Does this square polynomial system have a finite number of zeros? Wish to know if this square polynomial system has a finite number of zeros:
$$
\begin{align}
f_1(z,w)&=-\frac{976 w^2}{53}+\frac{178 w z}{71}+\frac{27 w}{94}+\frac{323 z^2}{84}+\frac{271 z}{67}+\frac{199}{28}\\
f_2(z,w)&=\frac{72 w^2}{89}-\frac{502 w z}{97}-\frac{44 w}{57}+\frac{73 z^2}{5}-\frac{369 z}{73}+\frac{483}{95}
\end{align}
$$
I can compute a (lexicographic-ordered) Groebner basis and by inspection conclude the ideal dimension is zero but I'm a little unsure of myself and was wondering if someone could help me confirm the dimension?
Thanks.
 A: Yes, you are right: the ideal $I = \langle f_1,f_2 \rangle \subset \mathbb{C}[z,w]$ is zero-dimensional, and the number of complex solutions to the system is $4$, which is finite.
Indeed, choosing the lexicographic ordering with $z > w$ and running Buchberger's algorithm — using $S(f,g)$ to denote the $S$-polynomial and $\operatorname{red}(f,G)$ to denote the remainder of $f$ after multivariate polynomial division by the list $G$ — we obtain:

*

*$f_3 = \operatorname{red}(S(f_1,f_2),[f_1,f_2])$,

*$f_4 = \operatorname{red}(S(f_1,f_3),[f_1,f_2,f_3])$,

*$f_5 = \operatorname{red}(S(f_1,f_4), [f_1,f_2,f_3,f_4])$.

All further $S$-polynomials reduce to zero (after division by $f_1,f_2,f_3,f_4,f_5$), and the leading monomials are
$$LM(f_1)=z^2, \quad LM(f_2)=z^2, \quad LM(f_3)=zw, \quad LM(f_4) = z, \quad LM(f_5) = w^4.$$
Checking for redundancy yields $\operatorname{red}(f_i,[f_4,f_5])=0$ for $i=1,2,3$, so $f_1,f_2,f_3$ are redundant, meaning $$G = \{f_4, f_5\}$$ is a Groebner basis of the ideal $I$ with respect to the lexicographic ordering with $z>w$.
The multivariate polynomial division algorithm with respect to $G$ implies that the quotient $\mathbb{C}[z,w]/I$ as a $\mathbb{C}$-vector space has a basis consisting of monomials not divisible by $LM(f_4)=z$ or $LM(f_5)=w^4$, so a $\mathbb{C}$-basis is $[1, w, w^2, w^3]$ of size $4$.
In the set of solutions, the four complex roots of $f_5(w)$ provide the values of $w$, and $f_4(z,w)=0$ being linear in $z$ yields each corresponding value of $z$, so there are four complex solutions (and $I$ is radical).
