Short answer: yes, your hypothesis is correct.
You can have a positive-definite inner product taking values in an ordered field $F$ without the induced norm $\|v\| = \sqrt{\langle v,v\rangle}$ existing inside $F$ . The usual passage from inner product to norm needs square roots of positive field elements; if those square roots are not in the field (e.g. $F = \mathbb Q$), the norm does not lie in $F$.
Let $F = \mathbb Q$ and $V=\mathbb Q$ as a $1$-dimensional vector space over $F$.
The function
$$\langle x , y\rangle = 2xy \in \mathbb Q$$
is bilinear, symmetric and $\langle x,x \rangle = 2 x^2 > 0$ for $x \neq 0$. So it is an inner product over $\mathbb Q$.
But for $v=1$ we have
$$\langle 1, 1 \rangle = 2,$$
and $\sqrt{2} \not\in \mathbb Q$. Hence no $\mathbb Q$-valued norm $\|v\| = \sqrt{\langle v,v \rangle}$ exists.
Many algebraic consequences of an inner product still hold over any ordered field: for instance the Cauchy–Schwarz inequality
$$\langle u, v\rangle^2 \le \langle u,u \rangle \langle v, v \rangle$$
can be proved algebraically (consider $\langle u-tv, u-tv\rangle$ as a quadratic in $t$. That inequality lives entirely in $F$.
Let $u, v\in V$. Consider the polynomial in $ t\in F$,
$$f(t) := \langle u-tv, u-tv \rangle = \langle u, u\rangle - 2t\langle u,v\rangle + t^2\langle v,v\rangle.$$
Since $\langle w, w \rangle\ \geq 0$ for all $w$, we have $f(t) \ge 0$ for every $t\in F$. For a quadratic $at^2+bt+c$ with $a>0$ to be nonnegative for all $t$ in an ordered field, its discriminant must be $\leq 0$. Here the discriminant is
$$\Delta = (-2\langle u,v\rangle)^2 - 4\langle v, v\rangle \langle u, u \rangle = 4\big(\langle u,v\rangle^2 - \langle u, u\rangle \langle v,v \rangle \big).$$
Nonpositivity of $\Delta$ gives
$$\langle u, v\rangle^2 - \langle u,u\rangle\langle v, v\rangle \le 0,$$
which is exactly Cauchy-Schwarz
$$\langle u,v\rangle^2 \le \langle u,u\rangle\langle v,v\rangle.$$
Everything in this argument uses only ordered-field inequalities and arithmetic. No square roots.
But to derive the triangle inequality $\|u + v \| \le \|u\| + \|v\|$ you need to take square roots.
If $F$ has the property that every positive element of $F$ has a square root in $F$, then the induced norm $\|v\| = \sqrt{\langle v, v \rangle}$ is well-defined in $F$ and all the usual metric/topological consequences follow.
There are two workarounds I can think of:
- You can always view $F$ as a subfield of a larger ordered field (e.g. embed $\mathbb Q \hookrightarrow \mathbb R$) and then take square roots there.
- Or treat $\langle v, v\rangle$ as a positive definite quadratic form on $V$ rather than as a norm-valued function (explained further down, below).
There are some things that work fine, apart from Cauchy-Schwarz. For example, you can orthogonalize just fine without square roots.
Given a finite set of linearly independent vectors $v_1,\dots,v_n$ in $V$, the usual Gram–Schmidt formulas
$$u_1 = v_1, \qquad u_k = v_k - \displaystyle\sum_{j=1}^{k-1} \frac{\langle v_k, u_j\rangle}{\langle u_j, u_j\rangle} u_j$$
produce vectors $u_1, \dots, u_n$ that are pairwise orthogonal. The only field operations used are addition, subtraction, multiplication and division by the nonzero $\langle u_j, u_j \rangle$.
What you cannot do inside $F$ is normalize these $u_j$ to unit length $\tilde u_j = u_j/\|u_j\|$, because $\|u_j\| = \sqrt{\langle u_j, u_j\rangle}$ may not lie in $F$.
Polarization and the paralellogram law also holds. Define $q(v) := \langle v, v \rangle$. For fields $\operatorname{char} F \neq 2$, the polarization identity holds:
$$\langle u, v\rangle = \frac{1}{2} \big(q(u+v) - q(u) - q(v) \big).$$
The parallelogram identity
$$\langle u+v ,u+v \rangle + \langle u-v, u-v \rangle = 2 \langle u, u\rangle + 2 \langle v, v\rangle$$
is also purely algebraic and holds in any field with $\operatorname{char} F \neq 2 $.
But specifying the quadratic form $q$ is equivalent to specifying the inner product (over fields of characteristic $\neq 2$). Thus an inner-product structure is the same data as a positive-definite quadratic form, and that makes many algebraic manipulations possible without referring to any square root operations.
