## Over an integral domain, the modules R and R/Tor R have the same rank

Let $R$ be an integral domain and let $M$ be a (left, unital) $R$-module. Recall that the rank of a module is the largest possible cardinality of a linearly independent subset. $\mathsf{Tor}\ M$ is the set of all $m \in M$ such that $rm = 0$ for some nonzero $r \in R$.

1. Prove that $\mathsf{Tor}\ M$ has rank 0.
2. Prove that $M$ and $M/\mathsf{Tor}\ M$ have the same rank.

Let $S \subseteq \mathsf{Tor}\ M$ be a nonempty set. Note that for each $m \in S$, there exists a nonzero element $r \in R$ such that $rs = 0$. In particular, $S$ cannot be linearly independent. Thus $\mathsf{Tor}\ M$ has rank 0.

Let $S = \{m_i\}_I \subseteq M$ be a linearly independent set of maximal order. Note in particular that no element of $S$ can be in $\mathsf{Tor}\ M$ (each subset of $S$ is linearly independent, in particular the singleton subsets are linearly independent.) We claim that the subset $S + \mathsf{Tor}\ M \subseteq M/\mathsf{Tor}\ M$ is linearly independent. To see this, suppose $\sum r_i m_i \in \mathsf{Tor}\ M$ for some $r_i \in R$. Then there exists a nonzero element $a \in R$ such that $\sum ar_im_i = 0$ in $M$. Since $S$ is linearly independent, we have $ar_i = 0$ in $R$. Since $R$ is an integral domain, and $a \neq 0$, we have $r_i = 0$ for all $i$. Thus $S + \mathsf{Tor}\ M$ is linearly independent. We claim also that $S$ and $S + \mathsf{Tor}\ M$ have the same cardinality. It suffices to show that for all $i$ and $j$, $m_i - m_j \notin \mathsf{Tor}\ M$. To that end, suppose $a \in R$ such that $am_i - am_j = 0$. Since $m_i$ and $m_i$ are linearly independent, $a = 0$. Thus $S$ and $S + \mathsf{Tor}\ M$ have the same cardinality. In particular, the rank of $M/\mathsf{Tor}\ M$ is at least as great as the rank of $M$.

Conversely, let $S = \{m_i\}$ and suppose $S + \mathsf{Tor}\ M$ is a linearly independent subset of maximal order. We claim that $S$ is linearly independent in $M$. To see this, suppose $\sum r_i m_i = 0$. Now $\sum r_i \overline{m_i} = 0$ in $M/\mathsf{Tor}\ M$. Thus $r_i = 0$, and so $S \subseteq M$ is linearly independent. Certainly the cardinality of $S$ is at least as great as the cardinality of $S + \mathsf{Tor}\ M$. Thus the rank of $M$ is at least as great as the rank of $M/\mathsf{Tor}\ M$.

Hence $M$ and $M/\mathsf{Tor}\ M$ have the same rank.