<s>[INST] Complete solution for a system of polynomial equations A research question led to the following system of polynomial equations.
\begin{align*}
  & x^3 - x^2y + x^2z + x^2 - xy + xz + y^2 - yz - y = 0 \\
  & x^3z - 3x^2y + 3xy^2 + xz^2 - y^3 + y^2 - 2yz = 0 \\
  & x^4 - x^3y - 2x^3z + x^3 + x^2yz - x^2y + x^2z + xy^2 + xyz - xy - xz^2 - yz^2 = 0 \\
  & x^4z + x^4 - x^3y + x^2yz - x^2y + 2x^2z^2 + x^2z - x^2 - xy^2z + xy^2 + xy - y^2z + z^3 - z = 0
\end{align*}
I used the poly_system function in SymPy (a Python library for symbolic mathematics) to solve this system, and it returned the following solutions:
$$[(-1, 0, 0), (0, 0, -1), (0, 0, 0), (0, 0, 1), (0, 1, 0), (1, 1, 0), (1, 2, 1), (2, 3, 0), (3, 6, 1)]$$
This was better than I had hoped, but I am concerned that the function may fail to find all solutions. How can I be certain that no other solutions exist?
 [/INST] Using Maple, I took a "plex($x,y,z$)" Groebner basis of the ideal generated by the left sides of your equations.  The result was
$$ \eqalign{&{z}^{5}-{z}^{3}
,\cr
&y{z}^{3}-y{z}^{2}
,\cr
&{y}^{2}{z}^{2}-{z}^{4}-{y}^{2}z-y{z}^{2}+yz+{z}^{2}
,\cr
&{y}^{3}z-7\,{z}^{4}-8\,{y}^{2}z+5\,y{z}^{2}+7\,yz+7\,{z}^{2}
,\cr
&{y}^{4}-18\,{z}^{4}-4\,{y}^{3}-23\,{y}^{2}z+27\,y{z}^{2}+3\,{y}^{2}+21
\,yz+18\,{z}^{2}
,\cr
&-3\,{z}^{4}+6\,x{z}^{2}+{y}^{3}-4\,{y}^{2}z+7\,y{z}^{2}-3\,{z}^{3}-4\,
{y}^{2}-yz+3\,{z}^{2}+3\,y+3\,z
,\cr
&6\,{z}^{4}+6\,xyz-{y}^{3}+{y}^{2}z-7\,y{z}^{2}+3\,{z}^{3}+4\,{y}^{2}-2
\,yz-6\,{z}^{2}-3\,y-3\,z
,\cr
&2\,x{y}^{2}-{y}^{3}+{y}^{2}z-{z}^{3}-2\,xy-2\,xz+y+z
,\cr
&3\,{z}^{4}+12\,{x}^{2}z-2\,{y}^{3}+5\,{y}^{2}z-11\,y{z}^{2}+6\,{z}^{3}
+8\,{y}^{2}-7\,yz-3\,{z}^{2}-6\,y-6\,z
,\cr
&6\,{x}^{2}y-{y}^{3}+{y}^{2}z+2\,y{z}^{2}-3\,{z}^{3}-6\,xy-6\,xz-2\,{y}
^{2}+4\,yz+3\,y+3\,z
,\cr
&-{z}^{4}+4\,{x}^{3}-{y}^{2}z+5\,y{z}^{2}-4\,{z}^{3}+4\,{x}^{2}-8\,xy+y
z+{z}^{2}+4\,z
\cr
}$$
From the first basis element, all solutions must have $z \in \{-1, 0, 1\}$. 
If $z = -1$, the second basis element tells us $y=0$, and then the sixth gives us $x=0$.
If $z = 0$, the fifth basis element tells us $y \in \{0,1,3\}$.  If $y=0$ we get $x=0$ or $x=-1$ from the 11th basis element.  If $y=1$ we get $x=0$ or $x=1$ from the 10th basis element.  If $y=3$ we get $x=2$ from the 8th basis element.
If $z=1$, the fourth basis element gives us $y = 0$, $2$ or $6$.  If $y=0$, the 6th basis element gives us $x=0$.  If $y=2$, the 6th basis element gives us $x=1$.  If $y=6$, the 6th basis element gives us $x=3$.
Thus the listed solutions are the only ones.
</s>