Let $R$ be an integral domain and let $M$ be finitely generated module over $R$. Prove that
$\operatorname{rank}(M) = \operatorname{rank}(M/\operatorname{Tor}(M))$
What I know is if $M$ is finitely generated by $r$ elements then $\operatorname{rank}(M) \le r$. There is a surjective $R$-module homomorphism $M\to M/\operatorname{Tor}(M)$. How to proceed this problem?
Solution 1.
The rank is the largest size of a linearly independent subset of an $R$ module $M$. If we have linearly independent elements $[m_1],\dotsc,[m_n]$ in a quotient module $M/U$, it is trivial to verify that also $m_1,\dotsc,m_n$ are linearly independent in $M$, so $$\mathrm{rank}(M/U) \leq \mathrm{rank}(M)$$ holds anyway. Now assume that $U$ only consists of torsion elements (this includes the most extreme case $U = \mathrm{Tor}(M)$). We will prove $$ \mathrm{rank}(M/U) \geq \mathrm{rank}(M).$$ It is enough to show that for linearly independent elements $m_1,\dotsc,m_n$ in $M$ also their images $[m_1],\dotsc,[m_n]$ are linearly independent in $M/U$. So assume $r_1,\dotsc,r_n \in R$ satisfy $r_1 [m_1] + \cdots + r_n [m_n] = 0$ in $M/U$. By definition, this means that $r_1 m_1 + \cdots + r_n m_n \in U$. Since by assumption $U$ consists of torsion elements, there is some non-zero $s \in R$ such that $s(r_1 m_1 + \cdots + r_n m_n) = 0$. In other words, $s r_1 m_1 + \cdots + s r_n m_n = 0$. The assumption tells us $s r_1 = \cdots = s r_n = 0$. Also, $s \neq 0$ and we work in an integral domain. Hence, $r_1 = \cdots = r_n = 0$, and we are done.
Solution 2.
We use tensor products. The rank of an $R$-module $M$ is the dimension of the $Q(R)$-vector space $M \otimes_R Q(R)$. If $U \subseteq M$ is a submodule that consists of torsion elements, then the exact sequence of $R$-modules $$0 \to U \to M \to M/U \to 0$$ induces the exact sequence of $Q(R)$-modules $$U \otimes_R Q(R) \to M \otimes_R Q(R) \to M/U \otimes_R Q(R) \to 0,$$ using that the tensor product is right exact. But $U \otimes_R Q(R)$ vanishes: For every $u \in U$ there is some non-zero $r \in R$ with $ru = 0$, and then $u \otimes 1 = u \otimes r \cdot \frac{1}{r}= ru \otimes \frac{1}{r} = 0$. (More generally, the tensor product of a torsion module and a divisible module vanishes.) So the exact sequence simplifies to $M \otimes_R Q(R) \cong M/U \otimes_R Q(R)$, so the dimensions are the same.
The 2nd solution shows that the claim also holds when $M$ is not assumed to be finitely generated (and the 1st solution can be adapted to show this well).