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Finite Abelian algebras are fully dualizable

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 Added by Pierre Gillibert
 Publication date 2015
  fields
and research's language is English




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We show that every finite Abelian algebra A from congruence-permutable varieties admits a full duality. In the process, we prove that A also allows a strong duality, and that the duality may be induced by a dualizing structure of finite type. We give an explicit bound on the arities of the partial and total operations appearing in the dualizing structure. In addition, we show that the enriched partial hom-clone of A is finitely generated as a clone.



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115 - Pierre Gillibert 2015
A finite algebra $bA=alg{A;cF}$ is emph{dualizable} if there exists a discrete topological relational structure $BA=alg{A;cG;cT}$, compatible with $cF$, such that the canonical evaluation map $e_{bB}colon bBto Hom( Hom(bB,bA),BA)$ is an isomorphism for every $bB$ in the quasivariety generated by $bA$. Here, $e_{bB}$ is defined by $e_{bB}(x)(f)=f(x)$ for all $xin B$ and all $fin Hom(bB,bA)$. We prove that, given a finite congruence-modular Abelian algebra $bA$, the set of all relations compatible with $bA$, up to a certain arity, emph{entails} the whole set of all relations compatible with $bA$. By using a classical compactness result, we infer that $bA$ is dualizable. Moreover we can choose a dualizing alter-ego with only relations of arity $le 1+alpha^3$, where $alpha$ is the largest exponent of a prime in the prime decomposition of $card{A}$. This improves Kearnes and Szendrei result that modules are dualizable, and Bentz and Mayrs result that finite modules with constants are dualizable. This also solves a problem stated by Bentz and Mayr in 2013.
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