Without using the fact that submodules of free modules over PIDs, how can one go about showing that the real numbers don't form a free Z module

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    $\begingroup$ Do you know how to show that a free $\Bbb Z$-module is not divisible? On the other hand, $\Bbb R$ is divisible. $\endgroup$ – Ayman Hourieh Jan 27 '14 at 21:49
  • $\begingroup$ Minor tidbit, "submodules of free modules over PIDs" is not a fact, it is a collection of things (namely, of various modules). Perhaps you forgot to write something about a fundamental theorem. $\endgroup$ – anon Jan 28 '14 at 5:04

(Adding my comment as an answer to remove the question from the unanswered questions list.)

Show that a free $\Bbb Z$-module is not divisible. (Hint: Show that $\bigcap_{n=1}^\infty nF = 0$ if $F$ is free.)

Note that $\Bbb R$ is divisible.


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