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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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    $\begingroup$ No. Let $n = 1$, $D = \mathbb{Z}$, $L = 2 \mathbb{Z}$. $\endgroup$ – Zhen Lin Aug 4 '11 at 9:53
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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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