What are the irreducible components of the algebraic set $V(y^2-xz,z^2-x^2y)$ in $\mathbb{A}^3_K$? I'm stuck on a exercise from my class notes of Commutative Algebra.Exercise goes as follows:

What are the irreducible components of the algebraic set $V(y^2-xz,z^2-x^2y)$ in $\mathbb{A}^3_K$? Here I'm just letting $K$ be an algebraically closed field.

I've tried to solve this problem and I've found that one irreducible component will be $V(y,z)$ but I'm unable to solve it completely. I guess this problem requires a t"trick" and I'm unable to catch that trick. I solve such problems in the following way: 
Normally, what I do is take the equations determining an algebraic set $V(I)$, and usually one of them factors so that $V(I)$ decomposes as $V(J_1)\cup V(J_2)\cup\cdots$ or something. After breaking things down enough, I can eventually find that $K[x,y,z]/J_i$ is an integral domain, so $J_i$ is prime, and $V(J_i)$ is irreducible. Any hints/ideas?
 A: In general , you can use 
primary decomposition in Macaulay2.
In your case,
The primary decomposition of your ideal is $(y^2  - xz, x^2 y - z^2 , x^3 - yz)$ and $(z, y)$ . This means that their intersection is your ideal $(y^2-xz,z^2-x^2y)$ and both ideals are primary.
Then you can take radical of both ideal to find the irreducible components. The radical of both ideals remain the same so the above two ideals are prime and their the prime ideals of the irreducible components.
A: Okay, so you have already found out that $I_1 = (y,z)$ is one irreducible component of $I=(y^2-xz,z^2-x^2y)$. Then to get rid of $I_1$ in $I$, one can compute $I:(g) = \{ f \in k[x,y,z] \mid rg \subset I \}$ for various $g$. It is a fact that all irreducible components of $I$ occurs in this fashion.
So let's compute $I:(z)$. 
By nothing that $x^2(y^2-xz)+y(z^2-x^2y)=yz^2-x^3z=z(yz-x^3)$ is in $I$, we see that $yz- x^3 \in I:(z)$. It is also clear that $I \subset I:(z)$. In fact it is true that that $I:(z) = I+(yz-x^3)=(y^2-xz,z^2-x^2y,yz-x^3)$, but we don't need this.
We have found a candidate for another component if $I$. So we need to check that this is in fact a prime ideal. I'm not completely sure how to do this without a computer, we can note that the ideal $I:(z)$ is given by the minors of 
$$
\begin{bmatrix}
z & y & x \\ x^2 & z & y
\end{bmatrix}
$$
And such ideals are very often prime. 
