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Commutative rings, modules, ideals, homological algebra, computational aspects, invariant theory, connections to algebraic geometry and combinatorics.
6
votes
1
answer
1k
views
If some powers of polynomials are linearly independent, does it imply higher powers are also...
Let $P_1,\dotsc,P_k$ be polynomials. Assume they are pairwise non-proportional (i.e., any two of them are linearly independent). Suppose $N$ is a power such that $P_1^N,\dotsc,P_k^N$ are linearly inde …
5
votes
Accepted
Ideals: If $\langle f_1,f_2 \rangle = \langle g_1,g_2 \rangle$, then $\langle f_1-\lambda,f_...
The answer to your questions is no. The ideals $\langle x, y(1-xy) \rangle$ and $\langle x, y \rangle$ are equal, and maximal; but
$$ \langle x-\lambda, y(1-xy)\rangle \neq \langle x-\delta, y-\epsilo …
3
votes
Accepted
Modulo $x^2 + y^2 - 1$, is every homogeneous polynomial that is a square of a polynomial, ne...
Yes. Will Sawin's answer uses a topological fact about the unit circle in $\mathbb{R}^2$. Here is an answer replacing $\mathbb{R}$ with any field of characteristic not $2$.
First, working modulo $x^2+ …
2
votes
Accepted
When a sum of the ideals is radical
(Adapting earlier comment into an answer.) Assume for simplicity that $X$ and $Y$ are irreducible, or in general take an irreducible component of each. Let $Z$ be an irreducible component of the inter …
5
votes
Accepted
Naive question on tensor product
A trivial counterexample is $m=m'=0$. Perhaps more interesting is a situation where $b$ does not belong to the image of $A$, but some multiple or power of $b$ does. Say $M = B = \mathbb{C}[x]$ and $A …
10
votes
Accepted
When is an algebra defined by generators and relations finite-dimensional and satisfies Poin...
Say $I = (f_1,\dotsc,f_l)$ is the ideal generated by the $f_i$. The $f_i$ are homogeneous; let’s add an assumption that none of the $f_i$ are constant (degree zero). The following conditions are equiv …
2
votes
Veronese and Segre
One way to think of the Segre map $\mathbb{P}^{n_1} \times \dotsm \times \mathbb{P}^{n_s} \to \mathbb{P}^N$ is as the map corresponding to multiplication of linear forms. By this I mean, for each $i=1 …
4
votes
Accepted
Condition for a monomial to belong to a particular ideal
Such a value $d$ does not necessarily exist. For $J=(xy,xz)$ we have $(xyz)^{d-1} \notin J^d$ for any $d$. It’s even worse (but maybe a little degenerate) if $J$ is principal.
A necessary and suffici …
0
votes
Accepted
Relation between Hilbert function and complete intersection ideals
The sum of values of the Hilbert function is equal to the vector space dimension of the algebra:
$$
\sum_{i=0}^{\infty} HF(T/I,i) = \dim(T/I).
$$
In this case, the vector space $T/I$ has a basis cons …
3
votes
Sum of initial ideals
Some authors write $\operatorname{in}_<(f)$ to mean the largest term of $f$, and other authors write $\operatorname{in}_<(f)$ to mean the smallest term of $f$. (Some authors say "leading term".) In yo …
2
votes
How to use Hilbert series to count combinatorial objects?
See:
Mordechai Katzman, Counting monomials
This paper presents two enumeration techniques based on Hilbert functions. The paper illustrates these techniques by solving two chessboard problems.
2
votes
Accepted
Under which conditions on the homogeneous ideal $ I $, the quotient ring $ \mathbb{C} [X_0, ...
$\mathbb{C}[x_1,\dotsc,x_n]/I$ is regular if and only if the affine variety $V(I)$ is smooth. When $I$ is homogeneous,a $V(I)$ is a cone (with vertex at the origin). The only way for a cone to be smoo …
3
votes
Betti numbers of a Cohen-Macaulay Module in small projective dimension
Yes, using Boij-Söderberg theory. Your Betti table is a convex combination of precisely two pure tables, each determined by Herzog-Kuhl, and the values of $\beta_{1,j}$ let you find the coefficients o …
1
vote
Accepted
Is this algorithm for primary decomposition correct?
(Just making an answer out of the above comment, with a small modification.) I think this will have problems if the ideal's generators don't factor at all. For example, for an ideal like
$$
(y^2 - x …
1
vote
Is being a Frobenius algebra a rare condition for local algebras?
Here is an emphatically "no-brainer", and partial, answer, just for the commutative case.
In the commutative case each such algebra $A$ is a quotient of a polynomial ring $R = k[x_1,\dotsc,x_r]$ ($k$ …