Consider a vector space $V$, with a symmetric bilinear form $\cdot$ , over a field $\mathbb F$ with characteristic not $2$.
Suppose, for any $x\in V$, if $x\cdot x=0$, then $x=0$. What is this property called? I might say that the form is definite, or perhaps formally positive-definite in analogy with formally real fields (where any sum of nonzero squares is nonzero). Indeed if $\mathbb F=\mathbb R$ then this is equivalent to the usual meaning of definiteness.
But it's not equivalent for other ordered fields. For example, the rational form $\mathbb Q^2\ni(x_1,x_2)\mapsto(x_1,x_2)^2=x_1^2-2x_2^2$ has this property since $\sqrt2$ is irrational, even though the form is indefinite according to the order on $\mathbb Q$. And in general $\mathbb F$ needn't be ordered, so "positive-definite" is meaningless, though the above property is still meaningful.
Similarly, I might say that the form is semidefinite, or formally positive-semidefinite, if this condition holds: For any $x\in V$, if $x\cdot x=0$, then $x$ is in the degenerate subspace $V^\perp\subseteq V$ (also called the "radical" of $V$), meaning $x\cdot y=0$ for all $y\in V$. Another characterization of this is that $V$ contains no "hyperbolic planes": 2D subspaces where the quadratic form is like $(x_1,x_2)^2=2x_1x_2$ or $x_1^2-x_2^2$. This is equivalent to the usual meaning of semidefiniteness when $\mathbb F=\mathbb R$.
Are there standard terms for these properties? If not, what do you suggest?
Commenters have reminded me of "anisotropic" for the first property (though the term doesn't make sense to me; see Etymology of the word "isotropic"). The second property remains.
Here's another description of those properties, in terms of the origin $O=\{0\in V\}$, the radical $R=V^\perp=\{x\in V\mid\forall y\in V, x\cdot y=0\}$, and the null cone $C=\{x\in V\mid x\cdot x=0\}$. It's clear that $O\subseteq R\subseteq C\subseteq V$. The form is nondegenerate when $O=R$. The form is anisotropic when $O=C$. What is the form when $R=C$?
