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Let $R$ be a ring (not necessarily commutative) with multiplicative identity.

A $R$-module $M$ is called free if $M$ has a linearly independent generating set $\beta\subseteq M$.

That is, for any $m\in M$, $$m=\sum_{i=1}^{n} r_i b_i$$ for some $n\in \Bbb{N}^+$ and $r_1, r_2,...,r_n\in R$ and $b_1, b_2,...,b_n\in \beta$ and if $\sum_{i=1}^{n} r_i b_i=0$, where $b_1, b_2,..., b_n$ are distinct, then $r_1=r_2=\cdots =r_n=0$.

I can't feel the sense of the "free".

My guess is that the coefficients in the linear combination of the basis elements is "arbitrary". But it seems not to be a satisfactory explanation. Because in the non-free $\Bbb{Z}$-module $\Bbb{Z}_2\oplus \Bbb{Z}_3$, the coefficients of the generating set $\{(1,0),(0,1)\}$ is also arbitrary.

bfhaha
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  • "Free" as in there are no dependencies among the generators. Are you familiar with any other "free" algebraic structures? – Erick Wong Jul 09 '15 at 06:40
  • Yeah. Like free group, free abelian group. I know that free abelian group is just the free $\Bbb{Z}$-module. The sense of the free group is easy to know. But the free module... – bfhaha Jul 09 '15 at 06:43
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    In that case I assume you're familiar with the concept of group presentations like $\langle a,b \mid a^2, b^n, abab\rangle$ for a dihedral group. A free group like $\langle a,b \rangle$ has no identities to weigh it down: it consists exactly of all combinations of $a$s and $b$s and nothing less. Likewise the $R$-module is free because there are no relations between the generators. – Erick Wong Jul 09 '15 at 06:48
  • Thanks! @ErickWong Your answer is also good. – bfhaha Jul 09 '15 at 07:03

1 Answers1

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Free is in the sense of "having no non-trivial relations", or alternatively, in the sense of coming "for free" (i.e. without any additional work necessary) from a set.

For example, in the $\mathbb{Z}$-module $\mathbb{Z}/3\mathbb{Z}$, even though $\{\overline{1}\}$ is a generating set, it has the property that $$3\cdot\overline{1}=0\cdot\overline{1}$$ whereas in a free $\mathbb{Z}$-module $M$, if $\{m_\alpha\}$ is a generating set then the only way $\sum_{i=1}^n a_im_i=0_M$ to be true for some $a_i\in\mathbb{Z}$ and $m_i\in \{m_\alpha\}$ is if every $a_i=0$.

In the other sense, given a set $S$, we can form the free $\mathbb{Z}$-module on $S$ by simply saying $$F(S)=\{\textstyle\sum_{i=1}^na_is_i:a_i\in\mathbb{Z},s_i\in S\}$$ whereas to make a non-free module requires the extra act of imposing relations between such elements.

Free modules are just one example of a free object (Wikipedia link). Take a look at the math.SE threads What are Free Objects? and definition of a free object in a category.

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Zev Chonoles
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