Buchberger algorithm with three variables

Question

Consider $>_{degrevlex}$ with $x>y>z$ in $\mathbb{C}[x,y,z]$. Let $f_{1} = y^2 - xz$, $f_{2} = x^2y - z^2$, $f_{3} = x^3 - yz$.

Show (manually) that $\lbrace f_{1}, f_{2}, f_{3}\rbrace$ is a Gröbner basis of the Ideal $\langle f_{1}, f_{2}, f_{3}\rangle$.

So I know that I need to find the S-polynomials, but I am a little confused on how to go from there, when I got three and not two variables.

$f_{4} = S(f_{1}, f_{2}) = x^3z - yz^2$

$f_{5} = S(f_{1}, f_{3}) = x^4z - y^3z$

$f_{6} = S(f_{2}, f_{3}) = -y^2z + xz^2$

I could only find examples and explanations for two variables, so any hints, explanations or similar examples would be greatly appreciated.

Answer 1

You've correctly computed all the S-polynomials. Maybe you want to flip the sign of each of them, depending on your conventions (definition of S-polynomial). Anyway, the sign is ultimately irrelevant.

The next step is to reduce each of these $S(f_i,f_j)$ with respect to the tuple $F=(f_1,f_2,f_3)$. The linked explanation is for any number of variables. For example, to start reducing $f_4$ with respect to $F$, you start with $p:=f_4$ and search for an $f_i \in F$ with $LM(f_i) \mid LM(p)$. Here $i=3$ works, so we can subtract a multiple of $f_3$ from $p$ to cancel the leading monomials (and hence make $p$ "smaller" with respect to the monomial ordering), namely we set $p :=p-zf_3$ and $q_3:=z$. What we get is $p=0$, so we are done: $f_4$ reduces to zero with respect to $F$ (in some books this is denoted by $f_4 \xrightarrow{F} 0$). Now you can continue with $f_5$ and $f_6$.

When they all reduce to zero, you have a Groebner basis. This is Buchberger's criterion.