A nonzero PID submodule has a maximal coordinate ideal and a primitive pivot

Some coordinate functional has a nonzero value on $N$. Fix one such nonzero value ideal $I_0$. By [L1], a nonzero generator of $I_0$ has only finitely many divisor classes, so only finitely many principal ideals can contain $I_0$. Choose a maximal value ideal among them, write it as $\psi (N)=(a)$ with $a\mathrm{\ne }0$, and choose $v\in N$ with $\psi (v)=a$.

For any coordinate functional $\phi$, let $d$ generate $(a,\phi (v))$ and choose $r,s$ with $d=ra+s\phi (v)$. The functional $r\psi +s\phi$ takes $v$ to $d$, so its value ideal contains $(d)\supseteq (a)$; maximality in step 1.1 forces $(d)=(a)$, hence $a\mid \phi (v)$.

Divisibility of every coordinate of $v$ gives $v=a{e}_1$ for some $e_1\in M$. Since $a=\psi (v)=a\psi (e_1)$ and $a\mathrm{\ne }0$, cancellation gives $\psi (e_1)=1$, so $e_1$ is primitive.

The ideal generated by the coordinates of $e_1$ in a basis of $M$ is $\{\phi (e_1):\phi \in M^{\ast }\}$, so it does not depend on the basis, and $\psi (e_1)=1$ makes it all of $R$. Fix a basis $f_1,\dots ,f_n$ of $M$ and let $b_1,\dots ,b_n$ be the coordinates of $e_1$ in it. For $j\ge 2$ let $d$ generate $(b_1,b_j)$. If $d=0$ then $b_1=b_j=0$ and nothing is done; otherwise write $d=r{b}_1+s{b}_j$ and put $A=\left(\begin{array}{cc}r& s\\ -b_j\mathrm{/}d& b_1\mathrm{/}d\end{array}\right)$, whose entries lie in $R$ and whose determinant is $(r{b}_1+s{b}_j)\mathrm{/}d=1$, so $A^{-1}$ has entries in $R$ as well. Replacing the basis pair $(f_1,f_j)$ by the pair whose coordinates are the columns of $A^{-1}$ again gives a basis of $M$, in which the coordinates of $e_1$ at $1$ and $j$ are the entries of $A(b_1,b_j{)}^{\mathsf{T}}=(d,0{)}^{\mathsf{T}}$ and the other coordinates are unchanged. Doing this for $j=2,\dots ,n$ in turn clears the coordinates $2,\dots ,n$, so the resulting basis $g_1,\dots ,g_n$ has $e_1=c{g}_1$ with $(c)$ the coordinate ideal, which is $R$. Hence $c$ is a unit and $e_1,g_2,\dots ,g_n$ is a basis of $M$.

Put $u_j=g_j-\psi (g_j)e_1$ for $2\le j\le n$. The passage from $e_1,g_2,\dots ,g_n$ to $e_1,u_2,\dots ,u_n$ is triangular with $1$ on the diagonal, so the latter is again a basis of $M$, and each $u_j$ lies in $\ker \psi$. If $m={\lambda }_1{e}_1+{\sum }_{j\ge 2}{\lambda }_j{u}_j$ lies in $\ker \psi$, applying $\psi$ gives ${\lambda }_1=0$; hence $u_2,\dots ,u_n$ is a basis of $\ker \psi$, which is therefore free of rank $n-1$, and zero when $n=1$. Every $m\in M$ is $\psi (m)e_1+(m-\psi (m)e_1)$ with the second summand in $\ker \psi$, and $R{e}_1\cap \ker \psi =0$ because $\psi (r{e}_1)=r$, so $M=R{e}_1\oplus \ker \psi$. For $w\in N$ one has $\psi (w)\in \psi (N)=(a)$, say $\psi (w)=ra$, and then $w=rv+(w-rv)$ with $w-rv\in N\cap \ker \psi$; also $Rv\cap \ker \psi =0$, since $\psi (rv)=ra$ and $a\mathrm{\ne }0$ in the domain $R$. Hence $N=Ra{e}_1\oplus (N\cap \ker \psi )$. When $a$ is a unit, $v$ and $e_1$ generate the same submodule and the second decomposition reads $N=R{e}_1\oplus (N\cap \ker \psi )$.