Let $\mathbb{F}$ be an arbitrary field of characteristic $\mathrm{char}(\mathbb{F})\neq 2$ and $V$ a $\mathbb{F}$-vector space. The definition of a quadratic form I am used to is a map $\varphi\colon V\to\mathbb{F}$ with the property $\varphi(\lambda v)=\lambda^{2}\varphi(v)$ such that
$$\beta_{\varphi}(v,w):=\frac{1}{2}(\varphi(v+w)-\varphi(v)-\varphi(w))$$
is bilinear. Now, clearly, $\beta_{\varphi}$ is symmetric and $\beta_{\varphi}(v,v)=\varphi(v)$, by definition. In the case $\mathbb{F}=\mathbb{R}$, I have seen the following claim:
Proposition 1: If $\varphi\colon V\to\mathbb{R}$ with the property $\varphi(\lambda v)=\lambda^{2}\varphi(v)$ satisfies the paralellogram law
$$\varphi(v+w)+\varphi(v-w)=2\varphi(v)+2\varphi(w)$$
then it is a quadratic form, i.e. there exists a symmetric bilinear form $\beta_{\varphi}$ such that $\beta_{\varphi}(v,v)=\varphi(v)$.
This is essentially a similar statement as the Jordan-von Neumann theorem, which asserts that a norm $\Vert\cdot\Vert$ on $V$ admits a inner product $\langle\cdot,\cdot\rangle$ inducing it if and only if the paralellogram law
$$\Vert v+w\Vert^{2}+\Vert v-w\Vert^{2}=2\Vert v\Vert^{2}+2\Vert w\Vert^{2}$$
holds. Now, my question is the following:
Is Proposition 1 also true for arbitrary fields (excluding $\mathrm{char}(\mathbb{F})=2$ of course)?
I was not able to find a proof of this statement. I tried to mimik the proof of Jordan-von Neumann, but all the proofs I know of this result are somehow specific to $\mathbb{R}$. The idea of the proof is usually to define
$$\beta_{\varphi}(v,w):=\frac{1}{2}(\varphi(v+w)-\varphi(v)-\varphi(w))$$
Then clearly $\beta_{\varphi}(v,v)=\varphi(v)$ and $\beta_{\varphi}$ is symmetric. The only thing to show is that $\beta_{\varphi}$ actually defines a bilinear form.
Any literature on this is appreciated.
Best Answer
Without positivity, this is not even true for $\mathbb F=\mathbb R$. See this wonderful post.
We show some more general results. It has been shown by P. Quinton that $(\cdot, \cdot):=\beta_{\varphi}$ is bi-additive. In particular it follows $$(-u, v)=(u,-v)=-(u,v)$$ Consider $$\begin{cases} (\lambda(u+v), \lambda(u+v)=\lambda^2(u+v, u+v) & (1)\\ (\lambda(u-v), \lambda(u-v))=\lambda^2(u-v,u-v) & (2)\end{cases}$$
Consider $(1)-(2)$, and after additive expansion, we get $4(\lambda u, \lambda v) =4 \lambda^2(u, v)$ that is $$(\lambda u, \lambda v)=\lambda^2(u, v)$$
where $u, v$ are not necessarily equal.
Now we show $(\lambda u, u)=\lambda (u, u)$. This follows from $$((1+\lambda)u, (1+\lambda)u)=(1+\lambda)^2(u, u)$$
After the expansion and cancellation, $2(\lambda u, u)=2\lambda (u, u)$.
Now we can show that in the very special case $\dim V = 1$, $\beta_{\varphi}$ is bilinear. And since the post has constructed examples for $\dim V=2$, we know the statement also fails for all $\dim V\ge 2$.
Let $V=\text{Span}\{e\}$, then assume $a\not=0$, $$(ae,be)=a^2(e, \frac{b}{a}e)=a^2(\frac{b}{a}e, e)=a^2\frac{b}{a}(e,e)=ab(e,e)$$ In other words, $(\cdot, \cdot)$ is completely determined by $(e,e)$ alone.