Is it possible to define a "direct product" of two algebras with different signatures? For example, boolean lattice $\otimes$ monoid? Perhaps we need to take some quotient to make sense of it (e.g Q is direct product of two groups with standard congruence)?
direct product of different algebras?
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abstract-algebra
universal-algebra
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0What kind of object do you want the result to be and what kind of relation do you want it to bear to the original objects? – 2012-03-15
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0I would like see if [idempotent] semiring, or relation algebra can be constructed this way from lattice and monoid. – 2012-03-15
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0You can try to define them in terms of functors from a product category, but there is nothing within universal algebra that I know of that would let you do that. – 2012-03-15
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0@Arturo you often post how Q is constructed as a set of pairs (p,q) where $p \in N$ and $q \in N$ with equivalence relation `(p,q)~(r,s) <-> ps-qr=0`. What is this construction algebraically; it is not quite direct product N by N over some congruence? – 2012-03-15
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2@Tegiri: You can view the construction as a functor from a subcategory of $\mathbf{Ring}\times\mathbf{Monoid}$ to $\mathbf{Ring}$ (the localization functor), but this is not a universal-algebra construction in the same sense as products, quotients, etc. (And that should be $p\in\mathbb{Z}$, $q\in\mathbb{Z}-\{0\}$). Precisely what I said: for *specific* instances, you can define such a construction in terms of functors from an appropriate product category, but I don't see how you can view this as a "general algebra" construction of algebras of different signatures. – 2012-03-15
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1There's a generic definition of a "tensor product" of two monads, which algebraically amounts to the theory obtained by taking the disjoint union of the two signatures, the disjoint union of the axioms, and adding a suitable commutativity axiom for each pair of operations from the two different signatures. There are also weaker notions. – 2012-03-15