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The following is a theorem from Hungerford's Algebra, Section II.1:

Theorem 1.6. If $F$ is a free abelian group of finite rank $n$ and $G$ is a nonzero subgroup of $F$, then there exists a basis $\{x_1,\ldots,x_n\}$ of $F$, an integer $r$ ($1\leq r\leq n$) and positive integers $d_i,\ldots,d_r$, such that $d_1\mid d_2\mid\cdots\mid d_r$, and $G$ is free abelian with basis $\{d_1x_1,\ldots,d_rx_r\}$.

The theorem (and its proof) naturally generalizes to Euclidean domains. My question is, does this hold for modules over PIDs? I.e.,

If $R$ is a PID, $F$ is a free $R$-module of finite rank $n$ and $M$ is a nonzero submodule of $F$, then there exists a basis $\{x_1,\ldots,x_n\}$ of $F$, an integer $r$ ($1\leq r\leq n$) and elements $d_i,\ldots,d_r\in R$, such that $d_1\mid d_2\mid\cdots\mid d_r$, and $M$ is free with basis $\{d_1x_1,\ldots,d_rx_r\}$?

I'm not familiar with PIDs that are not Euclidean domains. Can anyone prove or disprove this? Thanks in advance!

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  • $\begingroup$ Yes, this works for PIDs. It's the Smith Normal Form algorithm. $\endgroup$ – Lord Shark the Unknown Mar 25 '18 at 12:46
  • $\begingroup$ @LordSharktheUnknown Thanks! I'll look into that. $\endgroup$ – Colescu Mar 25 '18 at 12:50

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