Inner Product Spaces over Finite Fields

1) The action of $F$ on $\mathcal{V}$ that you defined does not turn $\mathcal{V}$ into a vector space. If it did, we would need to have $$ V=bV=(1+a)V=1V+aV=V+V=0. $$

2) We usually restrict the notion of an inner product to vector spaces over a field, where the elements of the field (=the scalars) can be ordered. Most notably we want to have a set of positive elements. You hopefully remember that one of the axioms of an inner-product demands that the inner product of a non-zero vector with itself should be positive. Well, what are the positive elements of a finite field? Here we have a problem. Normally we would want to declare $1$ as a positive element. But also we want the sum of two positives to be positive, so $1+1=2$ should be positive. But in the field of four elements $1+1=0$, so $0$ should be positive?? By a similar argument non-zero vectors in a vector space over a finite field often have zero 'inner product' with itself. Doesn't look too good, does it? [Edit] As Theo points out, the reason for this is that the inner product is used to define a length of a vector (and a metric in the space), so we need to be able to do square roots. As it happens, in your field all the elements do have square roots (all the finite fields of characteristic 2 share this property), but the problem with the positive elements persists.[/Edit]

3) The closest thing to an inner product on a vector space over a finite field (or other non-ordered field) is a bilinear symmetric form. It comes together with an associated quadratic form. These are well studied objects in algebra.

4) The kind of conjugations you seem to talk about are also well studied in algebra. Many number fields have several such symmetries, and they are called automorphisms of the field. Look up field theory (or Galois theory) to learn more. Even your field of 4 elements has a non-trivial one. The mapping $F:0\mapsto, 1\mapsto 1, a\mapsto b, b\mapsto a$ has the following nice properties:

$$ F(x+y)=F(x)+F(y)\qquad F(xy)=F(x)F(y) $$

for all $x,y$ in your field.

(I'll not address your concrete example, sorry!)

Actually you need $K$ to be a $*$-division ring in order to make sense of a generalization of inner product space. You can always equip every division ring with the trivial conjugation operation $*$ that maps every element to itself. (This is not very interesting, but it shows that the need for a conjugation operation does not exclude any division rings from consideration.)

But the notion of a $*$-division ring $K$ is enough to define a $K-$ vector space $H$ as a $K-$module, and also define what a nondegenerate hermitian form on $H$ should be.

I can't tell you if this notion has any applications. I can tell you instead that it does not have any applications in quantum mechanics (one major area of applications of Hilbert spaces), because there is a very strong theorem that states that assuming some basic axioms that are needed for quantum mechanics in infinite dimensions, K has to be the real numbers, the complex numbers or the quaternions.

For more details please visit this blog post: Soler's theorem, nCafe.