How does an inner product $V \times V \to F$ induce a vector norm $V \to \mathbb{R}?$

Question

Let $\mathbb{F}=\left(F,+_F,\cdot_F\right)$ be a field and $\mathbb{V}=\left(V,\mathbb{F},+,\cdot\right)$ be a vector space. Let $\langle\cdot,\cdot\rangle : V \times V \to F$ be an inner product. If $F \neq \mathbb{R},$ then how can we say that the $\left\Vert x \right\Vert = \sqrt{\langle x,x \rangle}$ is a valid vector norm$?$ Because a vector norm is a map $\left\Vert\cdot\right\Vert : V \to \mathbb{R}$.

Should we modify the definition of the induced norm to be $\left\Vert x \right\Vert = \sqrt{\left|\langle x,x \rangle\right|},$ where $\left|\cdot\right| : F \to \mathbb{R}$ is a pre-fixed norm$?$

Answer 1

For those who say that "an inner product induces a norm" define the term "inner product" to apply specifically to the case in which $\Bbb F \in \{\Bbb R,\Bbb C\}$. In those cases, the definition $\|x\| = \sqrt{\langle x, x \rangle}$ is perfectly fine.

With this convention, a bilinear map $V \times V \to \Bbb F$ would be called a "scalar product" or a "bilinear form" as opposed to an "inner product".

Answer 2

This is a quote from Wikipedia

There are various technical reasons why it is necessary to restrict the basefield to R and C in the definition. Briefly, the basefield has to contain an ordered subfield in order for non-negativity to make sense,[4] and therefore has to have characteristic equal to 0 (since any ordered field has to have such characteristic). This immediately excludes finite fields. The basefield has to have additional structure, such as a distinguished automorphism. More generally any quadratically closed subfield of R or C will suffice for this purpose, e.g., the algebraic numbers or the constructible numbers. However in these cases when it is a proper subfield (i.e., neither R nor C) even finite-dimensional inner product spaces will fail to be metrically complete. In contrast all finite-dimensional inner product spaces over R or C, such as those used in quantum computation, are automatically metrically complete and hence Hilbert spaces.