Under Choice, a submodule of an arbitrary-rank free module over a PID is free

Fix a basis $(e_a{)}_{a\in I}$ of $F$ and well-order $I$. Put $F_{\le a}=\mathrm{span}\{e_b:b\le a\}$, $F_{<a}=\mathrm{span}\{e_b:b<a\}$, and $N_{\le a}=N\cap F_{\le a}$. The coordinate projection $p_a:F\to R$ is linear by uniqueness of basis expansions.

Its image $J_a=p_a(N_{\le a})$ is an ideal: $p_a(x)-p_a(y)=p_a(x-y)$ and $r{p}_a(x)=p_a(rx)$ for $x,y\in N_{\le a}$ and $r\in R$. Write $J_a=({\rho }_a)$. For each $a$ with $J_a\mathrm{\ne }0$, the set of pairs $(\rho ,u)$ with $J_a=(\rho )$, $u\in N_{\le a}$, and $p_a(u)=\rho$ is nonempty. AC selects such a pair $({\rho }_a,u_a)$ simultaneously for these indices. In particular ${\rho }_a\mathrm{\ne }0$. Let $S=\{a:J_a\mathrm{\ne }0\}$.

We verify the hypothesis of transfinite induction for the assertion that $N_{\le a}$ is spanned by the $u_b$ with $b\in S$, $b\le a$. Suppose the assertion holds at every $b<a$ and take $x\in N_{\le a}$. If $p_a(x)=0$, put $y=x$. Otherwise $a\in S$, and $p_a(x)=r{\rho }_a$ for some $r\in R$; put $y=x-r{u}_a$. In both situations $y\in N\cap F_{<a}$. If $y\mathrm{\ne }0$, its finite support has a greatest element $b<a$, so the assertion at $b$ expresses $y$ in the required earlier generators. If $y=0$, its expression is the empty sum. Restoring $r{u}_a$ when present proves the assertion at $a$.

For a finite relation ${\sum }_{a\in E}{r}_a{u}_a=0$ with distinct $a\in S$, suppose some coefficient is nonzero and take the greatest such index $a$. Every $u_b$ with $b<a$ has zero $a$ coordinate. Applying $p_a$ gives $r_a{\rho }_a=0$. Since $R$ is a domain and ${\rho }_a\mathrm{\ne }0$, this forces $r_a=0$, contrary to its selection. Thus all coefficients vanish.

Transfinite induction now proves the assertion at every $a$. Each nonzero $x\in N$ has a greatest support index and hence belongs to one $N_{\le a}$; therefore the $u_a$ span $N$. This also covers a limit initial segment: every one of its finite supports lies in a smaller principal initial segment, so no additional generator is needed at a limit cut.

Spanning and independence make $(u_a{)}_{a\in S}$ a basis. If $N=0$, every $J_a=0$ and this is the empty basis; if $I=\mathrm{\varnothing }$, also $F=N=0$. In rank one the same construction is simply the zero ideal or its single nonzero generator. These possibilities require no choice from an empty fiber.