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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 $P:=\left<a,b\right>$ 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?

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What you wrote as definition of $\,Rad(U)\,$ seems to be incomplete: all the elements in U s.t. for all the elements in U, their product $\,x.y\,$...**what?** Is zero? – DonAntonio Jun 25 '12 at 12:27
of course "is zero"... sorry (edited). – born Jun 25 '12 at 12:31
up vote 2 down vote accepted

Proving the implication '$\Leftarrow$' is equivalent to proving that $$ (a,a)(b,b) = (a,b)^2 \Rightarrow \dim(P)<2 \text{ or } P \text{ degenerate }$$ But assuming the identity on the left you might want to try and look at the element $$(b,b)a -(a,b)b \in V$$ (this element shows up naturally when trying to manipulate the identity into something saying that a single inner product is equal to $0$).

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looking at this element (let's call it $c=(b.b)a-(a.b)b$) i get $c.a=0$ and $c.b=0$ yielding that $c$ is in the radical of $P$. This implies that $c$ is zero or $P$ is degenerate. In the later case we are done. If $c$ is zero, we have that $(b.b)a-(a.b)b=0$ which is a linear combination of $0$ in $P$. So either $\dim(P)<2$ or $(b.b)=0$ and $(a.b)=0$. – born Jun 25 '12 at 14:10
So either $\dim(P)<2$ or ($(b.b)=0$ and $(a.b)=0$). The later implies that $b\in\mathop{\rm rad}$. Now if $b=0$, $\dim(P)<2$ and if $b\neq 0$, $P$ is degenerate. Thank you very much! – born Jun 25 '12 at 14:17

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