Timeline for Irreducibility of Analytic Sets
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Jan 2, 2011 at 2:02 | comment | added | Georges Elencwajg | (continuation) And if you just write $V=V_1\cup V_2$ without bothering whether the $V_i$'s are irreducible, the statement is false: just add a smooth point $s$ of $V$ to each of $V_1$ and $V_2$ and look at $V= W_1 \cup W_2 $ with $W_i=V_i \cup \{s\}$. The point $s$ is in $W_1 \cap W_2 $ and yet is a smooth point of $V$. I have modified your question in my answer below in order to take these remarks into account. | |
| Jan 2, 2011 at 1:36 | comment | added | Georges Elencwajg | Dear unknown, your statement (and Griffiths-Harris's) should be made more precise. Indeed, if $V$ is reducible, it can be be decomposed into irreducibles but there might be more than two irreducible components. Actually there might be infinitely many such components.For example, think of a comb i.e. in $\mathbb C^2$ the union of the horizontal $x$-axis and the vertical lines with integral first coordinate (to be continued) | |
| Jan 1, 2011 at 20:37 | answer | added | Georges Elencwajg | timeline score: 11 | |
| Jan 1, 2011 at 19:07 | comment | added | Poincare-Lelong | Can you elaborate? | |
| Jan 1, 2011 at 15:51 | comment | added | Thierry Zell | Product rule?... | |
| Jan 1, 2011 at 15:42 | history | asked | Poincare-Lelong | CC BY-SA 2.5 |