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[parent] example of free module (Example)

Clearly from the definition, $ \mathbb{Z}^n$ is free as a $ \mathbb{Z}$-module for any positive integer $ n$.

A more interesting example is the following:

Theorem 1   The set of rational numbers $ \mathbb{Q}$ do not form a free $ \mathbb{Z}$-module.
Proof. First note that any two elements in $ \mathbb{Q}$ are $ \mathbb{Z}$-linearly dependent. If $ x=\frac{p_1}{q_1}$ and $ y=\frac{p_2}{q_2}$, then $ q_1p_2x-q_2p_1y=0$. Since basis elements must be linearly independent, this shows that any basis must consist of only one element, say $ \frac{p}{q}$, with $ p$ and $ q$ relatively prime, and without loss of generality, $ q>0$. The $ \mathbb{Z}$-span of $ \{\frac{p}{q}\}$ is the set of rational numbers of the form $ \frac{np}{q}$. I claim that $ \frac{1}{q+1}$ is not in the set. If it were, then we would have $ \frac{np}{q}=\frac{1}{q+1}$ for some $ n$, but this implies that $ np=\frac{q}{q+1}$ which has no solutions for $ n,p\in\mathbb{Z}$ , $ q\in\mathbb{Z}^+$, giving us a contradiction. $ \qedsymbol$



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Cross-references: contradiction, solutions, implies, without loss of generality, relatively prime, basis, linearly independent, rational numbers, integer, positive
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This is version 2 of example of free module, born on 2003-07-29, modified 2003-07-30.
Object id is 4534, canonical name is ExampleOfFreeModule.
Accessed 1935 times total.

Classification:
AMS MSC: 13C10 (Commutative rings and algebras :: Theory of modules and ideals :: Projective and free modules and ideals)
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