Torsion Elements and Rank (12.1.1)

Dummit and Foote Abstract Algebra, section , exercise :

Let $M$ be a module over the integral domain $R$.
(a) Suppose $x \neq 0$ is a torsion element of $M$. Show $x$ and $0$ are "linearly dependent." Conclude the rank of $\text{Tor}(M)$ is $0$, so that in particular any torsion $R$-module has rank $0$.
(b) Show $\text{rank} M = \text{rank} M/\text{Tor}(M)$.

Proof: (a) By definition we have $rx=0$ for nonzero $r$. Therefore there exist no candidates for linearly independent sets.

(b) Let $n = \text{rank} M$ and $m = \text{rank} M/\text{Tor}(M)$. ($≥$) Let $\overline{x_1},...,\overline{x_m}∈M/\text{Tor}(M)$ be linearly independent. If $m > n$ then we can arrange for a linear dependence$$r_1x_1+...+r_mx_m = 0$$which clearly extends to give a linear dependence in the quotient. ($≤$) Let $x_1,...,x_n∈M$ be linearly independent. Assume that their images in the quotient are linearly dependent, so that$$r_1x_1+...+r_nx_n∈\text{Tor}(M)$$and by definition of $\text{Tor}(M)$ we obtain a nonzero annihilator to obtain a contradictory linear dependence.$~\square$