Timeline for Hilbert Nullstellsatz and Non-Complete Fields [closed]
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Dec 28, 2011 at 15:06 | history | undeleted |
Andreas Blass François G. Dorais |
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| Dec 24, 2011 at 9:14 | history | deleted |
user6976 Andy Putman Kevin Buzzard |
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| Dec 19, 2011 at 13:21 | comment | added | Emil Jeřábek | This question is a partial duplicate of mathoverflow.net/questions/32559/… . | |
| Dec 19, 2011 at 5:55 | history | closed |
Martin Brandenburg Felipe Voloch Andy Putman Gjergji Zaimi Alain Valette |
too localized | |
| Dec 18, 2011 at 23:58 | comment | added | Karl Schwede | Matsumura's Commutative Ring Theory handles this by looking at geometric points (which it calls algebraic points). | |
| Dec 18, 2011 at 19:13 | answer | added | Qiaochu Yuan | timeline score: 5 | |
| Dec 18, 2011 at 6:55 | answer | added | J.L. Nelson | timeline score: 4 | |
| Dec 17, 2011 at 21:43 | comment | added | Robert Kucharczyk | @Mahdi: Well, let's just wait for him to explain what he actually means, because at least for my taste the current formulation can be interpreted in many different ways, yielding many different answers. | |
| Dec 17, 2011 at 21:40 | comment | added | Mahdi Majidi-Zolbanin | @Robert: Not scheme theoretically, of course! I think in the question variety does not mean scheme. That's how I interpreted it. | |
| Dec 17, 2011 at 21:34 | comment | added | Robert Kucharczyk | @Mahdi: well, that exactly depends on what you understand by "variety". If you work in scheme theory (as your notation $\mathbb{A}_{\mathbb{R}}^2$ suggests), then they are not the same. Their sets of $\mathbb{R}$-valued points are. | |
| Dec 17, 2011 at 21:31 | comment | added | Mahdi Majidi-Zolbanin | In $\mathbb{A}^2_{\mathbb{R}}$ polynomials $x^2+y^2$ and $x^4+y^4$ define the same variety. | |
| Dec 17, 2011 at 21:28 | comment | added | Robert Kucharczyk | Also you have to make precise your use of the term "variety", since this is used differently by different authors. But with any of the usual definitions, your statement of Hilbert's Nullstellensatz is wrong. | |
| Dec 17, 2011 at 21:23 | comment | added | Robert Kucharczyk | You certainly mean "algebraically closed" instead of "complete". | |
| Dec 17, 2011 at 21:20 | history | asked | Jean Delinez | CC BY-SA 3.0 |