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I'm reading Kenneth Hoffman's Linear Algebra, Ed2.

In $\S5.5$ it talks about Module and Vector Spaces:

(1) If $K$ is a commutative ring with identity, a module over $K$ ( or a $K$-module) is an algebraic system which behaves like a vector space, with $K$ playing the role of the scalar field.

(2) A basis for the module $V$ is a linearly independent subset which spans (or generates) the module. This is the same definition which we gave for vector spaces; and, the important property of a basis $\mathscr B$ is that each element of $V$ can be expressed uniquely as a linear combination of (some finite number of) elements of $\mathscr B$.

(3) The reader is well aware that a basis exists in any vector space which is spanned by a finite number of vectors. But this is not the case for modules.

(4) Definition. The $K$-module $V$ is called a free module if it has a basis. If $V$ has a finite basis containing $n$ elements, then $V$ is called a free $K$-module with $n$ generators.

(5) Definition. The module $V$ is finitely generated if it contains a finite subset which spans $V$. The rank of a finitely generated module is the smallest integer $k$ such that some $k$ elements span $V$.

(6) We repeat that a module may be finitely generated without having a finite basis.

(7) Theorem 5. Let $K$ be a commutative ring with identity. If $V$ is a free $K$-module with $n$ generators, then the rank of $V$ is $n$.

(8) From Theorem 5 we know that ‘free module of rank $n$’ is the same as ‘free module with $n$ generators.’

I'm quite lost here --

Q1: (3) and (6) seem are talking about the same thing, that a module $V$ could be spanned by a finite number of elements $\{\beta_1, \dots, \beta_n\}$, but $V$ might not have a finite basis. Does this mean that in module $V$ there might be some element $\alpha$, that it can be expressed as a linear combination of $\{\beta_1, \dots, \beta_n\}$, but in two different ways? (So this violates the "uniquely" requirement in the basis definition?) If so, could you pls give me some examples of such a $V$?

Q2: however, (8) says: ‘free module of rank $n$’ is the same as ‘free module with $n$ generators.’ Seems this conflicts with (3) and (6)?

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Q1: Yes. As an example take $V=\mathbb{Z}/2\mathbb{Z} = \{ [0],[1]\}$ as a $\mathbb{Z}$-module. Then you have $$[1] \cdot 1 = 3 \cdot [1] = 5 \cdot [1] = \cdots$$ are all different representations of $[1]$ of the form $\lambda \cdot [1]$ with $\lambda \in \mathbb{Z}$.

Q2: No. (8) means that IF a module is free and is generated by $n$ elements, then every other basis has $n$ elements. (8) is not talking about modules which are not free. (3) and (6) just say that there exist some non-free modules.

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  • $\begingroup$ thank you so much, now I get it! $\endgroup$ – athos Mar 23 '15 at 18:42

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