Timeline for What does it mean geometrically that an element in a domain is irreducible?
Current License: CC BY-SA 3.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 7, 2011 at 11:23 | answer | added | François Brunault | timeline score: 7 | |
| May 6, 2011 at 10:16 | answer | added | Sándor Kovács | timeline score: 10 | |
| May 6, 2011 at 7:57 | comment | added | Georges Elencwajg | Dear A.Rex, thanks for the LaTex edits. As for the purely linguistic edits, I agree that your words are definitely more standard, but for some reason I should like to maintain my idiosyncratic lexical choices: English is not my mother tongue, and I don't mind if it shows...Anyway, thank you for your interest. | |
| May 6, 2011 at 7:22 | comment | added | aorq | @Georges Elencwajg: I've taken the liberty of editing your question to change some LaTeX, mostly changing $\lt\dotsb\gt$ to $\langle\dotsb\rangle$ (the actual brackets might be harder to read, but the spacing with $\lt\dotsb\gt$ is wrong because LaTeX thinks the symbols are relational rather than left/right delimiters). I've also made $\operatorname{Spec}$ upright and changed a couple words. Let me know if you don't like these changes and I won't change your posts again. | |
| May 6, 2011 at 7:19 | history | edited | aorq | CC BY-SA 3.0 |
\langle and \rangle, upright Spec, some words
|
| May 5, 2011 at 22:14 | answer | added | user13113 | timeline score: 6 | |
| May 5, 2011 at 18:10 | comment | added | roy smith | the problem for me seems to be how to give a geometric description of a principal hypersurface. on a real circle, are they unions of divisors of degree 2? | |
| May 5, 2011 at 13:26 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
added an "edit"
|
| May 4, 2011 at 17:09 | comment | added | Karl Schwede | You are right, I was being dumb. Here's a correct statement (hopefully). For what it's worth, if R is normal and local then I think that $f$ is irreducible if and only if the associated divisor $\text{div}(f)$ cannot be written as a sum of effective non-zero Cartier divisors. This is basically the geometric phrasing of François's statement in the local case. | |
| May 4, 2011 at 16:19 | comment | added | Georges Elencwajg | Dear Karl, unless I am misunderstanding something, this doesn't seem to be correct. Consider the real circle $x^2+y^2=1$ evoked in my question. The element $x\in \mathbb R[x,y]$ is irreducible. However its divisor $div(x)$ is the sum of the two divisors corresponding to the two points with coordinates $(0,1)$ and $(0,-1)$ (since the circle is a regular variety, there is no distinction between Cartier and Weil divisors). | |
| May 4, 2011 at 12:36 | comment | added | Georges Elencwajg | Dear François, you are too modest: this is an interesting remark, which shows that the hypersurface $V(f)$ does not scheme-theoretically strictly contain a principal hypersurface . If you feel so inclined, I'm sure you can develop this into an answer that I'd be happy to upvote, since it would definitely be a step in the geometric direction. Anyway, thanks for your comment. | |
| May 4, 2011 at 11:55 | comment | added | François Brunault | An element is irreducible iff the ideal it generates is maximal among the principal ideals. But from the point of view of schemes we are considering all ideals not just principal ones, so it isn't clear there should be a geometric characterization... | |
| May 4, 2011 at 11:27 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
corrected spelling of "used"
|
| May 4, 2011 at 11:07 | comment | added | Georges Elencwajg | Dear Qiaochu: no, it is (alas) not obvious at all. Take my question as a mixture of optimism and bias toward geometry... | |
| May 4, 2011 at 10:54 | comment | added | Qiaochu Yuan | Is it obvious that irreducibility ought to have a geometric meaning? | |
| May 4, 2011 at 10:53 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
Added "for example"
|
| May 4, 2011 at 10:47 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
Added that irreducible is easy concept. Corrected spelling mistake.
|
| May 4, 2011 at 10:05 | history | asked | Georges Elencwajg | CC BY-SA 3.0 |