Timeline for Higher dimensional Bezout via Hilbert polynomials: a reference
Current License: CC BY-SA 3.0
20 events
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| Oct 7, 2014 at 15:10 | answer | added | David E Speyer | timeline score: 2 | |
| Oct 7, 2014 at 14:52 | comment | added | David E Speyer | I came here last night planning to ask exactly this question! Thanks aglearner and Sándor. My conclusion after reading the answers is that I'll probably prove the result in $\mathbb{P}^2$, explain the connection to Cohen-Macaulayness in general but not give a full general proof. | |
| Jan 6, 2014 at 7:07 | answer | added | aglearner | timeline score: 7 | |
| Jan 6, 2014 at 7:02 | history | edited | aglearner | CC BY-SA 3.0 |
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| Mar 5, 2013 at 23:03 | comment | added | roy smith | there are some wonderful answers here. In addition to using them to give a rigorous proof, you may want to give some motivation for why the theorem is expected to be true. I.e. the idea that intersection numbers should only go down under specialization, and that transverse ones should remain constant when varied, may lead someone to believe the problem reduces to the case of intersecting unions of hyperplanes, where it is clear. Actually this is Shafarevich's argument, i.e. his intersection numbers are constant under linear equivalence, and every divisor is equiv. to a union of hyperplanes. | |
| Mar 5, 2013 at 22:20 | history | edited | aglearner | CC BY-SA 3.0 |
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| Mar 5, 2013 at 22:16 | vote | accept | aglearner | ||
| Mar 4, 2013 at 15:30 | comment | added | aglearner | Abdelmalek, thank you for these references! It looks to me indeed that Van der Waerden proves Bezout using only Nullstelensatz, this is very nice :) But it will take me some time of course, to understand the proof. | |
| Mar 4, 2013 at 14:40 | comment | added | Abdelmalek Abdesselam | @aglearner: there are many places depending on the level of sophistication you want. There is the book by Gelfand, Kapranov and Zelevinsky. The book "Using Algebraic Geometry" by Cox Little and O'Shea. A quick intro is in the first edition of Modern Algebra by van der Waerden. | |
| Mar 4, 2013 at 13:44 | comment | added | aglearner | Abdelmalek, thank you for your remark. To my shame I don't know what is the multidimensional resultant. Could you tell where can I read about this? I would like to learn more about the approach you propose. | |
| Mar 4, 2013 at 13:41 | comment | added | aglearner | Sandor, I agree that claim 1 from your answer would suffice. Also, Manin's suggested proof of Bezout goes as follows: you calculate by induction the leading coefficient of the Hilbert polynomial of $A_r=k[x_0,....,x_n]/(f_1,...,f_r)$. When you make a step of induction and pass form the Hilbert polynomial of $A_r$ to the Hilbert polynomial of $A_{r+1}$ the easiest way to proceed would be to say that $f_{r+1}$ is not a zero divisor in $A_r$...Hartshorne proposes a more detailed analysis of what happens at this step (introducing multiplicities), but I don't see how to state all this with ease... | |
| Mar 4, 2013 at 13:34 | comment | added | Abdelmalek Abdesselam | @aglearner: How about using multidimensional resultants? If you consider Res(F_1,...,F_n,U) where U is the linear form defining a variable hyperplane, Bezout is just the statement that the above expression is a completely factorized polynomial in U of degree d_1...d_n. | |
| Mar 4, 2013 at 13:06 | comment | added | Sándor Kovács | @aglearner: I was actually wondering whether you really need that the $F_i$ form a regular sequence. I haven't seen Manin's notes, but it seems plausible that what you need is the statement about the dimension of $Z(f_1,\dots,f_r)$ (Claim 1 in my answer). After all, that's actually equivalent to the statement that it is a regular sequence, but it is easier to prove from the assumptions. Or more generally, if you can isolate what it is for which you need the sequence to be regular, it is possible that you can prove that relatively easier. | |
| Mar 4, 2013 at 6:26 | answer | added | Sándor Kovács | timeline score: 9 | |
| Mar 3, 2013 at 23:56 | comment | added | aglearner | Jack I agree with what you say and I agree that this it is probably more important. But this is also harder to state (comparing to Bezout's theorem). | |
| Mar 3, 2013 at 23:26 | comment | added | Jack Huizenga | It's easy to prove this if you know that every component of a hyperplane section of an irreducible variety has codimension at most 1. Obviously this is easier to show from some foundations than others, but it's probably a more important fact than Bezout. | |
| Mar 3, 2013 at 22:03 | comment | added | aglearner | Felipe, thanks. I am afraid that what you propose would not fit into my course... I have 4 more lectures to give and finished with Hilbert basis theorem last time. It will take me more than four lectures to go in details through section 1.7 of Hartshorne and this will kill my students (and myself I guess). So I am looking for some miraculous solution. If it does not exist I'll do just like Hasset and Manin (living this statement about regular sequence as an exercise :( ...) | |
| Mar 3, 2013 at 21:49 | history | edited | aglearner | CC BY-SA 3.0 |
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| Mar 3, 2013 at 20:08 | comment | added | Felipe Voloch | Hartshorne, Thm I.7.7 pg 53, applied $n$ times. | |
| Mar 3, 2013 at 18:39 | history | asked | aglearner | CC BY-SA 3.0 |