# Basis of a module

#### quasar987

Homework Helper
Gold Member
Why couldn't a module have a finite basis and an infinite one?

Related Linear and Abstract Algebra News on Phys.org

#### ThirstyDog

I think the reasoning goes something like this:

Consider bases for some R-module, say A,
$$\{x_{i}\}_{i=1}^{n} \mbox{ and } \{y_{i}\}_{i=1}^{\infty}$$
Everything in A can be written as a FINITE linear combination of elements from either basis. That means there is some natural number m with the property
$$x_{i} = \sum_{j=1}^{m} \alpha_{i}^{j} y_{j}, \forall 1 \leq i \leq n \mbox{ and } \alpha_{i}^{j} \in R$$
Every x can be written as a linear combination of y's from some finite subset.

Now consider an arbitrary element in A it can be written in terms of finite number of x's which can be written in terms of a finite number of y's thus everything can be written using a finite subset of the y's. In particular
$$y_{m+1} = \sum_{i=1}^{n} \beta_{i}x_{j} = \sum_{i=1}^{n}\sum_{j=1}^{m} \beta_{i}\alpha_{i}^{j} y_{j}$$

This proves that the infinite basis is not linearly independent therefore in fact not a basis. If you find a flaw in this working or have more question tell me.

#### quasar987

Homework Helper
Gold Member
Oh, very nice. Thank you!

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving