Given an integral domain $R$ and an $R$-module $M$, an element of $x$ of $M$ is said to be a torsion element of $M$ if there exists some nonzero $r\in R$ such that $r\cdot x=0$. The set of all torsion elements of $M$, denoted $Tor(M)$, is a submodule of $M$.
The rank of $M$ is defined to be the cardinality of the largest subset of $M$ which is $R$-linearly independent.
Prove that any module in which all elements are torsion has rank 0. Then prove that the rank of $M$ is equal to the rank of $M/Tor(M)$.
Proving the first part is easy once one realizes that any linearly independent subset cannot have any torsion elements. So, if every element is torsion, the largest linearly independent set is $\emptyset$, and so the rank is 0.
The second part is tripping me up. I have thought about two ways of doing it; I think the more rigorous way would be this:
Let $S\subseteq M$ be a linearly independent set with largest possible cardinality, so that its cardinality is the rank of $M$. Then define (in some sense; I'm finding it hard to do this part) the subset of $M/Tor(M)$ which has the same elements as $S$, but with all elements of $Tor(M)$ identified / "glued" together (that is my topology background showing). It might be something like $S'=\lbrace s+Tor(M):s\in S \rbrace$. Then, since $S$ cannot have any elements of $Tor(M)$ in it, $S$ and $S'$ have the same cardinality. If $S'$ is linearly independent, then we are almost done. The only part I'm also unsure that this $S'$ has the largest possible cardinality for a linearly independent subset of $M/Tor(M)$.
The second, more heuristic method I thought of is the following: intuitively, $M/Tor(M)$ should have rank less than or equal to the rank of $M$, since $M$ loses elements when we take its quotient. If $S$ is as above, then $S$ cannot lose any elements when we pass to the quotient module, since it can't have any torsion elements in it. Thus they in fact have the same rank.
Best Answer
Write $S+{\rm Tor}(M)=\{s+{\rm Tor}(M):s\in S\}\subset M/{\rm Tor}(M)$.
There is a map $S\mapsto S+{\rm Tor}(M)$ which sends subsets of $M$ to those of $M/{\rm Tor}(M)$. Then you can check that $S$ is linearly independent iff $S+{\rm Tor}(M)$ is linearly independent, and also that this assignment preserves cardinality. Then prove by contradiction that there can't be maximally sized linearly independent subsets $S\subset M$, $S'\subset M/{\rm Tor}(M)$ of different cardinality.