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The question about projective modules over a PID being free, has been posted here, and then closed because it duplicated a previous one.

However, that other posting addresses the more general result by which, in a PID, submodules of free modules are always free. Unfortunately, the more general result has a fairly advanced proof (it uses ordinals and transfinite induction rather than, say, Zorn). The case where the submodule is projective, however, should be much simpler. In fact, it is Exercise 13 a), Chapter 6, of Advanced Linear Algebra by Steven Roman. Moreover, this could be a route to finding a simpler proof of the more general theorem.

Could someone provide a hint on how to attack the exercise, i.e., how to prove that, in a PID, projective implies free?

Leandro Caniglia
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    Which edition of Advanced Linear Algebra are you using? I can't find that exercise in the third (most recent) edition. Maybe the exercise was removed because the author noticed that the non-finitely generated case was too difficult. – anankElpis Nov 29 '24 at 01:53
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    But at least in the current edition, Roman does prove the general fact about submodules of free modules in Theorem 6.5. The proof is fairly understandable in my opinion (and only uses well-orderings instead of transfinite induction). When dealing with non-finitely generated modules, I wouldn't expect something much simpler to exist. – anankElpis Nov 29 '24 at 01:56
  • @anankElpis Second edition. – Leandro Caniglia Nov 29 '24 at 13:12

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I don't see an easier approach than the mentioned general fact that submodules of free modules are free. Luckily, there are fairly elementary proofs for that, in particular it is not necessary to use ordinals and transfinite induction. One such proof can be found in Steven Roman, Advanced Linear Algebra, 3rd edition, Theorem 6.5. For completeness, let me write it (or rather a more concise version) down here.

Let $R$ be a PID, $M$ a free $R$-module and $S\subset M$. We may assume $M=(R^\kappa)_0$, the $R$-module of finitely supported functions from $\kappa$ to $R$ for some cardinal $\kappa$. Let $0<\alpha\leq \kappa$. Define $M_\alpha:=\{f\in S\mid \operatorname{supp}(f)\subseteq[0,\alpha]\}$. Then $I_\alpha:=\{f(\alpha)\mid f\in M_\alpha\}$ is an ideal, hence of the form $I_\alpha=\langle f_\alpha(\alpha)\rangle$ for some $f_\alpha\in M_\alpha$ since $R$ is a PID. Finally let $$B=\{f_\alpha\mid 0<\alpha\leq\kappa,\ f_\alpha(\alpha)\neq0\}.$$ We show that $B$ is a basis: If $\sum r_if_{\alpha_i}=0$ is a minimal non-trivial linear combination, then plugging in $\max(\alpha_i)$ yields a contradiction. Thus $B$ is linearly independent.

Assume $B$ does not span $S$, and for $f\in M$ denote by $i(f)$ the largest $\alpha$ with $f(\alpha)\neq0$. Since $\kappa$ is well-ordered, we can find $g\in S\setminus\langle B\rangle$ with $i(g)=:\alpha$ minimal. From $0\neq g(\alpha)\in I_\alpha$ it follows that $g(\alpha)=cf_\alpha(\alpha)$ for some $c\in R$, and $f_\alpha\in B$. But then $i(g-cf_\alpha)<\alpha$, hence $g-cf_\alpha\in\langle B\rangle$ by minimality of $\alpha$. Finally $g=(g-cf\alpha)+cf_\alpha\in\langle B\rangle$, contradiction.

anankElpis
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