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Let $(R, \mathfrak m)$ be a commutative local Gorenstein ring such that every minimal prime ideal of $R$ is contained in $\mathfrak m^2$. Then is it true that $R$ is an integral domain ?

If this is not true in general, what if we also assume $R$ is either reduced (has no non-zero nilpotent) or irreducible (has exactly one minimal prime) ?

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None of what you say may be true. Take $R=\mathbb{C}[[x,y,z]]/(x^2+y^2+z^2)^2$. Then it is irreducible, the only minimal prime is generated by $x^2+y^2+z^2$, whichis contained in the square of the maximal ideal and it is Gorenstein.

Similarly, take $R=\mathbb{C}[[x_1,x_2,x_3, y_1,y_2, y_3]]/ (\sum x_i^2)(\sum y_i^2)$. It is reduced and has all the other properties you want.

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  • $\begingroup$ Yes, I have thought a similar example as yours just now. $\mathbb{R}[[x,y]]/(x^2+y^2)^2$. $\endgroup$
    – Jian
    Oct 19, 2019 at 15:55

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