## Modules of fractions

Let be an integral domain with field of fractions . Let be a left unital -module, and let denote the set of all nonzero elements of . Define a relation on as follows: if and only if . This is clearly an equivalence relation; denote by the set of -equivalence classes.

- Prove that the operation on given by is well defined, and that is an abelian group. Prove further that the operation makes into a left -module.
- Prove that .
- Prove that in if and only if for some .
- Let be an abelian group. Prove that if and only if is torsion (that is, every element of has finite order).

- Suppose and . Then and , so that and . Adding these equations together, we see that . Thus . In particular, the operator on is well defined.
Since , is associative.

Since , is a left identity (similarly, a right identity). Moreover, note that for all .

Since , every element of has an additive inverse. So is a group with respect to .

Since , is an abelian group.

Now suppose . Then , so that (since is commutative). Thus , and so the action is well defined.

Now , , and . So is a left -module with respect to this action.

- Define by . Note that if , then , so that , and thus . That is, is well-defined. Moreover, we have , , and . Thus is -balanced, and so induces an -module homomorphism such that .
Now define by . If , then . So , so that . That is, is well-defined. Moreover, note that . Thus is a group homomorphism.

Now and . Thus and are mutual inverses, and so is a group isomorphism, hence an -module isomorphism. So .

- Suppose in . Then . In particular, we have . Conversely, suppose with and let be nonzero. Then .
- Suppose . In particular, every simple tensor is zero, so that by part (3) above, there exists a nonzero such that . Thus is torsion. conversely, if is torsion, then for every element , there exists nonzero so that . But then the simple tensor is zero, so that .

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