Unique representation of zero in terms of the basis of a free module

302 Views Asked by At

I am currently reading the book on abstract algebra by Dummit and Foote. They define a free module as follows (p.330):

An $R$-module $F$ is said to be free on the subset $A$ of $F$ if for every element $x$ of $F$, there exists unique elements $\alpha_1,\alpha_2,\dots,\alpha_n$ of $R$ and unique $a_1,a_2,\dots,a_n$ in $A$ such that $x=\alpha_1a_1+\alpha_2a_2+\dots+\alpha_na_n$, for some $n\in\mathbb{Z}^+$. In this situation we say $A$ is a basis or set of free generators for $F$ and the cardinality of $A$ is called the rank of $F$.

Now I am pretty confused by the unicity of the elements of $A$ in the representation of any module element $m\in F$. Take for instance the vector space $\mathbb{R}[x]$ over $\mathbb{R}$ (which is a free module, as all vector spaces are). This is an infinite-dimensional space with basis $A=\{1,x,x^2,\dots\}$, hence by the previous definition we should have that $0\in \mathbb{R}[x]$ has a unique representation in terms of a finite $\mathbb{R}$-linear combination of elements of $\mathbb{R}[x]$. But obviously we can express $0$ in many such combinations. For example, $0=0\cdot 1+0\cdot x^2=0\cdot x^5+0\cdot x^6$, etc. I feel like I am fundamentally misunderstanding something.

For context, I am trying to prove that for a given free $R$-module $M$ over a basis $A$, the definition given is equivalent to saying that $A$ generates $M$ and $r_1a_1+r_2a_2+\dots+r_na_n=0$ for some $r_i\in R$, $a_i\in A$ and $n\in\mathbb{Z}^+$ implies $r_1=r_2=\dots=r_n=0$. As I mentioned, it's the mention of unicity of the $a_1,a_2,\dots$ etc. in the definition that confuses me.

1

There are 1 best solutions below

2
On BEST ANSWER

But obviously we can express $0$ in many such combinations.

You're making an artificial distinction between the coefficients of $0\cdot 1+0\cdot x^2$ and $0\cdot x^5+0\cdot x^6$.

In fact, for an infinite dimensional space, the list of coefficients is always infinite, it's just that we usually omit the terms with zeros when writing a linear combination.

For the sake of your example, consider linear combinations of the form $\sum_{i=0}^\infty \alpha_i x^i$ (finitely supported, of course.) In both representations you suggested, the complete set of coefficients for both expressions is $\alpha_i=0$ for $i\in \{0,1,2,\ldots\}$.

And the unique representation for $x$ would be $\alpha_1=1$ and $\alpha_i=0$ for $i\in \mathbb N$ with $i\neq 1$ and so on.