If $I=\langle q\rangle$ then $\{q\}$ is a Gröbner basis for I.

Question

I've been reading Cox "Ideals, Varieties and Algorithms" and I've got stuck on this problem.

It looks easy but I don't know what to do.

I tried finding a contradiction on remainder being not equal to $0$ but I'm having doubts.

Answer 1

Hint

• The condition that each member of $\ I\ $ has remainder $\ 0\ $ with respect to $\ \{g\}\ $ looks like it might be one of the easiest to verify.
• For polynomials in two or more indeterminates, the proof that every element of $\ \langle g\rangle\ $ leaves a remainder of $\ 0\ $ with respect to $\ \{g\}\ $ is closely analogous to that for the case of univariate polynomials.