Determining irreducible components of $Z(y^4 - x^6, y^3 - x y^2 - y x^3 + x^4) \subset \mathbb A^2$ I'm trying to determine the irreducible components of the zero set $Z(I)$ for the ideal $I = (y^4 - x^6,\, y^3 - x y^2 - y x^3 + x^4)$, in the affine space $\mathbb A^2$ over an algebraically closed field $k$.
I can factor $I = (y^2 - x^3)(y^2  +x^3,\, y - x)$, hence $Z(I) = Z(y^2 - x^3) \cup Z(y^2 + x^3,\, y - x)$, but I'm stuck trying to prove these ideals are prime (and is the second one even prime? If not, how do I reduce it further?).

The key fact that we will use to find the irreducible components is that an algebraic subset of a variety is irreducible iff its coordinate ring is an integral domain. Therefore, consider the coordinate ring of the algebraic set $Z(y^2 - x^3)$: it is $\frac{k[x,y]}{(y^2 - x^3)}$, which is an integral domain, hence $Z(y^2 - x^3)$ is irreducible.
In order to apply this fact to the second algebraic set, it must be simplified further:
$$
Z(y^2 + x^3, y-x) = Z(x^2 + x^3, y-x) = Z(x^2(x+1), y-x) = Z(x^2, y-x) \cup Z(x+1, y-x). 
$$
Now, $\frac{k[x,y]}{(x+1,y-x)} \simeq \frac{k[x]}{(x+1)} \simeq k$ is an integral domain, so $Z(x+1,y-x)$ is irreducible. Finally, the coordinate ring of $Z(x^2, y-x)$ is $\frac{k[x,y]}{(x,y-x)} \simeq k$, which is an integral domain, so $Z(x^2, y-x)$ is irreducible, as well (it's important here that, when constructing the coordinate ring of $Z(x^2,y-x)$, one takes the radical of the ideal $(x^2,y-x)$ rather than the ideal itself).
Therefore, we conclude that the irreducible components of $Z(y^4-x^6, y^3-xy^2-yx^3+x^4)$ are 
$$
Z(y^2 - x^3) \textrm{ and } Z(x^2, y-x) \textrm{ and } Z(x+1, y-x).
$$
