 Anisotropic quadratic form

A quadratic form $q$ over a field $k$ is an anisotropic quadratic form if $q(v)=0$ forces $v=0$. An isotropic quadratic form instead has a nonzero zero vector. This is a property of the form together with its coefficient field, rather than just the displayed polynomial.^[ref]

For example, $q(x,y)=x^2-2y^2$ is anisotropic over $\mathbb{Q}$. If a nonzero rational solution had $y=0$, then $x=0$ too; otherwise $x/y$ would be a rational square root of two. Over $\mathbb{R}$, however, $(\sqrt{2},1)$ is a nonzero zero vector. The coordinate change $u=x+\sqrt{2}y$, $v=x-\sqrt{2}y$ gives $q=uv$, a hyperbolic plane of a quadratic form.

This example also separates anisotropy from positive definiteness: the rational form takes the positive value $1$ and the negative value $-2$, despite having no nonzero rational zero. In a Witt decomposition, an anisotropic remainder is what remains after all hyperbolic planes have been split off. Its Witt index is zero. In characteristic two, one must test vanishing of the quadratic form itself; properties of the associated polar bilinear form alone do not supply the same test.