Timeline for A version of Hilbert's Nullstellensatz for real zeros
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Apr 15 at 1:48 | comment | added | Iosif Pinelis | What edition are you referring to? | |
| Apr 14 at 21:44 | comment | added | KhashF | @IosifPinelis In my version, the result appears in Section 2 (Complex Manifolds) of Chapter 0, in a subsection called Submanifolds and Subvarieties. Also see this: math.stackexchange.com/questions/3461334/… | |
| Apr 14 at 21:27 | comment | added | Iosif Pinelis | It is nice to see that my guess that the irreducibility of $Q$ can be used to get the connectedness may be correct. However, in Chapter 0 of Griffiths & Harris I have only found something only on the local irreducibility of the analytic hypersurface that is the zero set of a holomorphic function (and this local irreducibility is again based on the Nullstellensatz!). | |
| Apr 14 at 20:50 | comment | added | KhashF | @IosifPinelis I amended my answer. | |
| Apr 14 at 20:50 | history | edited | KhashF | CC BY-SA 4.0 |
Clarification added
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| Apr 14 at 14:44 | comment | added | Iosif Pinelis | Oh, yes. Yet, can the irreducibility be also used to get the connectedness? | |
| Apr 14 at 14:33 | comment | added | KhashF | @IosifPinelis Isn't the irreducibility used once the problem is reduced to the complex Nullstellensatz? $Z_{\Bbb{C}}(Q)\subseteq Z_{\Bbb{C}}(P)\Rightarrow Q\mid P^r$ for some $r$. If $Q$ is irreducible, $Q\mid P^r$ implies $Q\mid P$. | |
| Apr 14 at 14:27 | comment | added | Iosif Pinelis | Thank you for your answer. It seems you have not explicitly used the irreducibility of $Q$. Maybe, it can be used to get the connectedness? | |
| Apr 14 at 14:17 | history | answered | KhashF | CC BY-SA 4.0 |