# Inner product with values in a finite field over space of finite functions

Suppose we have $$\mathbb{Z}_n$$ the group of residues modulo $$n$$ and $$\mathbb{F}_q$$ a Galois finite field with $$q$$ elements where $$q=p^m$$ with $$p$$ prime and $$m\in\mathbb{N}$$ and suppose $$n\vert q-1$$.

Suppose we consider the set $$ZF = \{f:\mathbb{Z}_n\to\mathbb{F}_q\vert\, f\text{ function}\}$$. We see that $$\#ZF=q^n$$ and that $$ZF$$ is a vector space. (a simple basis could be $$\{e_n(m) = \delta_{nm} \vert n,m\in\mathbb{Z}_n\}$$)

If we define the inner product in $$ZF$$ as $$\langle f,g\rangle = \sum_{k\in\mathbb{Z}_n}f(k)g(-k)$$ then we can show that it is in fact an inner product.

It is linear in the first argument ($$\langle \alpha(f+h),g\rangle=\alpha\langle f,g\rangle + \alpha\langle h,g\rangle$$)

It is symmetric since if we have $$f(m)g(-m)$$ for some $$m$$ then in a finite group we define $$-m:=h$$ so that $$h+m\equiv 0\mod n$$ but then $$f(h)g(-h)=f(-m)g(m)$$, and since we are summing over all $$\mathbb{Z}_n$$ the resulting sum is finite and then not dependent on the order of the elements hence the same for every $$f(m)g(-m)$$ and every $$f(-m)g(m)$$.

It is defined positive, in fact for the inner product to be zero $$f(m)f(-m)$$ must be zero for every m in $$\mathbb{Z}_n$$ (we are multiplying over $$F_q$$, there aren't two non-zero elements so that $$\alpha\cdot\beta=0$$ since it's a field and so an intergrity domain).

But now the dilemma, i knew that pre-Hilbert spaces are defined over vector spaces over $$\mathbb{R}$$ or $$\mathbb{C}$$. Can i say that $$ZF$$ is a pre-Hilbert space nonetheless? What's the catch, what did i do wrong?

In that case, how does the completeness (or closure) works in a finite case? Is it free from the finiteness (in the sense that a finite pre-Hilbert space is automatically a Hilbert space)?

EDIT: As pointed out in the comments i can't hope for definite positiveness. So it's not a pre-Hilbert space. What other property can i ask that's weaker and makes thing somewhat work?

• Say $p=5$ , $n=2$. Take $f(0)=2$, and $f(1)=1$. Then $\langle f, f\rangle = 2^2 +1 = 0$. – peter a g Feb 20 at 14:56
• Ok, asking for definite positiveness could be a bit too much then, already going from $\mathbb{R}$ to $\mathbb{C}$ we have to lose the symmetry. So maybe we have to lose that in finite fields, what other property can i use? – WhiteEyeTree Feb 20 at 15:23