Timeline for A Theorem in Intersection theory.
Current License: CC BY-SA 2.5
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 3, 2011 at 13:17 | vote | accept | Sagar Kolte | ||
| Jan 31, 2011 at 2:25 | comment | added | roy smith | It has been a long time, but as I recall there can be "embedded" components of the intersection which can contribute to the degree, so this question is essentially asking when those do not occur. One would expect them to occur only when either there is a component of lower than expected codimension, or else a "bad" component of expected codimension. This is meant to motivate Allen Knutson's answer. | |
| Jan 24, 2011 at 16:22 | comment | added | Sagar Kolte | Here is a related question: If the inequality in the quoted theorem is strict, does it imply that the intersection is not complete? | |
| Jan 21, 2011 at 16:30 | comment | added | Sándor Kovács | @Allen: Uh-huh! I tried figuring out how $3\geq 1\ast 2$ explains it... :) | |
| Jan 21, 2011 at 15:19 | comment | added | Allen Knutson | Oh dear, I see now that I was thinking about the inequality backwards, exactly because of this example. So $deg((X\cap Y)_{red}) \leq deg\ X\ deg\ Y \leq deg(X\cap Y)$, when the codimensions add up? | |
| Jan 21, 2011 at 7:26 | comment | added | Sándor Kovács | @Dave: I had the same thoughts regarding what had been asked... | |
| Jan 21, 2011 at 4:09 | comment | added | Dave Anderson | @Sagar, sorry, that was written in haste; I mean just the theorem you're quoting. | |
| Jan 21, 2011 at 3:42 | comment | added | Sagar Kolte | @Dave: Will you please state the "generalized Bezout's Theorem"? It will be a lot of help. | |
| Jan 21, 2011 at 1:09 | comment | added | Dave Anderson | ...In this case, a closely related question is: When is the cycle representing $X\cdot Y$ equivalent to the scheme-theoretic intersection $X\cap Y$? (And here's where the Cohen-Macaulay hypothesis does its work.) | |
| Jan 21, 2011 at 1:04 | comment | added | Dave Anderson | This is indeed a little confusing, I think because of language. The example Allen gives appears to be a counterexample to the statement from Fulton, since the degree (=multiplicity) of the (irreducible) scheme Z is greater than the product of the degrees of the intersecting schemes. However, I believe the statement the OP refers to is the more elementary fact that the underlying variety Z has degree at most equal to the product of the degrees of the intersecting varieties X and Y, a simple application of the generalized Bezout theorem. | |
| Jan 20, 2011 at 18:48 | comment | added | Allen Knutson | Sorry, Sagar: fixed. @Sándor: the intersection of $X$ and $Y$ should be $1*2$ points, not $3$. Note that $3 \geq 1*2$ is in agreement with the statement from [Fulton]. | |
| Jan 20, 2011 at 18:47 | history | edited | Allen Knutson | CC BY-SA 2.5 |
added 235 characters in body
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| Jan 20, 2011 at 17:38 | comment | added | Sagar Kolte | Thanks Allen, but I am a little confused, you say: "If the co-dimensions are all exactly n..." by this do you mean that the dimension of the intersection is zero? I am sorry if I have not given this enough thought. | |
| Jan 20, 2011 at 8:44 | comment | added | Sándor Kovács | Allen, I don't understand your non-example. Are you saying that $Y$ as a cycle is not the sum of its components? Or are you saying that the intersection of two random planes through a point is $1.5$ points? Finally, are you saying that the degree of the triple point is strictly less than the product of the degrees (=$2$)? I am totally confused. | |
| Jan 20, 2011 at 5:20 | history | answered | Allen Knutson | CC BY-SA 2.5 |