On varieties of rings whose finite rings are determined by their zero-divisor graphs. (Q2920384)

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scientific article; zbMATH DE number 6094195
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On varieties of rings whose finite rings are determined by their zero-divisor graphs.
scientific article; zbMATH DE number 6094195

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    16 October 2012
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    zero-divisor graphs
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    subdirectly irreducible finite rings
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    finite nilpotent rings
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    varieties of associative rings
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    polynomial identities
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    On varieties of rings whose finite rings are determined by their zero-divisor graphs. (English)
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    The zero-divisor graph \(\Gamma(R)\) of an associative ring \(R\) is the graph whose vertices are all nonzero zero-divisors of \(R\), and two distinct vertices \(x\) and \(y\) are joined by an edge if and only if \(xy=0\) or \(yx=0\). In the paper under review the authors continue the investigation of the first named author [\textit{A. S. Kuzmina}, ``On some properties of ring varieties, where isomorphic zero-divisor graphs of finite rings give isomorphic rings'', Sib. Èlektron. Mat. Izv. 8, 179-190 (2011)] where she considers varieties of associative rings \(\mathfrak M\) with the property that two finite rings \(R,S\in\mathfrak M\) with isomorphic zero-divisor graphs \(\Gamma(R),\Gamma(S)\) are isomorphic.NEWLINENEWLINE The main results of the present paper are:NEWLINENEWLINE (1) Let \(\mathfrak M\) satisfy a polynomial identity of the form \(xy+f(x,y)=0\), where \(f(x,y)\) has no monomials of degree \(\leq 2\). Then \(\Gamma(R)\cong\Gamma(S)\) implies \(R\cong S\) for all finite \(R,S\in\mathfrak M\) if and only if \(\mathfrak M\) is a subvariety of the variety generated by a finite number of algebras \(\langle a\mid a^2=p_ia=0\rangle\) and \(\mathbb Z_p\), where \(p_i\) and \(p\) are pairwise different primes such that \((p_i,p)\neq(3,2)\) for all \(p_i\).NEWLINENEWLINE Below \(\mathfrak M\) satisfies the condition that \(\Gamma(R)\cong\Gamma(S)\) implies \(R\cong S\) for all finite \(R,S\in\mathfrak M\).NEWLINENEWLINE (2) Let \(\mathbb Z_p\in\mathfrak M\) for some prime \(p\). Then any subdirectly irreducible finite ring \(R\in\mathfrak M\) is either isomorphic to \(\mathbb Z_p\) or is nilpotent and the polynomial identity \(q^2x=0\) holds for some prime \(q\).NEWLINENEWLINE (3) If \(\mathbb Z_2\in\mathfrak M\), then any subdirectly irreducible ring of order \(2^t\) in \(\mathfrak M\) is either isomorphic to \(\mathbb Z_2\) or is a nilpotent commutative ring satisfying \(2x=0\) and \(x^2=0\).
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