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The theorem in question is:

Theorem. Let $R$ be a principal ideal domain and let $F$ be a free $R$-module. Then any submodule $K\subseteq F$ is free.

This is proved in Hungerford theorem 6.1, and elsewhere using the axiom of choice and transfinite induction very explicitly. I have a feeling my proof is too simple, and so has to be very incorrect. It doesn't (explicitly) assume the axiom of choice, nor explicitly use transfinite induction.

Proof. Consider $F$ to be free over the set $X=\{x_i|i\in I\}$. That is, every element of $F$ can be written as a finite sum $\sum_j r_j x_j$. Denote $K_i=K\bigcap \langle x_i\rangle=\{r_i x_i|\sum_j r_j x_j\in K\}$.

First, I want to show each $K_i=\langle h_i\rangle$ for some $h_i\in K_i$. So consider the projection $p_i:K_i\to R$ with $p_i(r_i x_i)=r_i$ (which is well defined by the universal free property of $F$). This must have $\ker p_i=(0)$ (by the construction of the free module, $x_i$ is just a formal placeholder, and $r_i x_i=0$ if and only if $r_i=0$). Since $K_i$ is an $R$-module, $p_i(K_i)$ is an $R$-module and so is generated by a single element, $p_i(K_i)=\langle a_i\rangle$. Then $p_i:K_i\to \langle a_i\rangle$ is a one-to-one and onto homomorphism. Denote $h_i=p_i^{-1}(a_i)$. Then $K_i=\langle h_i\rangle$. Every nonzero principal ideal of an integral domain is isomorphic to the ring, so each $K_i$ is isomorphic to $R$ or $(0)$.

Next, consider the internal direct sum of all $K_i$, $\bigoplus^{i\in I} K_i$. Due to its construction by taking intersections, it is a subset of $K$. If we are given an arbitrary finite sum $\sum_i r_i x_i\in K$, each $r_i x_i$ must individually lie in some $K_i$, and so the whole thing is in the direct sum of the $K_i$. Thus $K=\bigoplus^{i\in I}\langle h_i\rangle\cong \bigoplus_{i\in I} R$ (where the last direct product is taken over all $i$ such that $K_i\neq (0)$), and so $K$ is free. $\square$

as far as I see, I have used:

  • the existence of the direct sum over arbitrary families indexed by $I$
  • the fact that an $R$-module is equal to some direct sum of $R$ if and only if it is free
  • that every nonzero principal ideal of an integral domain $R$ is isomorphic to $R$.

but the proof can't be correct! The proof in Hungerford runs a page long (where most proofs in Hungerford are a paragraph or a few lines!), and emphasizes the use of the axiom of choice. Where have I been led astray?

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It's not true that $K\cap\langle x_i\rangle=\{r_i x_i|\sum_j r_j x_j\in K\}$. For instance, an element of the form $ax_1+bx_2$ might be in $K$ even if $ax_1\not\in K$. You can see this in a really simple example by taking $R=\mathbb{Z}$, $F=\mathbb{Z}^2$, and $K$ to be generated by $(1,1)$. In this example $K\cap\langle x_i\rangle=0$ for all $i$, and it is not true that $K$ is the sum of the $K_i$.

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    $\begingroup$ Agh, I knew I should have got more sleep last night. And yep, the proof isn't repaired at all if $K_i$ is replaced with $\{r_i x_i|\cdots\}$, because then $K_i\subseteq K$ doesn't hold. Thank you! $\endgroup$ – user18862 Feb 13 '17 at 2:47
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EDIT: Eric Wofsey pointed out the fundamental error, which is ring theoretic. However, there is a step which appears to invoke choice (but see Eric's comments below), and I think it worthwhile to point it out.

Specifically, the following step smells like choice:

You state

Every nonzero ideal of an integral domain $R$ is isomorphic to $R$.

And from this you conclude that - since each $\langle h_i\rangle$ is isomorphic to $R$ - we have $$\bigoplus_{i\in I}\langle h_i\rangle\cong \bigoplus_{i\in I}R.$$ However, this appears to require you to choose an isomorphism $f_i: \langle h_i\rangle\rightarrow R$ for each $i\in I$; otherwise, how do you build the "large" isomorphism above?

It may be that there's a hidden uniformity here that lets you make this work without choice; as I said, ring theory isn't my specialty. But currently I think this step relies on choice.

Tl;dr: in general algebraic contexts, choice is needed to argue that "$A_i\cong B_i$ ($i\in I$)" implies "$\bigoplus A_i\cong\bigoplus B_i$" (or similar).

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  • $\begingroup$ This step seems to require choice for arbitrary $R$, but it is avoidable if $R$ is well-orderable (since all you need is to choose a generator of the image of $p_i$). $\endgroup$ – Eric Wofsey Feb 13 '17 at 2:41
  • $\begingroup$ @EricWofsey Quite right, but there's no assumption that $R$ be well-orderable that I see (although of course the error you observed is the fundamental one). $\endgroup$ – Noah Schweber Feb 13 '17 at 2:41
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    $\begingroup$ Right, of course. I was more musing along the lines that even seeing this place where choice could be needed, you still know there must be another error or overlooked use of choice since choice is needed even in the case that $R$ is well-orderable. $\endgroup$ – Eric Wofsey Feb 13 '17 at 2:45
  • $\begingroup$ @EricWofsey Stupid question: why is choice necessary even if $R$ is well-orderable? Off the top of my head I don't see this . . . $\endgroup$ – Noah Schweber Feb 13 '17 at 2:49
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    $\begingroup$ @AsafKaragila: The question is whether if $R$ is a well-orderable PID, "any submodule of a free $R$-module is free" requires choice. $\endgroup$ – Eric Wofsey Feb 13 '17 at 17:41

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