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From my textbook:

A free $R$-module is "A left $R$-module $F$ is called a free left $R$-module if $F$ is isomorphic to a direct sum of copies of $R$..."

I know that another definition of a free R-module is a module with a basis, but I don't know how to connect that definition with this one. Also, what does "copies of $R$" mean?

Thanks in advance

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It's easy to prove: think about vector spaces - how would you prove that n-dimensional vector space over $\mathbb R$ is isomorphic to n "copies" of $\mathbb R$ (i.e. direct sum)? –  rafaelm Jan 14 '13 at 23:04
Free modules always have basis. –  rafaelm Jan 14 '13 at 23:26
When $R$ is principal ideal domain (PID), we have more: Every submodule of free module is free! (This does not hold in general.) –  rafaelm Jan 15 '13 at 0:31

2 Answers 2

up vote 7 down vote accepted

The two definitions of free module that you mentioned are indeed equivalent (for a ring with $1$). Here is a sketch of the proof.

$I$ "copies" of $R$ mean direct sum: $$ \sum_{I} R = \{f \colon I \to R \text{ such that } f(i)=0 \text{ for all but finite indeces $i$}\},$$ and module operations are defined componentwise.

Let $M$ be left $R$-module with basis $X=\{x_i \colon i \in I\} \subseteq M$. Then for every element $x \in M$ there is unique presentation $x=r_1 x_{i_1} + \dots + r_k x_{i_k}$, for unique $k \in \mathbb N \cup \{0\}$, $x_{i_1},\dots x_{i_k} \in X$ and $r_1,\dots,r_k \in R\setminus\{0\}$. Define $\Phi(x)$ to be the function $I \to R$ such that $i_k \mapsto r_k$, and all other indices go to $0$.

Using uniqueness of the presentation of $x$ in the given basis, verify that $\Phi \colon M \to \sum_{I} R$ is $R$-module isomorphism.

EDIT: Intuitively, $\Phi$ is coordinatization with respect to the given basis. Argumentation is the same as in linear algebra, only notation is more complicated since basis may be infinite. Direct sum is "substructure" of direct product which is a way to generalize the notion of $n$-tuples to infinite case.

Conversely, one must see that direct sum $\sum_{I} R$ has basis as $R$-module. It is easy: basis constitutes of all functions $f \colon I \to R$ given with formula $$f(i) = \begin{cases} 1 &\colon i=k \\ 0 & \colon i \neq k \end{cases}$$ for some $k \in I$. Try to prove this.

EDIT2: There is one other definition of free module that is equivalent to these two. It involves some basic category theory. An $R$-module $M$ is free iff it is free object over some set X in the category of $R$-modules.

EDIT3: Feel free to ask some details if you need.

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Copies of R means each summand is the addition group of R.

Typically, a free algebraic object is one that can generated from a set without any relations (quotients) applied. When it's a free module, each element of that set is used as a basis vector, and the rest of the module can be generated from the definition of a module.

Coproducts of algebraic objects, like the direct sum, are typically defined from generators and relations as the object requiring the least extra relations to satisfy the definition of that type of object, and hence are used when forming the most free objects of a kind, or most general object satisfying a property.

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