Search Results
| Search type | Search syntax |
|---|---|
| Tags | [tag] |
| Exact | "words here" |
| Author |
user:1234 user:me (yours) |
| Score |
score:3 (3+) score:0 (none) |
| Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
| Views | views:250 |
| Code | code:"if (foo != bar)" |
| Sections |
title:apples body:"apples oranges" |
| URL | url:"*.example.com" |
| Saves | in:saves |
| Status |
closed:yes duplicate:no migrated:no wiki:no |
| Types |
is:question is:answer |
| Exclude |
-[tag] -apples |
| For more details on advanced search visit our help page | |
Results tagged with projective-varieties
Search options not deleted
user 88133
In algebraic geometry, a projective variety over an algebraically closed field $k$ is a subset of some projective $n$-space $\mathbb P^n$ over $k$ that is the zero-locus of some finite family of homogeneous polynomials of $n + 1$ variables with coefficients in $k$, that generate a prime ideal, the defining ideal of the variety
2
votes
Accepted
Secant variety to a zero-dimensional projective variety
I believe this would be a dual arrangement of a star arrangement.
A star arrangement is a union of subspaces defined as follows. Let $H_1,\dotsc,H_d$ be a collection of hyperplanes and fix an integer …
- 6,237
5
votes
Are "transverse" hyperplane sections of nondegenerate irreducible projectice varieties alway...
This is a bit of a folk theorem. Harris (Algebraic Geometry, Proposition 18.10 and Exercise 18.11) states it for general hyperplane sections, but actually proves it for all generically transverse hype …
8
votes
Accepted
Linear spaces secant to Veronese varieties
Here is an answer in terms of power sum decompositions of polynomials. A point $p \in \mathbb{P}^9$ corresponds to a homogeneous polynomial $P$ of degree $3$ in $3$ variable, defining a plane cubic. P …
- 6,237