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

Question

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?

Please help me in understanding this concept.

Thank you in advance.

The groups of singular chains, as used in algebraic topology, in most common topological spaces are free of uncountable rank over $\Bbb Z$. — Angina Seng, Apr 26 '18 at 16:20

score 2 · Answer 1 · answered Apr 26 '18 at 16:19

2

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: $$R[X_r]_{r\in\mathbf R}. $$

answered Apr 26 '18 at 16:19

Bernard

175,478

score 1 · Answer 2 · answered Apr 26 '18 at 16:20

1

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.

answered Apr 26 '18 at 16:20

Alekos Robotis

26,184

Then what is the direct sum to which the free module is isomorphic? – Arnab Chattopadhyay. Apr 26 '18 at 16:23
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$M\cong \bigoplus_{s\in S} Rs$. – Alekos 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 Chattopadhyay. 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 $a_i\in R$. – Alekos Robotis Apr 26 '18 at 16:54
Why all $a_i=0$? – Arnab Chattopadhyay. Apr 26 '18 at 16:56
That's the definition of being a basis. – Alekos Robotis Apr 26 '18 at 16:56
You have taken all finite linear combinations of the elements $S$ to form $M$. Right? – Arnab Chattopadhyay. Apr 26 '18 at 16:57
and you take $S$ to be any uncountable set. Right? – Arnab Chattopadhyay. Apr 26 '18 at 16:58
From these two facts how can you conclude that $S$ is a basis for $M$? You have to show the linear independence. – Arnab Chattopadhyay. Apr 26 '18 at 17:00
How can you presume that $S$ is linearly independent? It is not clear to me. – Arnab Chattopadhyay. Apr 26 '18 at 17:02
Check the linear independence using my comment above. That's the definition of linear independence, and you just have to check that for any such linear combination. (Hint: formal linear combinations are linearly independent by definition.) – Alekos Robotis Apr 26 '18 at 17:02
Suppose $x_1,x_2, \cdots , x_n$ be finitely many elements of $S$ with $\sum_i a_i x_i=0$ then how do I show that $a_i=0$ for all $i$? I am still thinking! – Arnab Chattopadhyay. Apr 26 '18 at 17:07
Look at the answer of Arturo Magidin in the following link https://math.stackexchange.com/questions/131265/module-generated-by-set-s – Arnab Chattopadhyay. Apr 26 '18 at 17:22
$S$ is a generating set for $M$ but may not be a basis for $M$. You have confused me. – Arnab Chattopadhyay. Apr 26 '18 at 17:24

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

2 Answers2