How is the reduced Groebner base of the ideal of an inconsistent polynomial system of equations?

Question

Given a polynomial system ($f_1=0,....,f_s=0$) and the ideal $I$ <$f_1,....,f_s$>.

How is the reduced Groebner base of $I$ in the case that the system has no solutions?

Answer 1

Saying that the system has no solutions means that the variety associated with your system of equations is the empty set, i.e., $\mathbf{V}(f_1,\dots,f_s) = \emptyset$.

The ideal associated with this variety is $\mathbf{I}(\emptyset) = \mathbb{C}[x_1,\dots,x_n]$, where $\mathbb{C}[x_1,\dots,x_n]$ denotes the set of polynomials on the complex field. By Hilbert's Nullstellensatz, this means that $\langle f_1,\dots,f_s\rangle = \mathbb{C}[x_1,\dots,x_n]$, for which the reduced Gröbner basis is clearly $f = 1$. Thus: $$ \langle f_1,\dots,f_s\rangle = \langle 1\rangle. $$ (Edited for clarity.)