How to interpret the square root of inner product over an arbitrary field? In an inner product space, the norm $\|x\| = \sqrt{\langle x,x\rangle}$ is induced. I realise I have almost always dealt with either real or complex numbers, so took the square root for granted.
Reading the Inner Product entry on
Wolfram (https://mathworld.wolfram.com/InnerProduct.html), it says &quot;A vector space together with an inner product on it is called an inner product space. This definition also applies to an abstract vector space over any field.&quot; This was said after introducing the axioms in the context of a space over $\mathbb{R}$.
I am confused because I do not know how to interpret the square root in an arbitrary field. I assume the obvious way would be to define it to be the element $a \in \mathbb{F}$ such that $a^2 = \langle x,x\rangle$. But the issue I am having is how we even know whether such an element exists in the field? Is this a standard result from ring theory?
My understanding has always been that inner product spaces (and normed spaces) are only defined over either the real or complex numbers. How do you construct them (or something equivalent) over some arbitrary field?
The claim doesn't make sense. On vector spaces $V$ over arbitrary field $k$ we have bilinear forms $b(x,y)$. When $k=\Bbb{C}$ we also look at sesquilinear forms, which means that the second argument is linear after applying an automorphism $\sigma$ of the field (the complex conjugaison). But then we can consider $V$ as a $k^\sigma$ vector space to make it linear, so assume that $b$ is truly linear.
$q(x) = b(x,x)$ is a quadratic form.
A first desirable property is that $b(x,y)=b(y,x)$ (when $char(k)\ne 2$ there is a one-to-one correspondence between quadratic forms and symmetric bilinear forms).
A second one is that $q(x)=0$ iff $x=0$. In that case $q$ is said anisotropic.
When $k$ is an ordered field there is a 3rd one: that $\forall x,q(x)\ge 0$. With the previous ones this is the definition of &quot;$b$ is an inner product&quot;. When it is the case then $\|x\|=\sqrt{q(x)}$ is some kind of norm (when $k$ is not a subfield of $\Bbb{R}$ then $\|x\|$ is not real valued so this is a bit different). Do you think we always have $\|x+y\| \le \|x\|+\|y\|$ ?
$\sqrt{q(x)}$ is an element of the algebraic extension of $k$ obtained by adding all the square roots of elements $\ge 0$, it is ordered too, through $\sqrt{a}\ge \sqrt{b}$ iff $a\ge b$, then applying the law of orders.
Note that real valued norms exist other any fields, for example $\|x\| = 0$ if $x_1=x_2=0$ and $=1$ otherwise is a real-valued norm over $k^2$ for any field, a norm for the trivial absolute value  $|a|_{tr}= 0$ if $a=0$ and $=1$ otherwise, such that $\|ax\|=|a|_{tr} \|x\|$.
You do not have a norm defined for an arbitrary inner product space. For instance,  $\langle x, y\rangle = x_1 y_1 - x_2 y_2$ is an inner product on $\mathbb{R}^2$.
The norm will be defined on euclidian spaces or hermitian spaces. There are extra conditions ( positivity) required.