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It is a fact that under the Axiom of choice, every submodule of a free module under a PID is free. A proof is available at https://math.stackexchange.com/a/162958/1389108.

My question is, does the implication go the other way? Suppose I have the axioms of ZF + the statement that every submodule of a free module under a PID is free. Does this imply the Axiom of choice?

(The intuition for this question is that "every vector space has a basis" is equivalent to the Axiom of choice. So I have suspicions that the same might be true for this module condition.)

Martin Brandenburg
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Dennis Chen
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  • Not sure if this works, but: this implies a subspace of a free vector space (i.e. vector space with a basis) is itself free. If we could injectively map any vector space into some free vector space, it would mean that any vector space has a basis, and this fact is known to be equivalent to AC. – lisyarus Sep 03 '24 at 21:12
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    Embedding a vector space into a free vector space is equivalent to the statement that for every non-zero vector there is a linear functional that maps this vector to something non-zero (or $1$), i.e., that there are enough linear functionals. This is not provable in ZF. – Martin Brandenburg Sep 03 '24 at 21:35
  • @MartinBrandenburg Oh, I see, thanks! – lisyarus Sep 03 '24 at 21:45
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    About my comment, sorry it's not equivalent, but $\implies$ holds, which suffices here. I mixed up $K^I$ with $K^{\oplus I}$. – Martin Brandenburg Sep 04 '24 at 00:06

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