Timeline for Flatness of normalization
Current License: CC BY-SA 4.0
7 events
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| Feb 6, 2020 at 21:02 | history | edited | Karl Schwede | CC BY-SA 4.0 |
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| Aug 1, 2011 at 22:23 | comment | added | Karl Schwede | Graham, I probably didn't say what I meant to above very clearly. I am not claiming either I think. What I am claiming is that $S \cong R \oplus M$, as an $R$-module, where $M$ has no $R$-summands. | |
| Aug 1, 2011 at 18:39 | comment | added | Graham Leuschke | Hi Karl - why exactly is $S$ indecomposable over $R$? Or are you saying that $S \cong M^n$ for a single indecomposable $M$? Either way, I'm confused. I think that if $f = xz-y^2$, then $R \cong k[[a^2,ab,b^2]] \subset k[[a,b]] = S$, and $S \cong R \oplus (a,b)R$. | |
| May 12, 2011 at 15:19 | history | edited | Karl Schwede | CC BY-SA 3.0 |
Added note about the direct summand conjecture; added 59 characters in body
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| May 12, 2011 at 14:15 | history | edited | Karl Schwede | CC BY-SA 3.0 |
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| May 12, 2011 at 14:04 | history | edited | Karl Schwede | CC BY-SA 3.0 |
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| May 12, 2011 at 13:58 | history | answered | Karl Schwede | CC BY-SA 3.0 |