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The fat point supported at $Z$ with multiplicity $k$ is the non-reduced scheme defined by $I(Z)^k$. This is often denoted $kZ$ in the fat points literature; it must not be confused with a zero-cycle. …

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 …

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.

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 …

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 …

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 …

(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 …

Some references include https://arxiv.org/abs/1309.5082 (see section 3) and Villareal, Monomial Algebras (2nd ed) 2015, Proposition 4.3.24-25.

It seems unlikely that the Betti numbers can be determined by the data you list (bigraded Hilbert function and ranks of multiplication maps). Consider the following possibility: $\dim M_{1,0} = \dim M …

This answer is just repeating comments, above, about a negative answer to conjecture (2). I'm unable to say anything for question (1). The conjecture (2) is not correct. As you note, $k(x) = k(x^3,x^ …

(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 …

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 $R = k[x_1,\dotsc,x_r]$ ($k$ …

$\mathbb{C}[x_1,\dotsc,x_n]/I$ 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 …

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 …

Let $\det_d = \det((x_{i,j})_{1 \leq i,j\leq d})$ be the determinant of a generic $d \times d$ matrix. Suppose $k \mid d$, $1 < k < d$. Can $\det_d$ be written as the determinant of a $k \times k$ mat …