In the following, we use Sage's terminology (see http://doc.sagemath.org/html/en/reference/polynomial_rings/sage/rings/polynomial/term_order.html).
Fix deglex or degrevlex on $k[x_1,...,x_n]$. Let $j\ge 2$ and $G$ be a Groebner basis of an ideal $I$. Is it possible that $G\cap k[x_j,...,x_n]\neq \emptyset$ but $G\cap k[x_j,...,x_n]$ is not a Groebner basis of $I\cap k[x_j,...,x_n]$ ?
I know the Elimination Theorem for lex Groebner bases. My question arises when I try to find examples establishing that deglex and degrevlex don't have elimination properties in the same way as lex. All the examples I have found have the following feature: $G\cap k[x_j,...,x_n]=\emptyset$ (where $G$ is w.r.t. deglex or degrevlex) but $I\cap k[x_j,...,x_n]\neq \emptyset$. For instance, consider
$I=\langle x^2+y+z-1,x+y^2+z-1,x+y+z^2-1\rangle\subseteq\mathbb{Q}[x,y,z]$. Then $G=\{x^2+y+z-1,y^2+x+z-1,z^2+x+y-1\}$ is a degrevlex Groebner basis of $I$ and therefore $G\cap\mathbb{Q}[z]=\emptyset$. However, $I\cap\mathbb{Q}[z]=\langle z^6-4z^4+4z^3-z^2 \rangle$.
I suspect the answer to my question is "yes", but I can't find an example. I would appreciate any help with this situation.
This can of course happen. Just take any Groebner basis with $G \cap k[x_j \dotsc, x_n] = \emptyset$ but $I \cap k[x_j \dotsc, x_n] \neq 0$. For instance one of your examples.
Then add some $f \in I \cap k[x_j \dotsc, x_n]$ to $G$, which does NOT generate $I \cap k[x_j \dotsc, x_n]$. Then the new $G$ has the property $G \cap k[x_j \dotsc, x_n] = \{f\}$ and this is of course not a Groebner basis of the intersection, since it does not generate it.