Let us consider the polynomial ring $A:=\mathbb{R}[x_1,x_2,\dots]$ and consider the family consisting of the empty set and of all the subsets of polynomials $G$ such that there exists an ideal of $A$ having $G$ as minimal Groebner basis. Is it true that if $B$ belongs to the above family, then also $B' \subsetneqq B$ does?


No, because the subset of a Gröbner basis isn't necessarily a Gröbner basis.1 Consider, for example, the ideal $I = (B')$, where $B' = \left\{x_1^2x_2 - x_1^3, x_2^3\right\}$. Note that $$x_1^5 = -x_1^2 \left(x_1^2 x_2 - x_1^3\right) - x_1\left(x_1^2 x_2 - x_1^3\right) x_2 - \left(x_1^2 x_2 - x_1^3\right)x_2^2 + x_1^2 x_2^3 \in I$$ but clearly $$x_1^5 \not \in \left(x_1^2 x_2, x_2^3\right),$$ meaning that $B'$ is not a Gröbner basis, minimal or otherwise. However, $$B = \left\{x_1^2x_2 - x_1^3, x_2^3, x_1^5\right\}$$ is a minimal Gröbner basis for $I$ (exercise left to the reader), and $B' \subsetneq B$. (I'm assuming $x_1 < x_2$ for my ordering in this example.)

1 In fact, that's the point of a Gröbner basis: you have enough generators for the ideal to know how to reduce everything modulo that ideal.

  • $\begingroup$ Assume that B is a Groebner basis for some ideal I. Take $B' \subsetneqq B$. Is it a Groebner basis for another ideal I'? This is the sense of my question. I edited my question in order to make clearer the sense and I apologize for the misunderstanding. $\endgroup$ – TheWanderer May 7 '18 at 16:08
  • 1
    $\begingroup$ @TheWanderer, in my example, $B'$ and $B$ generate the same ideal. $\endgroup$ – PersonX May 7 '18 at 16:11

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