• Journal of the Korean Mathematical Society
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• v.55 no.6
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• pp.1381-1388
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• 2018

Let ${\Delta}$ be a simplicial complex, $I_{\Delta}$ its Stanley-Reisner ideal and $K[{\Delta}]$ its Stanley-Reisner ring over a field K. Assume that ${\Gamma}(R)$ denotes the zero-divisor graph of a commutative ring R. Here, first we present a condition on two reduced Noetherian rings R and R', equivalent to ${\Gamma}(R){\cong}{\Gamma}(R{^{\prime}})$. In particular, we show that ${\Gamma}(K[{\Delta}]){\cong}{\Gamma}(K^{\prime}[{\Delta}^{\prime}])$ if and only if ${\mid}Ass(I_{\Delta}){\mid}={\mid}Ass(I_{{{\Delta}^{\prime}}}){\mid}$ and either ${\mid}K{\mid}$, ${\mid}K^{\prime}{\mid}{\leq}{\aleph}_0$ or ${\mid}K{\mid}={\mid}K^{\prime}{\mid}$. This shows that ${\Gamma}(K[{\Delta}])$ contains little information about $K[{\Delta}]$. Then, we define the squarefree zero-divisor graph of $K[{\Delta}]$, denoted by ${\Gamma}_{sf}(K[{\Delta}])$, and prove that ${\Gamma}_{sf}(K[{\Delta}){\cong}{\Gamma}_{sf}(K[{\Delta}^{\prime}])$ if and only if $K[{\Delta}]{\cong}K[{\Delta}^{\prime}]$. Moreover, we show how to find dim $K[{\Delta}]$ and ${\mid}Ass(K[{\Delta}]){\mid}$ from ${\Gamma}_{sf}(K[{\Delta}])$.