Welcome to MathOverflow

Question

3

1

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

edited Sep 13 2011 at 1:27

asked Sep 13 2011 at 0:30

Carl
127●4

1

Not that it's my business, but why Paul here and Thomas there? – Yemon Choi Sep 13 2011 at 0:32

2

Because my full name is Paul Thomas. Does it change anything? – Carl Sep 13 2011 at 0:44

1

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? – David White Sep 13 2011 at 1:16

1

@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... – Pete L. Clark Sep 13 2011 at 1:20

Thanks to both of you for the comments, I've specified my question. – Carl Sep 13 2011 at 1:27

Chris Godsil · Accepted Answer · 2011-09-13 01:19:57Z UTC

3

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).

answered Sep 13 2011 at 1:19

Chris Godsil
5,901●1●8●24

	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?? – Pete L. Clark Sep 13 2011 at 1:40
	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. – Joël Sep 13 2011 at 2:35
	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 $Z/qZ$ – Joël Sep 13 2011 at 2:43
	By the way, how do you do the nice blackboard Z and C? – Joël Sep 13 2011 at 2:43
	@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. – Pete L. Clark Sep 13 2011 at 2:48

show 7 more comments

Pete L. Clark · Answer 2 · 2011-09-13 01:16:12Z UTC

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.)

Double orthogonal complement of a finite module

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

2 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.