I'm currently working through a paper where the authors work with lattice ideals, and one of the propositions (Proposition 6, for which only a very short proof is given) seems to rely on a seemingly simple commutative algebra claim that I haven't been able to verify (or find a counterexample to).
Here is a simplified version of it, where I've tried to clean up the notation as much as possible:
Consider the polynomial ring $R=\mathbb{C}[\mathbf{x},\mathbf{t}]=\mathbb{C}[x_1,\ldots,x_n,t_1,\ldots,t_m]$.
Let $\mathbf{a}_1,\ldots,\mathbf{a}_m\in\mathbb{N}^n$ be tuples of non-negative integers that specify monomials $\mathbf{x}^{\mathbf{a}_1},\ldots,\mathbf{x}^{\mathbf{a}_m}\in \mathbb{C}[x_1,\ldots,x_n]$.
Let $J\subseteq R$ be the ideal generated by all elements of the form $t_i\mathbf{x}^{\mathbf{a}_j}-t_j\mathbf{x}^{\mathbf{a}_i}$ for $i,j\in\{1,\ldots,m\}$ with $i<j$.
Claim: $J:(x_1\cdots x_n\cdot t_1\cdots t_m)^\infty=J:(x_1\cdots x_n)^\infty$.
Observation: Since $$I:(x_1\cdots x_n\cdot t_1\cdots t_m)^\infty=(I:(t_1\cdots t_m)^\infty):(x_1\cdots x_n)^\infty,$$ it would suffice if we could show $J:(t_1\cdots t_m)^\infty=J$, which is what I've mostly focused on this far.
Example 1: If $m=2$, we get $J=\langle t_1\mathbf{x}^{\mathbf{a}_2}-t_2\mathbf{x}^{\mathbf{a}_1}\rangle\subseteq\mathbb{C}[\mathbf{x},t_1,t_2]$. In this case it is easy to see that $J:(t_1t_2)^\infty=J$.
A brute-force argument is this: If $f\in J:(t_1t_2)^\infty$, then there exists some $k\in\mathbb{Z}^+$ and some $h\in\mathbb{C}[\mathbf{x},\mathbf{t}]$ such that $f\cdot (t_1 t_2)^k=h\cdot (t_1\mathbf{x}^{\mathbf{a}_2}-t_2\mathbf{x}^{\mathbf{a}_1})$. Setting $t_2=0$ gives $0=h'\cdot t_1\mathbf{x}^{\mathbf{a}_2}$, where $h'=h(\mathbf{x},t_1,0)$. Clearly we must have $h'=0$, from which it follows that $t_2$ is a factor in $h$. Similarly, one can show that $t_1$ must be a factor in $h$. Repeated use of this argument gives that $(t_1t_2)^k$ is a factor in $h$, and we conclude that $f\in J$.
Example 2: If $m=3$, we get $$J=\langle t_1\mathbf{x}^{\mathbf{a}_2}-t_2\mathbf{x}^{\mathbf{a}_1},\:\:t_1\mathbf{x}^{\mathbf{a}_3}-t_3\mathbf{x}^{\mathbf{a}_1},\:\:t_2\mathbf{x}^{\mathbf{a}_3}-t_3\mathbf{x}^{\mathbf{a}_2}\rangle\,.$$ As before, we can observe that if $f\in J:(t_1t_2t_3)^\infty$, then there exists some $k\in\mathbb{Z}^+$ and $h_{12},h_{13},h_{23}\in\mathbb{C}[\mathbf{x},\mathbf{t}]$ such that $$f\cdot (t_1t_2t_3)^k=h_{12}\cdot(t_1\mathbf{x}^{\mathbf{a}_2}-t_2\mathbf{x}^{\mathbf{a}_1})+h_{13}\cdot(t_1\mathbf{x}^{\mathbf{a}_3}-t_3\mathbf{x}^{\mathbf{a}_1})+h_{23}\cdot (t_2\mathbf{x}^{\mathbf{a}_3}-t_3\mathbf{x}^{\mathbf{a}_2})\,.$$ The desired result would follow if we could show that the $h$'s can be chosen so that they are all divisible by $t_1t_2t_3$. But is that true? Can we be sure that they don't contain any terms that aren't divisible by $t_1t_2t_3$ but then somehow cancel out? Just substituting zeros into this expression didn't turn out to be very fruitful, so I probably need a more systematic approach...
Any ideas or pushes in the right direction would be much appreciated!