If we remove an element from a Gröbner basis $G$, is the new set $\overline G$ still a Gröbner basis of the new ideal generated by $\overline G$?

Question

Let $G=\{g_1,...,g_t\}$ be a Gröbner basis of $I$. If we remove $g_j$ from $G$ and let $\overline G=\{g_1,...,g_{j-1},g_{j+1},...,g_t\}$, is $\overline G$ still a Gröbner basis for the new ideal $\overline I=\langle g_1,...,g_{j-1},g_{j+1},...,g_t\rangle$ with respect to the same monomial order?

It is clear that $\overline I\subset I$. I have tried to use Buchberger’s Criterion but I have no idea how to proceed. Could anyone prove or give a counterexample? I will appreciate your help with this situation.

Answer 1

This is clearly false. If it were true, any generating set would be a Gröbner basis by the virtue of the following proof:

Just add some elements to get a Gröbner basis. Then remove these elements again. By your claim, you end up with a Gröbner basis, while the generated ideal has never changed during the whole process.

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For an easy counterexample take $G = \{x,y,x+y\}$ with $x>y$. If you remove $y$, you have no Gröbner basis anymore.

If you want a counterexample, where the new ideal is really a proper subset, take $G = \{x^2,y,x+y\}$.