Let $R$ be an integral domain and $M$ a finitely generated $R$-module. Show that $rank(M/Tor(M))$=$rank(M)$

The rank of a module over an integral domain is defined as the maximal number of linear independent elements of the module.

Suppose that $[x_1],..,[x_n]$ are linearly independent elements of $M/\mathrm{Tor}(M)$ and let $x_1,...,x_n$ be arbitrary lifts of these elements in $M$ - then clearly $x_1,...,x_n$ are linearly independent in $M$, for else if
$$\sum_{i} \lambda_i x_i =0$$
for some $\lambda_1,..,\lambda_n\in R$ then also
$$0=\pi(\sum_i \lambda_i x_i )  = \sum_i \lambda_i [x_i]\tag{$*$}$$
by linearity of the projection $\pi:M\to M/\mathrm{Tor}(M)$. This implies that the rank of $M/\mathrm{Tor}(M)$ is less than the rank of $M$.

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On the other hand if $x_1,.., x_n$ are linearly independent in $M$ we check that their projections $[x_1],...,[x_n]$ in $M/\mathrm{Tor}(M)$ are also linearly independent. Suppose that there are $\lambda_i\in R$ with:
$$0=\sum_i \lambda_i [x_i] = \pi(\sum_i \lambda_i x_i)$$
where we again used linearity of the projection $\pi$. The above equation means that $\sum_i \lambda_i x_i \in \mathrm{Tor}(M)$, hence there is some $r\in R$ with $r\cdot \sum_i \lambda_i x_i =0$, but then:
$$0=r\cdot\sum_i\lambda_i x_i = \sum_i (r\cdot\lambda_i) x_i$$
which by linear independence of the $x_i$ implies that $r\cdot \lambda_i=0$ for all $i$. But since $R$ is an integral domain you must have that then $\lambda_i=0$ for all $i$. Thus out of equation $(*)$ we may conclude $\lambda_i=0$ for all $i$, which means that $[x_1],..,[x_n]$ are linearly independent. Hence the rank of $M/\mathrm{Tor}(M)$ is larger than the rank of $M$.

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Take both parts together to get that the ranks are equal.