TITLE: 3-dimensional form contains a hyperbolic plane
QUESTION [0 upvotes]: Hello I need some advice for this problem:
Let $a,b,c \in F$ be such that $c=a^{2}+b^{2}$ and it is different from zero. Show that the 3 dimensional form $\phi=<1,1,-c>$ contains a hyperbolic plane.
I was thinking about this:
I know that $\phi=<1,1,-c>$ that means that there exist a basis $\{v,w, y\}$ such that $\phi(v,v)=1$, $\phi(w,w)=1$, $\phi(y,y)=-c$ and any combination of the element of the basis is zero.
With this I proved that $\phi$ is isotropic (taking the vector $av+bw+y$) also note that $\phi=<1>\perp<1,-c>$ and $<1>$ is anisotropic.
Now since $\phi$ is isotropic, it follows that $<1,-c>$ isotropic and therefore isometric to a hyperbolic plane.
I am not sure if my idea is correct, that is why I appreciate some advise about this idea.
Thank you for your time!

REPLY [1 votes]: I will assume that $F$ has characteristic different from $2$, to avoid the usual complications.
You are correct in saying that $\phi$ is isotropic, and that in itself implies that it contains a hyperbolic plane. On the other hand, it is not true in general that $\langle 1,-c\rangle$ is isotropic. This would mean that $c$ is a square; but there exist fields such that a sum of squares is not always a square: just take $\mathbb{Q}$ for instance, and $c=2$ (in general such fields are called non-pythagorean).
The flaw in your argument is that you assume that since $\phi = \langle 1\rangle + \langle 1,-c\rangle$ is isotropic and $\langle 1\rangle$ is anisotropic, then its orthogonal subspace must be isotropic. This is basically saying that a sum of anisotropic spaces must be anisotropic, which is clearly not true since any non-degenerate quadratic form is a sum of anisotropic subforms (for exmaple it is a sum of 1-dimensional forms).
To find an explicit hyperbolic plane, once you identified an isotropic vector $x$, just take any $y$ such that $\phi(x,y)\neq 0$, and the plane generated by $x$ and $y$ is hyperbolic (I let you figure out why).
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