Timeline for Local ring $(R,\mathfrak m)$ such that $\mathfrak m^2$ is the unique minimal ideal
Current License: CC BY-SA 3.0
5 events
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| Nov 27, 2012 at 16:59 | history | edited | Mohan | CC BY-SA 3.0 |
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| Nov 27, 2012 at 14:20 | comment | added | Mohan | You are right of course. To have a unique minimal ideal, the ring must be Gorenstein, and so there are not too many choices. | |
| Nov 27, 2012 at 6:01 | comment | added | Pham Hung Quy | 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 $(y)R$. | |
| Nov 25, 2012 at 5:32 | vote | accept | lina | ||
| Nov 22, 2012 at 17:23 | history | answered | Mohan | CC BY-SA 3.0 |