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Crossposted from math.stackexchange since I'm not getting any answer.

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 $V^{\perp} = \{w \in W \, : \, \forall v \in V,\, v \cdot w = 0 \}$ where "$\cdot$" is the dot product. Is it true that ${(V^{\perp})}^{\perp} = V$ for all $q \geq 2$? If not, when is it the case?

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    $\begingroup$ Not that it's my business, but why Paul here and Thomas there? $\endgroup$
    – Yemon Choi
    Commented Sep 13, 2011 at 0:32
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    $\begingroup$ Because my full name is Paul Thomas. Does it change anything? $\endgroup$
    – Carl
    Commented Sep 13, 2011 at 0:44
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    $\begingroup$ Be careful with your notation. I think it's safe to say most algebraists and algebraic topologists read $\mathbb{Z}_q$ as the $q$-adic numbers, not as $\mathbb{Z}/q\mathbb{Z}$. Because not everyone adopts this notation, I generally explicitly define my terms. Which do you mean? $\endgroup$
    – David White
    Commented Sep 13, 2011 at 1:16
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    $\begingroup$ @David: I agree with your injunction in general -- at least, when I write $\mathbb{Z}_q$ I mean the $q$-adic integers (although I know some algebraic topologists who take the other convention). But in this case the OP says he wants a "finite $\mathbb{Z}$-modules" which seems to imply that $\mathbb{Z}_q = \mathbb{Z}/q\mathbb{Z}$. Anyway, he also doesn't say what kind of integer $q$ is -- e.g., is it necessarily prime? In view of my answer below, that's a more serious question... $\endgroup$
    – Pete L. Clark
    Commented Sep 13, 2011 at 1:20
  • $\begingroup$ Thanks to both of you for the comments, I've specified my question. $\endgroup$
    – Carl
    Commented Sep 13, 2011 at 1:27

2 Answers 2

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The answer is yes. The easiest way for me is to appeal to character theory. If $\zeta$ is a complex $q$-th root of 1 then the map from $\mathbb{Z}_q^n$ to $\mathbb{C}$ given by $$ x \mapsto \zeta^{a^Tx},\qquad (a\in\mathbb{Z}_q^n) $$ is a character of the abelian group $W=\mathbb{Z}_q^n$. The set of characters obtained as $a$ varies over the elements of $W$ is the character group $W^*$ of $W$. If $V$ is a subgroup of $W$, then $V^\perp=V^*$ is isomorphic to the subgroup $(W/V)^*$ of $W^*$.

A convenient source for the relevant character theory is from our own KConrad: http://www.math.uconn.edu/~kconrad/blurbs/ (under characters of finite abelian groups).

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  • $\begingroup$ Are you perhaps mixing up inner products with values in $\mathbb{C}$ with inner products with values in $\mathbb{Z}/q\mathbb{Z}$? Or am I the one who is confused?? $\endgroup$
    – Pete L. Clark
    Commented Sep 13, 2011 at 1:40
  • $\begingroup$ I think Chris is right. He's talking about inner product with values in $Z/qZ$ which is identified with the group of $q$-roots of unity in $C^\ast$. Let me reformulate its proof. Letting G be the abelian group Z_q^n, then the scalar product of the question can be seen as a perfect pairing $G \times G \rightarrow C^\ast$ that identifies $G$ with its dual group in the sense of character theory. The question becomes: for a finite abelian group $G$, and a subgroup $V$, is $(V^\bot)^\bot=V$ and the answer is yes by standard theory of duality of abelian groups. $\endgroup$
    – Joël
    Commented Sep 13, 2011 at 2:35
  • $\begingroup$ Note that of course this is false if $Z/qZ$ is replaced by $Z$ (example $G=Z, V=2Z$). This is an example among others where the non-domain $Z/qZ$ behave in a more familiar way (that is more like a field) that the simplest non-field domain, namely $Z$. Here is another example with the same flavor: if a submodule of $(Z/qZ)^n$ is free, then it is direct summand. What make the things work is the (local) Artinian nature of the ring $Z/qZ$ $\endgroup$
    – Joël
    Commented Sep 13, 2011 at 2:43
  • $\begingroup$ By the way, how do you do the nice blackboard Z and C? $\endgroup$
    – Joël
    Commented Sep 13, 2011 at 2:43
  • $\begingroup$ @Joël: Thanks, I think I understand it now. BTW, you can use latex in the comments by enclosing it with single dollar signs as usual. $\endgroup$
    – Pete L. Clark
    Commented Sep 13, 2011 at 2:48
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Your question does not specify what $q$ is. But if $q$ is an odd prime -- so that you are talking about quadratic spaces over the field $\mathbb{F}_q$ of odd characteristic -- then the answer is yes.

In this case your inner product $\cdot$ is the bilinear form associated to the quadratic form $q(x_1,\ldots,x_n) = x_1^2 + \ldots + x_n^2$. This quadratic form is nondegenerate, so the result you want is Proposition 7 in these notes. (They are nothing so special: any sufficiently basic text on quadratic forms will contain this material.)

I am not really used to thinking about quadratic forms either in characteristic $2$ or over rings which are not domains, so if you're really interested in the case of $q$ not necessarily an odd prime, please say so, so that someone else can give a more complete answer. (But I will guess that the result is also true when $q$ is an odd prime power, for instance.)

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  • $\begingroup$ Yes, I recall learning this in a course I took on quadratic forms years ago. I don't think it works for composite $q$, but I can't be sure. $\endgroup$
    – David White
    Commented Sep 13, 2011 at 1:46

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