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 ac.commutative-algebra
Search options not deleted
user 605
Commutative rings, modules, ideals, homological algebra, computational aspects, invariant theory, connections to algebraic geometry and combinatorics.
23
votes
Accepted
How to show a set of polynomials is algebraically independent?
The polynomials are algebraically independent if and only if
$$
df_1 \wedge df_2 \wedge \cdots \wedge df_n
$$
is not identically zero. In other words, you have only to check that
one of the maximal …
- 8,828
11
votes
Accepted
When does the conormal bundle sequence split?
You are right. This is a result due to Van de Ven.
[A. Van de Ven, A property of algebraic varieties in complex projective spaces. In: Colloque Géom. Diff. Globale (Bruxelles, 1958), 151–152, Centre …
- 8,828
9
votes
which homogeneous polynomials split into linear factors?
There are Brill's equations. Look for them at the book by Gelfand, Kapranov, and Zelevinski. In general Brill's equations do not generate the ideal of totally decomposable polynomials, see this paper. …
- 8,828
8
votes
Are submersions of differentiable manifolds flat morphisms?
It is a consequence of Malgrange's preparation theorem for differentiable
functions that $C^{\infty}(M)$ is a faithfully flat $C^{\omega}(M)$-module ($C^{\omega}(M)$ is the sheaf of
analytic functi …
- 8,828
1
vote
Effective bound on "Jacobian rank" for (regular) planar algebraic curves
I believe such bound exist. Perhaps you can deduce it from effective versions of Hilbert's Nullstellensatz. Take a look at Kollár's paper "Sharp Effective Nullstellensatz". The wikipedia page on Hilbe …
- 8,828