This question is from my assignment of localization of modules and I need help with it.
Let A be a non-zero commutative ring. $S_0$ be the multiplicatively closed subset of non-zero divisors in A and $Q(A)= S_{0}^{-1}A $ be the total quotient ring of A. An A-module V is called a module with rank over A if $S_{0}^{-1} V$ is a free Q(A) module; we also say that V has rank over A and put $Rank_A V= Rank_{Q(A)} S_{0}^{-1} V$.
(a) Every free A-module V is an A-module with rank and in this case its rank is the rank of the free A-module V.
I have done this.
(b) If A is an integral domain, then Q(A) is the quotient field of A and hence every A-module has a rank and $Rank_A V= \dim_{Q(A)} S_{0}^{-1}V$.
How to construct basis of A-module in this case?
(c) If V is an A-module with rank , then $S_{0}^{-1} V$ has a Q(A)- basis of the type $x_i /1$ $i\in I$ and $x_i , i\in I$ is a maximal linearly independent (over A) family in V.
V is an A-module with rank means that $S_{0}^{-1} V$ is a free Q(A) module means there exists a basis. I have been given $x_i's$ are maximal Linearly independent set for V , it must also generate V as if it doesn't an element say v then { $v, x_i $} is a L I set , hence {$x_i's$} is not maximal LI set. So, this part is easy to complete and I have done it.
(d) Every finite torsion free A-module with rank is isomorphic to a A-submodule of a finite free A-module.
Well, any submodule of a free A-module will also be free and if in the basis of V there exists an element $a \in A$ which is not torsional free then I have av=0 , which will force v =0 which is a contradiction. Any submodule of V and V itself are torsion free but how should I prove that an isomorphism exists?
Thanks!
Best Answer
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 $V$.