Boolean Ring, From the Algebraic Geometry point of view.

Definition. Boolean Ring.

Let $R$$R$ be a ring, we call it a Boolean ring if for all $r\in R,r^2=r$$r\in R,r^2=r$.

Proposition. A Boolean Ring is a commutative Ring.

Proof.  $a+b=(a+b{)}^2=a^2+ab+ba+b^2=a+ab+ba+b$$a+b=(a+b)^2=a^2+ab+ba+b^2=a+ab+ba+b$, hence $ab+ba=0$$ab+ba=0$.

Proposition. The characteristic of Boolean Ring is 2.

Proof. $a+a=(a+a{)}^2=a^2+a^2+2a^2=0$$a+a=(a+a)^2=a^2+a^2+2a^2=0$

Example of Boolean Ring.

Let $2$$2$ be ${\mathbb{F}}_2=\mathbb{Z}/2\mathbb{Z}$$\mathbb F_2=\mathbb Z/2\mathbb Z$, ${\mathrm{Hom}}_{\mathrm{Set}}(X,2)$$\mathrm{Hom}_{\mathrm{Set}}(X,2)$ give you a Boolean Ring, which is isomorphic to $(P(X),\mathrm{\Delta },\cap )$$(P(X),\Delta,\cap)$

View $P(-)$$P(-)$ is a functor, for $f:X\to Y,P(f)=f^{-1}:P(Y)\to P(X)$$f:X\to Y,P(f)=f^{-1}:P(Y)\to P(X)$.

These two functors are from Set to Bool.

Easy to check that ${\mathrm{Hom}}_{\mathrm{Set}}(-,2)$$\mathrm{Hom}_{\mathrm{Set}}(-,2)$ is isomorphic to $P(-)$$P(-)$.

Example. In Measure Theory, sigma algebra is a Boolean Algebra.

Proposition. Let $B$$B$ be a Boolean Ring. Then every $b\in B$$b\in B$ is an endomorphism of $B$$B$ in Category of Ring.

Proof. $b(r+s)=br+bs,b(rs)=b^2(rs)=brbs=b(r)b(s).$$b(r+s)=br+bs,b(rs)=b^2(rs)=brbs=b(r)b(s).$

Proposition. Let $B$$B$ be a Ring. $\mathrm{\forall }b\in B$$\forall b\in B$, $b$$b$ is an endomorphism of $B$$B$ in Category of Ring $⟹$$\implies$ $B$$B$ is a Boolean Ring.

We only need to prove that if $b(sr)=b(s)b(r)$$b(sr)=b(s)b(r)$, then $B$$B$ is a Boolean Ring.

But it is obvious, let $s=r=1,b=b(1\cdot 1)=b(1)b(1)=b^2$$s=r=1,b=b(1\cdot 1)=b(1)b(1)=b^2$.

Example. Consider $(P(X),\mathrm{\Delta },\cap )$$(P(X),\Delta,\cap)$, then for any $A\in P(X),A\cap (-)$$A\in P(X),A\cap(-)$ is a ring homomorphism.

Via the natural transformation between $P(-)$$P(-)$ and ${\mathrm{Hom}}_{\mathrm{Set}}(-,2)$$\mathrm{Hom}_{\mathrm{Set}}(-,2)$, $A\cap (-)$$A\cap(-)$ correspond to $i_A^{\ast }$$i_A^*$, where $i_A:A\to X$$i_A:A\to X$ is the inclusion map.

From the Algebraic Geometric point of view, the $A\cap (-)$$A\cap(-)$ gives you the ${\mathrm{res}}_A^X$$\mathrm{res}_{A}^X$ over the sheaf.

The image of $A\cap (-)$$A\cap(-)$ is isomorphic to $(P(A),\mathrm{\Delta },\cap )$$(P(A),\Delta,\cap)$, which is the coordinate ring of $A$$A$.