# When does an inner product induce a norm?

When we consider a vector space $$V$$ over some field $$F$$, I know that when the $$F=\mathbb{R}$$ or $$=\mathbb{C}$$, by setting $$\|x\|=\left\langle{x,x}\right\rangle^\frac{1}{2}$$ we get a norm. However, since the inner product is a function with its image in $$F$$, what happens if we consider any $$V$$ over the rational numbers? For example, if we take $$\mathbb{Q}^2$$ over $$\mathbb{Q}$$ with the dot product, then $$v=(1,1)$$ has norm $$\sqrt{2}$$, which is not rational. How can one obtain a norm from a given inner product in such cases?

As you noted, to construct an an inner-product space we require the basefield $$F$$ of the vector space $$V$$ to be a quadratically closed subfield of $$\mathbb{R}$$ or $$\mathbb{C}$$ (i.e. every element of $$F$$ must have a square root in $$F$$).
However, a norm is a function $$n:V\rightarrow [0,+ \infty)$$, as opposed to an inner-product which is map $$\langle \cdot ,\cdot \rangle :V\times V\to F$$. In the case of the basefield $$\mathbb{Q}$$ therefore, we cannot construct an inner-product space from the dot product implied in your question, but we can construct a normed vector space from it.
An ordered field where $$a^2+b^2$$ is always a square is called a Pythagorean field. As you observe, not every ordered field is Pythagorean, but each ordered field has a Pythagorean extension. If you really want $$L^2$$-norms you could always extend your ground field to a Pythagorean extension field.