# Adjoint to the Hom functor in Boolean rigs

What I wanna ask is about analogies to the tensor product for commutative boolean rings. What I mean by commutative boolean ring is set with two operations, + and *, and two identities, 0 and 1, as is usual, with the usual properties that multiplication distributes over addition and the unusual characteristic that both operations are boolean.

An example would be the collection of subsets of some set $A$ where the union represents addition with identity $\emptyset$ and intersection multiplication with identity $A$

Considering a category of these i am wondering whether it is possible to define a left adjoint to the Hom-functor.

If so, can it be used to define topologies for topological rings in the sense that $op_*:R\otimes R\to R$ is continuous with respect to the proposed "tensor topology" on $R\otimes R$

The background stems from the fact that the Hom-tensor adjunction for abelian groups can be extended to any commutative monoid, including commutative boolean monoids. Further, unlike their non boolean counterparts the Hom-set for commutative boolean rings are commutative boolean rings with component-wise operations, which makes Hom an endofunctor and makes one wonder about adjoints.

I'm not sure that the proposed topology would actually differ from the product topology in anyway, but it would allow for the formulation of topological rings in a more consistent way.

Any thought would be appreciated.

$\bf {Edit1:}$

I apologize for the terminology above which is incorrect. To clarify: I mean a set with two operations, both commutative and idempotent, with two identities. Nothing else, no inverses of any kind, no "not" or any other lattice related operation.

It is a structure like a topology, not a topological space, but a topology, that's all.

It can be constructed from commutative idempotent monoids $(m^2=m$ for all $m\in M)$ using a Hom-tensor adjunction, like the one for abelian groups.

It can be extended, formally, to all commutative monoids "we only need commutativity to get an endofunctor $Hom_{\bf CMon}(A,-)$).

We can form a a free structure $F(M)=\bigoplus_{-1<n<\infty}M^{\otimes n}$ where $M^{\otimes 0}=\{0,1\}$ analogous to the free ring.

We can follow with a quotient by a congruence relation with $m\otimes m \sim m$ and $m\otimes n\sim n\otimes m$ for all $m\otimes m$ and $n\otimes m$ in $F(M)$ making both operations idempotent. I was initially wondering whether this procedure could be continued by making new free objects with new formal operations.

What brought this to mind was the desire to make a topology for the usual tensor product, coproduct, of normal rings.

$\bf Edit2:$ Properties above give us a structure with two identities and two operations, one which distributes over the other, which is very similar to a ring. If we consider a category of these objects where the morphisms are maps preserving identities and operations, like the ones for rings. If we look at two morphisms $f_1,f_2\in Hom(A,B)$ and define $f_1f_2$ as $f_1(a)f_2(a)$ for all $a\in A$ and $f_1+f_2$ as $f_1(a)+f_2(a)$ we get:

$(f_1f_2)(1)=f_1(1)f_2(1)=1*1=1$ $f_1f_2(a_1a_2)=f_1(a_1a_2)f_2(a_1a_2)=f_1(a_1)f_1(a_2)f_2(a_1)f_2(a_2)=f_1(a_1)f_2(a_1)f_1(a_2)f_2(a_2)=f_1f_2(a_2)f_1f_2(a_2)$

$(f_1+f_2)(1)=f_1(1)+f_2(1)=1+1=1$ $(f_1+f_2)(a_1+a_2)=....=(f_1+f_2)(a_1)+(f_1+f_2)(a_2)$

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No, the set of boolean ring homomorphisms is not automatically a boolean ring: componentwise addition destroys the property of preserving $1$. –  Zhen Lin Aug 2 '12 at 9:39
not when addition is idempotent –  user25470 Aug 2 '12 at 9:42
Then you aren't talking about boolean rings. You mean boolean lattices. Don't abuse $+$ for $\vee$. Even then we don't get a boolean lattice by componentwise operations: $\neg$ destroys the property of preserving $\top$. For that matter, the constant $\top$ map isn't a boolean lattice homomorphism. –  Zhen Lin Aug 2 '12 at 9:44
Apologies, i am not very familiar with lattices or the notation. The structure I'm referring to is the same as a topology where we have two operations, unions and intersections, both idempotent, and two identities, the empty set and to "total set". Nothing else –  user25470 Aug 2 '12 at 10:18
That's not even a boolean lattice – it's just a set with two possibly-incompatible semilattice structures. You must specify some compatibility conditions to even get a lattice. Note also that a topology is not a boolean lattice. You also have not addressed my objection that pointwise operations don't work. –  Zhen Lin Aug 2 '12 at 10:37