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Commutative rings, modules, ideals, homological algebra, computational aspects, invariant theory, connections to algebraic geometry and combinatorics.
19
votes
3
answers
787
views
What is the shortest polynomial divisible by $(x-1)(y-1)(x^2y-1)$
I am interested in polynomials with few terms ("short polynomials", "fewnomials") in ideals. A simple to state question is
Given an ideal $I\subset k[x_1,\dots,x_n]$, what is the shortest polynomial …
19
votes
4
answers
2k
views
What is the geometric object corresponding to a subalgebra in a polynomial ring
Many introductory texts on algebraic geometry set up some sort of algebra-geometry dictionary in which radical ideals correspond to varieties, and so on. I am wondering if there is a geometric way to …
12
votes
Accepted
Is an ideal generated by multilinear, irreducible, homogeneous polynomials of different degr...
One general fact that comes to mind: If an ideal $I\subset \mathbb{k}[x_1,\dots,x_n]$ contains an element of the form $f = gx_1 + h$ where $g,h$ don't use $x_1$, and $g$ is a nonzerodivisor mod $I$, …
12
votes
1
answer
985
views
Is an irreducible ideal in $R$ also irreducible in $R[x]$?
Let $R$ be a commutative Noetherian ring and $I\subset R$ an ideal that is irreducible in the sense that if $I = J_1 \cap J_2$, then $I=J_1$ or $I=J_2$. Is (the ideal generated by) $I$ irreducible in …
11
votes
1
answer
867
views
Proving that a variety is not (isomorphic to) a toric variety
Is there an algorithmic (or other) way to prove that a (projective)
variety is not isomorphic to a toric variety?
I'd be happy with an algebraic answer (for affine or projective varieties),
using the …
9
votes
1
answer
687
views
When and where did the term "module" enter commutative algebra?
Bruns/Herzog "Cohen-Macaulay-Rings" has a note in the notes for Chapter 1, saying roughly that after the influx of homological algebra into commutative ring theory, modules became popular objects (ins …
9
votes
Primary ideals of the polynomial ring
When you restrict to special classes like monomial or binomial ideals (those generated by polynomials with one (monomial) or two (binomial) terms) then combinatorial characterizations exist. For insta …
6
votes
Depth and dimension
In the paper "Eine Dualität zwischen den Funktoren Ext und Tor" (J. Algebra 11, 510–531) Ischebeck shows that if $A$ admits a finitely generated module $N$ of finite injective dimension, then the answ …
6
votes
Accepted
Proving that a variety is not (isomorphic to) a toric variety
The question of algorithmically deciding if an ideal is binomial after a (suitable, e.g. linear) automorphism of affine space is decidable and various algorithms are discussed in "When is a polynomial …
5
votes
1
answer
1k
views
Is there an algorithm to decide if an ideal contains monomials?
Let $I\subset k[x_1,\dots,x_n]$ be an ideal in a polynomial ring in commuting variables. Is there a procedure to decide if $I$ contains a monomial and possibly to find one?
Gröbner bases come to min …
4
votes
0
answers
187
views
A non-matroidal notion of dependence on a set of ideals
Assume we are given a set of ideals $I_1, \dots, I_s$ in a commutative polynomial ring. Let's define a subset indexed by $A\subseteq [s] = \{ 1,2,\dots, s\}$ as dependent if there exists an $a\in A$ s …
4
votes
Complexity of Groebner bases
This is not a complete answer, just some related thoughts. I don't really know about the size that you define; the following (and the Mayr-Meyer paper) are about the degree of the $g_i$ with coefficie …
3
votes
Algorithm to decide if ideal is principal
In the graded situation the concept of "minimal generators" is well-defined. Just think about the minimal generators as part of the minimal free resolution. Their number is the first total Betti num …
3
votes
Accepted
How to find the generic initial ideal?
You should apply a generic linear coordinate transform to the ideal and then compute the initial ideal. The matrices for which the result is the generic initial ideal is a (Zariski) open subset (Lemm …
3
votes
Height of ideal in graded ring
The answer follows from Theorem 1.5.8 of Bruns/Herzog "Cohen-Macaulay-Rings": The basic reason is that for any non-graded prime $p$ one has $\text{ht}(p/p') = 1$. To see this replace $R$ by $R/p'$ and …