2
$\begingroup$

Let $R = \mathbb{C}[x_1,…,x_n]$, let $J\subset R$ be a graded ideal, and consider the initial monomial ideal $\operatorname{in}(J)$ with respect to some term order. Suppose that we are given a linear subspace $V \subset \mathbb{C}\{x_1,\ldots,x_n\}$ such that $R/J$ and $R/\operatorname{in}(J)$ are both free of finite rank as a module over $\operatorname{Sym}(V)$.

Now suppose that we have a finite set $S$ of monomials that constitute a basis for $R/\operatorname{in}(J)$ as a module over $\operatorname{Sym}(V)$. Does $S$ also constitute a basis for $R/J$ as a module over $\operatorname{Sym}(V)$?

(Note that this is easy if $V=0$. Note also that $R/J$ being a free $\operatorname{Sym}(V)$-module does not follow from $R/\operatorname{in}(J)$ being a free $\operatorname{Sym}(V)$-module; it is an important part of the hypotheses that BOTH are free.)

Example: Let $J = \langle xy + xz + yz\rangle \subset \mathbb{C}[x,y,z]$, with $\operatorname{in}(J) = \langle xy\rangle$. Let $V = \mathbb{C}\{x-y, y-z\}$, and let $S = \{1, x\}$. Then $R/\operatorname{in}(J)$ is a free $\operatorname{Sym}(V)$-module with basis $S$, and the same is true for $R/J$.

$\endgroup$

1 Answer 1

1
$\begingroup$

The answer is no.

Let $R = \mathbb{C}[x,y,z]$, $J = \langle y+z, (x+y)z\rangle$, $V = \mathbb{C}\{x+y+z\}$, and $S = \{1,x\}$. Then $\operatorname{in}(J) = \langle y, xz\rangle$, and $S$ is a $\operatorname{Sym}(V)$-basis for $R/\operatorname{in}(J)$. The algebra $R/J$ is also free over $\operatorname{Sym}(V)$, but $S$ is not a basis because $(x+y+z)\cdot 1 = x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.