Is it possible that $G_j(=G\cap k[x_j,...,x_n])$ generates $I_j$ where $G$ is a Groebner basis of $I$ but $G_j$ fails to be a Groebner basis of $I_j$?

Question

In the following, I use Sage's terminology (see http://doc.sagemath.org/html/en/reference/polynomial_rings/sage/rings/polynomial/term_order.html). Fix a non-lex order on $k[x_1,...,x_n]$. Let $j\ge 2$ and $G$ be a Groebner basis of an ideal $I\subseteq k[x_1,...,x_n]$. Is it possible that $G\cap k[x_j,...,x_n]$ generates the ideal $I\cap k[x_j,...,x_n]$ of $k[x_j,...,x_n]$ but $G\cap k[x_j,...,x_n]$ is not a Groebner basis of $I\cap k[x_j,...,x_n]$ ?

I suspect the answer to my question is "yes". This question is related to a previous one I've asked: Is it possible that $G_j(=G\cap k[x_j,...,x_n])\neq \emptyset$ (where $G$ is a Groebner basis of $I$) but $G_j$ is not a Groebner basis of $I_j$?, but I fail to find a line of reasoning or an example again. Thank you.

Answer 1

Again, this is of course possible. Take any generating set, that is not Groebner basis, for instance (graded lex order, $x>y$)

$$J=(x,x^2+y).$$

Note that $J=(x,y)$.

Now add another variable $z$ with $x>y>z$ and add some polynomials to get a Groebner basis of an ideal in $k[x,y,z]$:

$$I=(x,x^2+y,z,y+z)=(G).$$

The leading terms of the generators are $x,x^2,z,y$, i.e. $G$ is a Groebner basis.

On the other hand we have $I=(x,y,z)$, i.e. $I \cap k[x,y]=J$. Thus $G \cap k[x,y]$ generates $I \cap k[x,y]$, but it is not a Groebner basis.