Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $m < n$ ?

share|improve this question
6  
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

1 Answer 1

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 $1$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.