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Normally the Hilbert symbol over a field $\mathbb{F}$ is defined for $a,b\in\mathbb{F}^*$ as follows:

$$ (a,b)=\begin{cases}1,&\text{ if }z^2=ax^2+by^2\text{ has a non-zero solution }(x,y,z)\in \mathbb{F}^3;\\-1,&\text{ else.}\end{cases}$$

I am wondering if anything keeps me from generalizing this definition to a ring $R$ (i.e. replacing the field $\mathbb{F}$ by a ring $R$). I'm mainly thinking about commutative rings that have a multiplicative neutral element $1$.

Another obvious generalization is to take $a,b\in\mathbb{F}$ or $R$ and not $\mathbb{F}^*$ or $R^*$, also allowing to look at equations with non-unit coefficients. Does this make any sense and is there any literature that defines the Hilbert-Symbol also for rings?

Some of the nice properties like symmetry ($(a,b)=(b,a)$), $(a,c^2)=1$ or $(a,-a)=1$ still hold for rings, but I am wondering if this thoughts are even in the spirit of the Hilbert symbol. The fact that I didn't find a lot of writings where the Hilbert symbol is defined over rings lets me think that I am somewhat not getting the main idea. Is there a reason one does not make this obvious generalization?

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Your list of "nice properties" is missing the nonobvious property $(a,bc) = (a,b)(a,c)$, which in fact is going to be false for a general field. Did you really see the Hilbert symbol defined for a general field in this way? The Hilbert symbol as you describe it only "works" for a field $F$ such that for any quadratic extension $E/F$, ${\rm N}_{E/F}(E^\times)$ has index 2 in $F^\times$. Like $F$ being the real or $p$-adic numbers. Definitely not $F = {\mathbf Q}$; generally the Hilbert symbol over a field would be a quaternion algebra (more or less). –  KCd Jan 1 '13 at 8:57
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