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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 $R/J$.

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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 $(x+y+z)\cdot 1 = x$.

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