most recent 30 from http://mathoverflow.net 2013-05-24T15:21:17Z http://mathoverflow.net/feeds/question/81937 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/81937/orthogonal-complement-in-mathbbz-qn Stephen 2011-11-26T05:50:57Z 2011-11-26T06:06:07Z

<p>Let $W$ be the finite $\mathbb{Z}$-module obtained from $\mathbb{Z}_q^n$ with addition componentwise where $\mathbb{Z}_q$ is the integers mod $q$. Let $V$ be a submodule of $W$. Let $V^{\perp} = \{w \in W : \forall v \in V \quad w \cdot v = 0 \}$ where $w\cdot v = w_1v_1 + \ldots + w_nv_n$. Is it true that ${(V^{\perp})}^{\perp} = V$ for all $q \geq 2$?</p> <p>According to Wikipedia, this holds for finite dimensional inner product space, but I wish to know whether it holds in $\mathbb{Z}_q^n$ where $\cdot$ isn't an inner product.</p>

http://mathoverflow.net/questions/81937/orthogonal-complement-in-mathbbz-qn/81938#81938 Greg Kuperberg 2011-11-26T06:06:07Z 2011-11-26T06:06:07Z

<p>Yeah, it's true. Since $\mathbb{Z}/q$ is a principal ideal ring, there is an extension of the Euclidean algorithm to matrices that puts any matrix in Smith normal form. It means that after an automorphism of $(\mathbb{Z}/q)^n$, any submodule $V$ can be put into a standard form in which it is generated by vectors of the form $d_k e_k$, where $e_k$ is a standard basis vector, $d_k$ is a divisor of $q$, and each $k$ only appears at most once. In that case you can check directly that $(V^\perp)^\perp$ is no larger than $V$.</p> <p>(I'm taking the question in the more interesting case in which $q$ might not be prime.)</p>