Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Suppose that $M\subseteq \mathbb{Z}^n$ is a module such that $\mathbb{Z}^n/M$ is free and $S\subseteq \mathbb{R}^n$ is a bounded, symmetric (around $0$) convex set. Let $M'$ be the module generated by $S\cap M$.

Question: Is $\mathbb{Z}^n/M'$ free?

I think it is free if the following is true: for any $x\in M\setminus M'$ and $a\neq 0$, it holds that $ax\not\in M'$. (Basically this is saying that $M/M'$ is free)

Is this true?

It seems true to me, but I haven't found a proof yet...

share|improve this question
add comment

2 Answers 2

up vote 1 down vote accepted

This is true if $M$ has rank at most $2$ but not beyond that. For a counterexample in rank $r \geq 3$, choose coordinates so that $M$ is the body-centered cubic lattice, that is, the subgroup of ${\bf Z}^r$ consisting of all-even and all-odd vectors; and let $S$ be the $l^1$ ball of radius $2$, that is, $$ S = \{ (x_1,\ldots,x_r) \in {\bf R}^r : \sum_{i=1}^r \left|x_i\right| \leq 2 \}. $$ Then $S \cap M$ spans $(2{\bf Z})^r$, which is a proper finite-index subgroup of $M$ (it does not contain $(1,1,\ldots,1)$).

share|improve this answer
    
But in this case, ${\bf Z}^n/M$ is not free? –  user35375 Sep 1 '13 at 17:16
    
The construction assumes that ${\bf Z}^n/M$ is free. You could even take $n=r$ and then change coordinates to identify ${\bf Z}^r$ with $M$, e.g. by using $2e_1,\ldots,2e_{r-1}$ and $\sum_{i=1}^r e_i$ as the new basis vectors. –  Noam D. Elkies Sep 1 '13 at 19:18
    
For $r=3$, doesn't it hold that ${\bf Z}^3/M\simeq {\bf Z}_2 \oplus {\bf Z}_2$, which isn't free? –  user35375 Sep 1 '13 at 20:06
    
Sorry this seems to be confusing you: for $r=3$, $M$ EQUALS ${\bf Z}^3$, just with a different choice of coordinates that makes it easy to see what $S$ is doing. –  Noam D. Elkies Sep 1 '13 at 20:36
add comment

I assume that you mean $S$ is symmetric with respect to the origin. Otherwise, a segment $[(0,1),(2,0)]$ in ${\mathbb R}^2$ with standard ${\mathbb Z}^2$ lattice gives a counterexample.

One does not need $S$ symmetric though, only $S\ni 0$, with the following argument. For any $x\in M$ with $ax\in S$ we have $[0,ax]\subseteq S$, so $x\in S$. Thus $M'$ is saturated in $M$ and the quotient is free.

share|improve this answer
    
I'm not following... To show that $M'$ is saturated in $M$, we need to show that if $ax\in M'$ then $x\in M'$. Why is this true? Why can't there be some $x\in M$, $x\not\in S$ but $3x\in M$? –  user35375 Sep 1 '13 at 15:57
    
Sorry, I meant that $x\in M$, $x\not\in S$ but $3x\in M'$. Is it possible that $3x$ can be written as the linear combination of some vectors in $M\cap S$ but $x$ can't? –  user35375 Sep 1 '13 at 16:25
add comment

Your Answer

 
discard

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