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I am looking for a counterexample to a simple question about proper sub-modules. The book I am reading mentions the following theorem but implys that there are pathological examples related to the theorem when one considers non-commutative rings.

Let $D$ be a principal ideal domain, let $n\in \mathbb{Z}$ and let $D^{(n)}$ denote a free $D$-module of rank $n$.

Theorem: If $L$ is a submodule of $D^{(n)}$ then $L$ is a free $D$-module of rank $m \leq n$

Question: If $L$ is proper submodule of $D^{(n)}$ must the rank of $L$ satisfy $m < n$ ?

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No. Let $n = 1$, $D = \mathbb{Z}$, $L = 2 \mathbb{Z}$. – Zhen Lin Aug 4 '11 at 9:53
@Zhen Thank you! – user7980 Aug 4 '11 at 9:56
@ZhenLin Please consider converting your comment into an answer, so that this question gets removed from the unanswered tab. If you do so, it is helpful to post it to this chat room to make people aware of it (and attract some upvotes). For further reading upon the issue of too many unanswered questions, see here, here or here. – Julian Kuelshammer Sep 11 '13 at 17:36
up vote 1 down vote accepted

Proper submodules can still be relatively "big". For instance, $2 \mathbb{Z}$ is a proper submodule of $\mathbb{Z}$, but both are free $\mathbb{Z}$-modules of rank $1$.

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