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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 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 $1$.2012-08-02
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    not when addition is idempotent2012-08-02
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    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.2012-08-02
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    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 else2012-08-02
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    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.2012-08-02
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    You seem to be missing the point. It is true that $\textbf{CMon}$ is a closed category. But that has nothing to do with your "structures like topology", because the category of your "structures" does not embed as a _full_ subcategory. (What's the point of having an extra binary operation if you're not going to preserve it?)2012-08-02
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    If you are genuinely interested in "structures like a topology, but not a topological space", then you should look at locale theory (a.k.a. pointless topology). Joyal and Tierney give an exposition of the algebraic properties of these objects in their memoir, _An extension of the Galois theory of Grothendieck_. In particular, they discuss tensor products.2012-08-02
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    pointless topology is a rather fun name, ill keep that in mind and look it over after i've grasped galois theory. I'm sorry if this has been frustrating for you and i am grateful for you taking the time to look at it. I'm not sure why its important with a full subcategory thou, the rings are not a full subcategory of the groups either, but they are still interesting. If we can continue this adjunction with new idempotent operations indefinitely i imagine it would somehow relate to the difference in diversity of philosophy and math, but that's pure speculation2012-08-02
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    Your latest edit gives a definition that is almost the definition of a distributive lattice. Why not just say that you are thinking about distributive lattices? Nonetheless, you _still_ haven't shown that pointwise operations work – and let me be blunt, they don't. The constant $0$ map is not a homomorphism because it doesn't preserve $1$. So your pointwise "operation" – even if it were well-defined – would not have an identity element.2012-08-02
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    Very true, anyway thanks for the pointless topology tip, looking into that now2012-08-02

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