Are minimal Groebner bases minimized bases?

Question

With "minimal Groebner basis" I mean, fixed an ordering, a Groebner basis $G$ such that any proper subset of $G$ is no more a Groebner basis for the ideal $I(G)$ generated by $G$.

With "minimized basis" I mean a basis $B$ such that any proper subset of $B$ is no more a basis for $I(B)$.

So, can exist a minimal Groebner basis which is not minimized?

I can't find any contradiction but neither a counterexample.

Answer 1

A minimal lex Groebner basis which is not "minimized" is given by $\{ x^2 + y, xy - y, y^2 + y \}$ in $\mathbb{Q}[x,y]$.

This is clearly a minimal lex Groebner basis, but $$y^2 + y=y\cdot(x^2+y) - (x+1)\cdot(xy-y)$$ so the third element is superfluous as a generator of the ideal.

I found it by a computer search with SageMath, followed by manual simplification.