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?
Best Answer
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.