# Dual codes over finite commutative rings

Let $R$ be a finite commutative ring and $C$ be a linear code of length $n$ over $R$, i.e., an $R$-submodule of $R^n$. We define $C^{\perp}$ by

$C^\perp=\{v\in R^n~|~[v,w]=0~~\text{for all}~w\in C\}$

where $[v,w]=\sum_{i=1}^nv_iw_i\in R$ is the Euclidean inner product. I read in "Algebraic Coding Theory Over Finite Commutative Rings"(Steven T.Dougherty) that $(C^\perp)^\perp=C$ for any finite commutative ring $R$ and any linear code $C$ over $R$. Clearly it follows that $(C^\perp)^\perp\supseteq C$. But I cannot show the converse inclusion. Could you tell me the proof? If the author gets wrong and such property does not hold in general, I would like to know a counterexample.

I have already known that this property holds for any finite quasi-Frobenius ring $R$ and any submodule $C\subseteq R^n$. See, for example, [J.A. Wood. Duality for Modules over Finite Rings and Applications to Coding Theory]. But I do not know whether this property holds for any finite commutative ring $R$ and any linear code $C$ over $R$.

You're right. It's not true, for example, using the ring $R=F_2[x,y]/(x^2,xy,y^2)$, which is not quasi-Frobenius.
If you take $C=(x)\subseteq R^1$, then $C^\perp=(x,y)$ and $C^{\perp\perp}=(x,y)$. A ring which is quasi-Frobenius cannot suffer this problem.
You could make an example for any length $n$ over $R$ by using $(x)\times\{0\}\times\ldots\times\{0\}$ with the same effect. The perp should be $(x,y)\times R\times\ldots\times R$, and the perp of that certainly contains $(x,y)\times\{0\}\times\ldots\times\{0\}$.
You could also try $C=\oplus_{i=1}^n(x)$, but I'm not sure how to compute the results. $C^\perp$ certainly at least contains $\oplus_{i=1}^n(x,y)$, but it also contains things like even weight vectors having $1$'s and $0$'s only. If you have a computer algebra system, you might check $(x)\times (x)$ to see what the perp and double-perp are.
One possibility is that the author actually only intended to talk about proper quotients of $\mathbb Z$, all of which are quasi-Frobenius.
• In fact, if $n=1$ is allowed for in this theorem, the double perp condition would imply that the underlying ring is necessarily quasi-Frobenius (a commutative Noetherian ring satisfying the double annihilator condition is Frobenius) – rschwieb Oct 6 '17 at 16:30