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A commutative ring $R$ with $1$ is said to be arithmetical if for all ideals $a , b$ of $R$, $ a ∩ ( b + c ) = ( a ∩ b ) + ( a ∩ c )$ or the localization $R_m$ is a uniserial ring for every maximal ideal $m$ of $R$. Now let $x\in R$, where $R$ is an arithmetical ring. Is there condition under which $\frac{R}{ann_R(x)}$ is a local ring, where $ann_R(x)=\{r\in R: rx=0\}$?

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    $\begingroup$ Other than the trivial one? It seems whether or not $ann_R(x)$ is contained in exactly one maximal ideal depends heavily on $R$ and $x$. What is the connection with the ring being arithmetical? $\endgroup$
    – rschwieb
    Commented May 4, 2018 at 15:19
  • $\begingroup$ when $R$ in uniserial, it is true. I want to consider the local case. $\endgroup$
    – Warner
    Commented May 4, 2018 at 15:23

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