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There already exists an answer here on Stack Exchange to the question of why projective modules are called projective modules. I would now like to find out the answer to the question "Why are free modules called free modules?"

The way I have long thought about the extent to which having a basis makes the modules free is the sense that, as in a vector space, you can pick a given "direction" in $R^n$, and that "direction" is then uniquely and unambiguously determined. You can "go on and on forever" on it, and while you might end up where you started, like in the case of a free module over $\mathbb{Z}_{m}$, $m \in \mathbb{N}$, you never actually "leave" the direction in which you started going in. In that sense, you are "free" in that "directions" are "independent".

It struck me however earlier today that this might not be the "canonical" reason for why free modules are said to be free.

Is there some "canonical" explanation for the term?

Look forward to your answers.

StormyTeacup
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3 Answers3

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I was under the impression that (1) free modules are called free modules, for same reason that free groups are called free groups, and (2) that free groups are called free groups because there are no relations among the group elements (besides trivial things you would get from the group laws, such as associativity). Free modules are free modules because they are free of relations among the module elements.

user729424
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    That is another good take on it. Luckily being "free of any relation" is exactly synonymous to "I can pick freely the image on the basis and extend it to a morphism", although it is sort of a pun, since the word "free" is understood in two different ways in these sentences. That makes the terminology really nice! – Thibaut Benjamin Feb 14 '20 at 17:55
  • Just to clarify, again, are all free modules free groups? – StormyTeacup Feb 14 '20 at 20:53
  • @StormyTeacup Free groups are groups with elements that don't satisfy any relations, except the group laws. Free Abelian groups are groups with elements that don't satisfy any relations, except the group laws and commutativity. From this, I believe you can show that the only free group that is also a free Abelian group is $\Bbb{Z}$ (and perhaps also the trivial group). As for modules, every Abelian group can be made into a $\Bbb{Z}$-module in one and only one way, and an Abelian group will be a free Abelian group iff it is a free $\Bbb{Z}$-module. – user729424 Feb 14 '20 at 21:31
  • @ThibautBenjamin I'm glad you liked my answer. I thought your answer was helpful too! – user729424 Feb 14 '20 at 21:32
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The way I see it (it might not be exactly the original reason, but must be close enough), is that a free object (module, group, vector space) is called like that because of its universal property.

Saying that a module $M$ is free, means can be formulated by the following property : There exists a set $X$ (of generators, or basis) of elements of $M$ such that for all other module $N$, an homomorphism of modules $M\to N$ is the same as just a function from $X\to N$. This property has 2 important parts : firstly, a morphism is completely determined by its values on $X$, but more importantly, given any way to associate to every values of $X$ an element of $N$ defines a valid morphism.

The second property gives the answer to the question "how free am I to chose values for elements of $X$, so that I can define a module homomrophism?" - the answer being : you are free to chose in any way you want.

Thibaut Benjamin
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Maybe it comes from the universal property of extension is the sense that each mapping $\phi_0: B\rightarrow N$ from a basis $B$ of a free module $M$ to another module $N$ can be extended to a module homomorphism $\phi:M\rightarrow N$ that extends $\phi_0$.

This is only valid for free modules and this extension property of an arbitrary mapping $\phi_0: B\rightarrow N$ is the ''freedom'' only a free module has.

Wuestenfux
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  • This seems reasonable, but, just to clarify, then, we're talking about the basis being a set of elements $B={b_1 , b_2 , b_3 , \dots }$, and $\phi_0$ mapping the $b_i$ to different elements in $N$ as $b_1 \mapsto n_1$, $b_2 \mapsto n_2$, etc., $n_1 , n_2, \dots \in N$, correct? – StormyTeacup Feb 14 '20 at 15:16