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UniversalGB -- whether the generators for an ideal form a universal Gröbner basis


Let $I \subseteq R = k[x_1, \ldots, x_n]$ be an ideal. A set of generators $\mathcal G$ for $I$ is a universal Gröbner basis for $I$ if it is a Gröbner basis for $I$ with respect to any monomial order on $R$. The default value is always UniversalGB=>false.

Set UniversalGB to true if it is known that the given generators for your ideal form a universal Gröbner basis. In this case, we can avoid computing Gröbner bases as geometric vertex decompositions preserve universal Gröbner basis. That is, if $\{ y^{d_i}q_i + r_i \mid i = 1, \ldots, s \}$ is a universal Gröbner basis for an ideal $I$, then $\{ q_1, \ldots, q_s \}$ and $\{ q_i \mid d_i = 0 \}$ are universal Gröbner bases for $C_{y,I}$ and $N_{y,I}$ in $k[x_1, \ldots, \hat y, \ldots, x_n]$, respectively.


If a universal Gr\"obner basis is not given, the intersection condition ${\rm in}_y(I) = C_{y,I} \cap (N_{y,I} + \langle y \rangle)$ via the results of [KR, Lemmas 2.6 and 2.12], which looks at the degree of $y$ in the reduced Gr\"obner basis of $I$. In general, a universal Gr\"obner basis is not reduced, so the intersection condition must be checked explicitly. So, although providing a universal Gr\"obner basis will speed up computing the ideals $C_{y, I}$ and $N_{y, I}$, it may take longer to verify the intersection condition.

See also

Functions with optional argument named UniversalGB :

For the programmer

The object UniversalGB is a symbol.