Let $E$ be finitely generated. Then $E/E_{\text{tor}}$ is free. There exists a free submodule $F$ of $E$ such that $E$ is a direct sum $E = E_{\text{tor}}\oplus F$. The dimension of such a submodule $F$ is uniquely determined.
In the proof of this theorem in Lang (p.147, Theorem 7.3), they said $x\mapsto ax$ is injective. I think is nontrivial to me so I tried to prove it (All the notations follow the proof in Lang): Let $x\in M$ such that $ax=0$. Let $x = \sum_{i=1}^m b_iy_i$ then $0=ax = \sum_{i=1}^mb_iay_i = \sum_{i=1}^mb_ia_1\cdots \hat{a_i}\cdots a_ma_iy_i = \sum_{i=1}^mb_ia_1\cdots\hat{a_i}\cdots a_m\sum_{j=1}^nc_{j}^{(i)}v_j = \sum_{j=1}^n(\sum_{i=1}^mb_ia_1\cdots\hat{a_i}\cdots a_mc^{(i)}_j)v_j\iff \sum_{i=1}^mb_ia_1\cdots\hat{a_i}\cdots a_mc^{(i)}_j = 0\ \forall j=1,...,n$
where $a_iy_i = \sum_{j=1}^n c_j^{(i)}v_j,\ c_j\in R$. In other words,
$$\begin{pmatrix} c_1^{(1)}& \cdots& c_1^{(m)} \\ \vdots & \ddots & \vdots \\ c_{n}^{(1)}&\cdots& c_n^{(m)} \end{pmatrix} \begin{pmatrix} b_1\hat{a_1}\cdots a_{m} \\ \vdots\\ b_{m}a_1\cdots\hat{a_m} \end{pmatrix} = 0$$
Here's where I'm stuck. From this, how can I conclude $b_i = 0$ for all $i$?
