This particular question was asked in my Abstract Algebra Exam and I couldn't solve it in examination hall. Now, I thought I should try it again but I was unable to.
Prove that every free module F over an arbitrary integral domain with identity is torsion-free but the converse is false.
Module is given free means that a basis set exists (say X).To prove that it is Torsional free i need to show that every element except identity has infinite order OR Assuming it to be not torsional free and then proving that it can't be free module.
Let $f \in F$. Then I am at loss of ideas how it doesn't have any non-identity element of finite order.
On the other hand, let converse holds. So, Every torsion-free module over an arbitrary integral domain is free. But what is the contradiction here?
I am at loss of ideas over this and would very much appreciate help.
Every free module over an arbitrary integral domain with identity is torsion-free.
Let $F$ be such a D-module, where $D$ is an integral domain, i.e. $D$ is commutative, has identity, and no non-zero divisors.
By Characterization of free D-modules ( any $R$-modules with identity actually ), $F$ is isomorphic to a direct sum of copies of the D-modules $D$ , say : $$F\cong \oplus_{i\in I} D$$ Then any element $x$ of $F$ that is not the neutral element can be identified with $\phi(x)= \sum_{i\in I} d_i\neq 0$ where the $d_i\in D$ ( and at least one of the $d_i$'s is not zero ), and for any $r\in D: $ $r\phi(x)= r(\sum_{i\in I} d_i)= \sum_{i\in I} rd_i= 0 \implies rd_i=0 \ \forall i\in I$ ( the former zero is the zero of the direct sum ) thus as $D$ has no zero divisors : $$r\phi(x)= 0\ \implies \forall i\in I:\ r= 0 \ or\ d_i= 0\ \implies r=0 $$ ( as we established one of the $d_i$'s at least is non zero )
This means that $\phi(x)$ has an infinite order hence as $\phi$ was our isomorphism also $x$ must have infinite order as desired.
The converse is false, think about an example of a torsion free module over $Z$ (an integral domain) which is not free.