The following problem might be as simple as it looks.
Let $R$ be an integral domain, $M$ a free $R$-module of rank $2$, and $\{m_{1},m_{2}\}$ a basis for $M$. Suppose that $$ x = s_{1}m_{1}+s_{2}m_{2}\ \ (s_{1},s_{2} \in R, s_{1} \neq 0) $$ and $(c)=(s_{1},s_{2})$ as ideals of $R$. Prove that we can choose an $R$-basis $\{ n_{1},n_{2} \}$ for $M$ so that $x=cn_{1}$.
Since, $s_i \in (c)$ we can write $s_i=cr_i$ for $i=1,2$ and $r_1 \ne 0$ as $s_1 \ne 0$. Set $n_1:= r_{1}m_{1}+r_{2}m_{2}$. Then $x=cn_1$. One sees that, $\{ n_{1},m_{2} \}$ is a linearly independent set because det$(n_1,m_2)=r_1 \ne0$ It follows that, $\{ n_{1},m_{2} \}$ is a basis of $M$ that we wanted to find.
Is proof above correct?
Any help would be much appreciated.
The proof is indeed faulty. However, the hunch that $n_1=r_1m_1+r_2m_2$ is part of a basis $\{n_1,n_2\}$ of $M$ is sound, it is only that the idea that we can take $n_2=m_2$ is too simple-minded.
Since $(s_1,s_2)=(c)$, we have $c = k_1s_1+k_2s_2$ for some $k_1,k_2\in R\,$. Substituting $s_i=c\mspace{1.5mu}r_i$ and canceling the factor $c$ (which we can do because $R$ is an integral domain), we obtain $1=k_1r_1+k_2r_2$. Now we define \begin{equation*} n_1 := r_1m_1 + r_2m, \quad n_2 := -k_2m_1+k_1m_2~, \end{equation*} which we rewrite as \begin{equation*} \begin{pmatrix} n_1 \\ n_2 \end{pmatrix} \,=\, \begin{pmatrix} r_1 & r_2 \\ -k_2 & k_1 \end{pmatrix} \begin{pmatrix} m_1 \\ m_2 \end{pmatrix} ~, \end{equation*} where the determinant of the conversion matrix is $1$: this proves that $\{n_1,n_2\}$ is a basis of $M$.