set $S$ of generators ( or the foo freely generated by the set $S$) is a foo $F$ with the following properties:

  • $S$ is a subset of the underlying set of $F$.
  • Whenever $G$ is a foo and $f\colon S\to G$ is a map from $S$ to the underlying set of $G$, there exists one and only one foo-homomorphism $\phi\colon F\to G$ such that the restriction of $\phi$ to $S$ (viewed as a set-map and not a homomorphism) equals $f$.

To go into more detail what "underlying set" and "homomoprhism viewed as set-theoretical map" mean, one should learn about category theory, where thse concepts are introduced via "forgetful fucntors".

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I would add: There isn't always a notion of free. For example, there is no free objects (except on the empty set) in the category of fields of characteristic zero. $\endgroup$ – Thomas Andrews Aug 16 '14 at 22:23
  • 3
    $\begingroup$ We need more people championing the corner of the noble foo. I salute you foo fighters! $\endgroup$ – Dan Rust Aug 16 '14 at 22:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.