Modules with rank and localization

Solution 1:

There are quite a few misunderstandings in what you say, so let's go through them.

b) You do not need to construct a basis of an $A$-module here. All you need is that every vector space (e.g. module over the field $Q(A)$) has a basis.

c) No, the $x_i$ do not necessarily generate $V$. A set can be maximal linearly independent, without generating the module. For example regard $\mathbb{Z}$ as a module over itself. Then $\{2\}$ is a maximal linearly independent set. Thus $\{2,3\}$ is not linearly independent: $2\times 3-3\times 2=0$. However $3$ is not in the span of $2$.

In fact you do not need to show the $x_i$ generate $V$ here. You are not "given" the $x_i$ are maximal linearly independent. You need to deduce it:

Any $v\in S_0^{-1}V$ has the form: $$v=s^{-1}\left(\sum_i a_i x_i) \right),$$ with $s\in S_0$ and $a_i\in A$. Thus $$vs-\left(\sum_i a_i x_i) \right)=0,$$ so $\{x_i|i\in I,v\}$ is not linearly independent over $A$.

d) No a submodule of a free module is not necessarily free. Consider $\mathbb{Q}\times \mathbb{Q}$ as a module over itself. Then neither factor of $\mathbb{Q}$ is free.

The point here is that because $V$ is torsion free, it embeds in $S_0^{-1}V$, which has a basis $\frac {x_i}1$ over $Q(A)$. As $V$ is finitely generated, all its generators are contained in the span of the $\frac {x_i}s$, for some fixed $s\in S$. It remains only to show that the $\frac {x_i}s$ are linearly independent over $A$, which follows from the fact that they are linearly independent over $Q(A)$.

Thus we can conclude that over $A$, the $\frac {x_i}s$ span a free $A$-module, which contains $V$.