I whould like to prove the following statement:
Lemma: Let $(V,Q)$ be a nondegenerate quadratic vectorspace over a field $\mathbb{F}$ and $a,b\in V\setminus\{0\}$. Then for $P:=\left$ it holds that $(a\cdot a)(b\cdot b)\neq (a\cdot b)^2\Leftrightarrow \mathop{\rm dim}(P)=2\text{ and } P \text{ is nondegenerate}$
Explanation of Terminology: Just to be sure that we are speaking of the same objects and that I'm not making any mistakes in the definitions, here are all definitions in detail: With quadratic vectorspace, I mean quadratic module over a field. Here $Q:V\rightarrow \mathbb{F}$ is a quadratic form, meaning that $\forall a\in\mathbb{F}, x\in V$ we have that $Q(ax)=a^2Q(x)$ and that the map $\varphi: V\times V\rightarrow\mathbb{F}$ defined by $\varphi(x,y) = Q(x+y)-Q(x)-Q(y)$ is bilinear. The expression $a\cdot b$ is defined as $a\cdot b:=\frac{\varphi(a,b)}{2}$. The notation $\langle a,b\rangle:=\mathbb{F}a + \mathbb{F}b$ is the subspace generated by $a$ and $b$. Being nondegenerate for a subspace $U\subseteq V$ means that the radical of $U$ which is defined as $\mathop{\rm rad}(U):=\{ x\in U\mid \forall y\in U,\ \ x\cdot y=0 \}$ is trivial i.e. $\mathop{\rm rad}(U)=\{0\}$ (in this context, if $U\neq V$, the pair $(U,Q\mid_{U})$ is a quadratic space too and $u\cdot v$ for $u,v\in U$ is defined the same as above).
Progress so far: I managed to prove the direction "$\Rightarrow$" by assuming that $\mathop{\rm dim}(P)\neq 0$ or that $P$ is degenerate and showing that the inequality on the left is then an equality.
Remark about the truth of this statement: I extracted this lemma from a proof of theorem 2 of Chapter IV in Jean-Piere Serre's "A Course in Arithmetic" (page 30) and am therefore not quite sure if I got all the requirements right. So there is a chance that the above statement is wrong. Can you give me a counterexample then?