If an ideal is made up by polynomials with disjoint variable parts, then those polynomials form a Grobner Basis.

Question

I've been learning symbolic computation over the summer (just independent learning) and I'm at the section of my book about Grobner bases. There's an exercise I'd like to see a proof of, but have not been able to answer. Here it is:

Suppose that $I = (f_1,\dots,f_n)$ where $$ variables(f_i) \cap variables(f_j) = \emptyset $$ for $i\neq j$. Show that $\{f_1,\dots,f_n\}$ is a Grobner basis for $I$.

If anyone could help, it'd be much appreciated.

Answer 1

Just follow Buchberger's algorithm. Every step includes the next element of the generators, no elimination occurs because the variables are disjoint (step 3), and syzygies are all zero for the same reason (step 4).