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?