Timeline for Complete Intersection
Current License: CC BY-SA 3.0
13 events
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| May 1, 2015 at 10:59 | comment | added | Bil | I have edited the question by the help of the comments. | |
| May 1, 2015 at 10:54 | history | edited | Bil | CC BY-SA 3.0 |
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| May 1, 2015 at 7:41 | comment | added | user26857 | @Bil I've thought that $P/I\simeq K[Y]$ means that $P/I$ is isomorphic to a polynomial ring, but it seems you conceded now that $P/I=K[\bar y_1,\dots,\bar y_m]$ which trivializes the question. | |
| Apr 30, 2015 at 19:14 | comment | added | Mohan | @Bil See my paper MohanKumar, N. Complete intersections. J. Math. Kyoto Univ. 17 (1977), no. 3, 533–538. The result is more general. | |
| Apr 30, 2015 at 18:59 | comment | added | Bil | @JasonStarr thanks for your comment, as well. Actually those $y_1,...,y_m$ are not random. Their equivalence classes form a base of the cotangent space $m/m^2,$ where $m\subset P/I.$ I believe (not sure) that $\{\bar{y}_1,...,\bar{y}_m\}$ form a generating set for $P/I.$ | |
| Apr 30, 2015 at 18:27 | comment | added | Bil | @Mohan thanks for the comment. Would you please write the reference for dimension n>5? | |
| Apr 30, 2015 at 9:19 | comment | added | Jason Starr | There seems to be a hypothesis built into the way you pose the problem. Just to make this explicit: are you assuming that there exist a subset $\{y_1,\dots,y_m\}$ of the set of coordinates $\{x_1,\dots,x_n\}$ such that the images $\{\overline{y}_1,\dots,\overline{y}_m\}$ in $P/I$ are generators? If so, then certainly $I$ is a complete intersection generated by $n-m$ polynomials of the form $x_j - g_j(y_1,\dots,y_m)$ for $x_j\not\in \{y_1,\dots,y_m\}$. | |
| S Apr 30, 2015 at 7:45 | history | suggested | user 1 |
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| Apr 30, 2015 at 7:38 | review | Suggested edits | |||
| S Apr 30, 2015 at 7:45 | |||||
| Apr 30, 2015 at 2:45 | comment | added | Mohan | In general smooth subvarieties are NOT complete intersections. The least you need is that the module of Kahler differentials is stably free and in this case if the codimension is large enough, it is. I do not think one knows whether $\mathbb{A}^2\subset\mathbb{A}^5$ is in general a complete intersection. Instead of 5, for any $n>5$ it is. | |
| Apr 29, 2015 at 22:42 | review | First posts | |||
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| Apr 29, 2015 at 22:40 | comment | added | Bil | I would like to note that the polynomials are not homogenous, they are just quadratic. | |
| Apr 29, 2015 at 22:38 | history | asked | Bil | CC BY-SA 3.0 |