When $\mathfrak m^2$ is the unique minimal ideal in a local ring $(R,\mathfrak m)$?
Note that in this case $\mathfrak m^3=0$ in $R$. Furthermore assume that $\operatorname{char}(R)$ is finite.
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3$\begingroup$ Umm...how about Z/p^3Z? $\endgroup$– David CorwinCommented Nov 22, 2012 at 15:19
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1 Answer
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All of them can be described as follows. Let $(A,m)$ be any local ring (commutative). By going modulo $m^3$ we may assume that $m^3=0$. Now $m^2$ is a vector space over $k=A/m$ and let $I\subset m^2$ be any codimension one $k$-subspace. Then $I$ is an ideal in $A$ and $A/I$ will have the property you need and any such looks like this.
As Pham pointed out, my answer was not correct. So, here is another attempt. Clearly, we may assume that $R$ is graded, since $m^3=0$. Thus, $R=k\oplus V\oplus k$ where $V\cong m/m^2$ and the last $k=m^2$. To make sure that $m^2$ is the unique maximal ideal, the only condition we need to assume seems to be the natural multiplication map $S^2V\to k$ is non-degenerate. So, all such rings seem to come from starting with a vector space $V$ and a non-degenerate symmetric bilinear form on $V$, which gives a commutative ring structure on $R$ as above.
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$\begingroup$ In your answer $\mathfrak{m}$ is a minimal ideal but may not be the unique minimal ideal. For example: $A = k[x, y]/(x,y)^3$, $k$ be a field. Let $\mathfrak{m} = (x,y)A$ we have $\mathfrak{m}^2$ is a 3-dimensional $k$-vector space of basis $\{x^2, xy, y^2\}$. Follow your construction we may choose $R = A/(xy, y^2)$. So $R = k[x^3, xy, y^2]$. We can check that $\mathfrak{m}^2 = (x^2)R$ is a minimal ideal of $R$ but not uniquely. Another minimal ideal of $R$ is $(y)R$. $\endgroup$ Commented Nov 27, 2012 at 6:01
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$\begingroup$ You are right of course. To have a unique minimal ideal, the ring must be Gorenstein, and so there are not too many choices. $\endgroup$– MohanCommented Nov 27, 2012 at 14:20