What are the irreducible components of $V(xy-z^3,xz-y^3,x-z^2+y)$? I was reading the question here, and trying to fill out msteve's answer. 
It's not clear to me how to break up $V(xy-z^3,xz-y^3,x-z^2+y)$ into irreducible components, of which there should be two I think, based on the other answers there. I saw that
$$
xy-z^3=xy-z(x+y)=xy-xz-yz=xy-y^3-yz=y(x-y^2-z)
$$
So it should break up as
$$
\begin{aligned}
V(xy-z^3,xz-y^3,x-z^2+y) &= V(y,xz-y^3,x-z^2+y)\cup V(x-y^2-z,xz-y^3,x-z^2+y)\\
&= V(x,y,z)\cup V(x-y^2-z,xz-y^3,x-z^2+y)
\end{aligned}
$$
Obviously $V(x,y,z)$ is irreducible, and I think $V(x-y^2-z,xz-y^3,x-z^2+y)$ should be as well, but I don't see an easy way to show its coordinate ring is a domain. Is there a trick?
 A: So I just put the system
$$x-y^2-z = x z - y^3 = x - z^2 + y = 0$$
into singular to see what happens:
> ring r = 0,(x,y,z),dp;
> poly p1 = x - y^2 - z;
> poly p2 = x*z-y^3;
> poly p3 = x - z^2 + y;
> ideal i = p1,p2,p3;
> ideal si = groebner(i);
> LIB "primdec.lib";
> list cpts = primdecGTZ(si);
> size(cpts);
3
> ideal i1 = cpts[1][2];
> ideal i2 = cpts[2][2];
> ideal i3 = cpts[3][2];
> ideal si1 = groebner(i1);
> ideal si2 = groebner(i2);
> ideal si3 = groebner(i3);
> dim(si1);
0
> dim(si2);
0
> dim(si3);
0
> si1;
si1[1]=z
si1[2]=y
si1[3]=x
> si2;
si2[1]=y-z+1
si2[2]=2x-1
si2[3]=2z2-2z+1
> si3;
si3[1]=2z+1
si3[2]=2y-1
si3[3]=4x+1

Note that it's working over $\mathbb{Q}$ here instead of $\mathbb{C}$, so the second "component" is something that we'd consider to be two points if we were working over a field where $2z^2 - 2z + 1$ had a root.  Anyway, this system (at least working over $\mathbb{C}$) turns out to have four solutions, assuming I didn't make a mistake putting it into the computer.
In fact, in this case you could get the same answer from WolframAlpha:
http://www.wolframalpha.com/input/?i=x-y%5E2-z+%3D+0%2C+x+z+-+y%5E3+%3D0%2C++x+-+z%5E2+%2B+y+%3D+0

All right, but is there anything we can say that doesn't rely on computers?  Well, first, the equations are set up in such a way that it's easy to eliminate the variable $z$.  So the coordinate ring is given by
\begin{align*}
k[x,y,z]&/(x-y^2-z,xz-y^3,x-z^2+y) \\
&= k[x,y]/(x^2 - x y^2 - y^3, x - x^2 + 2 x y^2 - y^4 + y) 
\end{align*}
So we're looking at the intersection of two curves in the plane.  One of two things can happen here -- either the curves share a common component, in which case their intersection is positive-dimensional, or they do not, in which case they intersect in a finite set of points.  Now, it's not too hard to see that $x^2 - x y^2 - y^3 = 0$ is an irreducible curve (just suppose it factors and start thinking about what the constant and linear terms of the factors could possibly be and where you could go from there), so the only way these curves share a component is if $x^2 - x y^2 - y^3$ divides into $x - x^2 + 2 x y^2 - y^4 + y$.  But that's impossible, because the former has no linear terms and the latter does.  Consequently, the intersection is a finite collection of points.
