Generalization of inner product I was wondering if there was a widely accepted generalization of inner product spaces where the inner product look something like $\langle\bullet  , \bullet\rangle:V\times V \to \mathbb{F}$, where $\mathbb{F}$ does not have to equal to $\mathbb{R}$ or $\mathbb{C}$. Could you define a meaningful inner product space over $\mathbb{Q}$ or a finite field?
 A: Yes it is called a bilinear form. A form because it takes value in the base field. You can guess what bilinear is.
https://en.wikipedia.org/wiki/Bilinear_form
A: Yes, you can do this. Choose a linearly independent basis for an as yet unconstructed vector space $V$. Define $V$ as the set of linear combinations of basis elements with coefficients drawn from any field $\mathbb{F}$ you like. (You'll need to make sure your basis elements can be multiplied by any of the fields' elements.) Now define your inner product by combining "the basis elements are orthonormal" with "this product is sesquilinear". (For self-conjugate fields such as $\mathbb{R}$, sesquilinearity is equivalent to bilinearity; see also @Bleuderk's answer.) This uniquely defines an inner product that does what you've asked.
To consider an example, there's a well-known inner product with respect to which Hermite polynomials are orthonormal. The linear combinations with strictly rational coefficients comprise a vector space of functions over $\mathbb{F}=\mathbb{Q}$. The inner product will be over $\mathbb{Q}$ too, since $a_i,\,b_i\in\mathbb{F}\to \sum_ia_ib_i\in\mathbb{F}$. (We need a bit of care with inner products that could be infinite due to convergence conditions failing, but that's a familiar complication for infinite-dimensional vector spaces over $\mathbb{R}$ or $\mathbb{C}$ too.)
A: This generalizes to a sesquilinear form on any module over a ring that admits an antiautomorphism.  See Sesquilinear form on Wikipedia.  The positive-definiteness of an inner product gets replaced by a constraint to a particular subset of the ring.
