Geometry over Boolean Ring

First, let me clarify that I have no idea about algebraic geometry.

These are some thoughts I captured while attending a presentation.

An analytic number theorist gave the first talk,

who began by introducing Modular Hyperbola, that is, $xy\equiv 1\mathrm{mod}q$$xy\equiv 1\mod q$

An algebraic geometer gave the second talk.

The speaker presented a polynomial and stated that it can be derived from any commutative ring. This made me wonder how it would work in a Boolean ring.

In propositional logic, we represent true and false as 1 and 0, respectively.

Consequently, "$p\wedge q$$p\wedge q$ " is true can be seen as $pq\equiv 1\mathrm{mod}2$$pq\equiv 1\mod 2$, which can be viewed as a Modula Hyperbola in ${\mathbb{F}}_2$$\mathbb F_2$.

For $p\vee q$$p\vee q$, observe that $p\vee q=\mathrm{\neg }(\mathrm{\neg }p\wedge \mathrm{\neg }q)=(p+1)(q+1)+1$$p\vee q=\neg(\neg p\wedge\neg q)=(p+1)(q+ 1)+1$

Therefore, let $p\vee q=(p+1)(q+1)+1=0,(p+1)(q+1)=1$$p\vee q=(p+1)(q+1)+1=0,(p+1)(q+1)=1$

$p\vee q$$p\vee q$ is false can be seen as $(p+1)(q+1)\equiv 1\mathrm{mod}2$$(p+1)(q+1)\equiv 1\mod 2$

If we view ${\mathbb{F}}_2\times {\mathbb{F}}_2$$\mathbb F_2\times \mathbb F_2$ as a Lattice, the partial order is given by $(a,b)⪯(c,d)$$(a,b)\preceq(c,d)$ iff $a\le c\wedge b\le d$$a\le c\wedge b\le d$

Easy to see that the solution of $p\wedge q=1:=pq\equiv 1\mathrm{mod}2$$p\wedge q=1:= pq\equiv 1\mod 2$ is $(1,1)$$(1,1)$, is the sup,

And the solution of $p\vee q=0:=(p+1)(q+1)\equiv 1\mathrm{mod}2$$p\vee q=0:=(p+1)(q+1)\equiv 1\mod 2$ is $(0,0)$$(0,0)$ is the inf.

Another interesting object is $S^1$$S^1$

Observe that in Boolean Ring, $x^2=x$$x^2=x$

Thus the unit circle $x^2+y^2=1\mapsto x+y=1$$x^2+y^2=1\mapsto x+y=1$ is a straight line!

For example, consider $(P(S),\mathrm{\Delta },\cap )$$(P(S),\Delta,\cap)$,

The point on $S^1$$S^1$ is $X\mathrm{\Delta }Y=S$$X\Delta Y=S$, that is, It's just $X$$X$ and $X^c$$X^c$!

It works on any Boolean Ring.

And any circle (Consider the $L_p$$L_p$ metric over ${\mathbb{R}}^n$$\mathbb R^n$) $x^p+y^p=1$$x^p+y^p=1$

And some polynomials like $\sum x_i^j=0\mapsto \sum x_i=0$$\sum x_i^j=0\mapsto\sum x_i=0$