I think the following two claims can be proved directly by Buchberger’s criterion, but I lack confidence in my mathematical ability. I have tried to google but can not find a reference to them. Could anyone tell me whether they are true or false?
Let $G=\{f_1,\ldots,f_m,f_{m+1},\ldots,f_s\}$ be a Gröbner basis of $I\subseteq k[x_1,\ldots,x_n]$. If $f_1,\ldots,f_m$ involve only the variables $x_{i_1},\ldots,x_{i_l}$ and $f_{m+1},\ldots,f_s$ involve only the remaining variables $\{x_1,\ldots,x_n\}\setminus\{x_{i_1},\ldots,x_{i_l}\}$, then $\{f_1,\ldots,f_m\}$ is a Gröbner basis of the ideal $\langle f_1,\ldots,f_m\rangle\subseteq k[x_1,\ldots,x_n]$.
Let S be a subring of $k[x_1,\ldots,x_n]$. If $G$ is a Gröbner basis of an ideal $I\subseteq S$, then $G$ is also a Gröbner basis for the ideal of $k[x_1,\ldots,x_n]$ generated by $G$.
If claim 1 is true, my third question is:
- Are there any other circumstances where after removing some elements of a Gröbner basis $G$ the set $\overline G$ of remaining elements is still a Gröbner basis for the ideal generated by $\overline G$? A parallel might be the elimination property of elimination orders.
Thank you.
This is true. The initial terms of $f_i$ and $f_j$ for $i<m+1$ and $j \geq m+1$ will be relatively prime, since $f_i$ only involves variables $x_{i_1}, \ldots, x_{i_l}$ and $f_j$ involves only the other variables. By the "sharpened Buchberger criterion," one only needs to check pairs of polynomials where the leading terms are not relatively prime. Then, the computation of whether or not $\{f_1, \ldots f_s\}$ is a Groebner basis in the large ring is equivalent (exactly the same steps will be taken) to checking if $\{f_1, \ldots, f_m\}$ is a Groebner basis in $k[x_{i_1}, \ldots, x_{i_l}]$ and separately, if $\{f_{m+1}, \ldots, f_s\}$ is a Groebner basis in $k[\{x_1,\ldots, x_n\}\backslash\{x_{i_1}, \ldots, x_{i_l}\}]$.
This is also True, for the same reason as 1: your computation for the Buchberger criterion would follow exactly the same steps in both cases.