# Can a basis of a free module contain uncountably many elements?

I am studying free modules. The definition of a free module given by our instructor is as follows $:$

An $R$-module $F$ is said to be free on a set $S \subset F$ if every element $x \in F$ can be written as the finite sum $x=\sum_i a_ix_i$ where $x_i \in S$ for all $i$. In this case $S$ is said to be a basis for the free module $F$.

If $S$ is a finite set of $n$ elements then it has been proved in the lecture note that

$$F \simeq R^n.$$

Also if $S$ is countably infinite then

$$F \simeq \underset {\alpha \in \Bbb N} {\bigoplus} R_{\alpha}.$$

where $R_{\alpha} = R$ for each $\alpha$. But no comment is given when a basis $S$ for the free module $F$ contains uncountably infinite number of elements. I can't also find any online source which describes this fact clearly. So a question naturally came into my mind which is - "Can a basis of a free module ever contain uncountably many elements?" If the answer to this question is affirmative one then in this case what is the direct sum to which $F$ is isomorphic as $R$-modules?

• Yes${{{{{{}}}}}}$. – Lord Shark the Unknown Apr 26 '18 at 16:13
• Can you provide me some examples? – Arnab Chatterjee. Apr 26 '18 at 16:18
• The groups of singular chains, as used in algebraic topology, in most common topological spaces are free of uncountable rank over $\Bbb Z$. – Lord Shark the Unknown Apr 26 '18 at 16:20

An example of such a module: consider a commutative ring $R$ and the ring of polynomials in indeterminates indexed by a non-countable set, for instance: $$R[X_r]_{r\in\mathbf R}.$$
Sure. Take $S$ to be an uncountable set and define the free $R-$module $M$ on $S$ by the set of finite $R-$linear combinations of elements of $S$. For example, if $x_1,\ldots, x_n$ are elements of $S$ and $a_1,\ldots, a_n$ are elements of $R$, elements of $M$ look like $$\sum_{i=1}^n a_ix_i.$$ The "length" of the linear combination can be as large as you want, so long as it's finite. Infinite combinations don't make sense here because we don't know what it means to take a limit in the absence of some extra structure.
• $M\cong \bigoplus_{s\in S} Rs$. – Antonios-Alexandros Robotis Apr 26 '18 at 16:24
• I forgot to ask you a question.The question is $:$ "how can you show in this case that $S$ is a basis for $M$?" – Arnab Chatterjee. Apr 26 '18 at 16:52
• Show that $\sum_{i=1}^n a_i x_i=0$ for $x_i\in S$ implies $a_1=\cdots=a_n=0$ for $a_i\in R$. – Antonios-Alexandros Robotis Apr 26 '18 at 16:54
• Why all $a_i=0$? – Arnab Chatterjee. Apr 26 '18 at 16:56