Fix some algebraically closed field $k$ and let $X$ be the affine variety given by the ideal $I=(z^2-xy,xz-z)$, how can I describe the irreducible components of $I$? I know that there is a bijection between the irreducible components and the corresponding prime ideals but this doesnt help me.
Also, a variety $X$ is irreducible if $I(X)$ is a prime ideal, I tried to use this to find the irreducible varieties but I didnt manage, any hints?
Here are a few hints. I leave out some details.
First off, note that $xz-z=z(x-1)$. Thus we must have $$ I = (z^2-xy,xz-z) = (z^2-xy,z) \cap (z^2-xy,x-1). $$
(Why?) Furthermore, we see that $(z^2-xy,z)=(xy,z)=(x,z) \cap (y,z)$.
Here's a geometric look at the problem. First of all, write:
$$X=V(z^2-xy)\cap V(z(x-1))$$
Notice that $V(z(x-1)) = V(z)\cup V(x-1)$ i.e. the union of two hyperplanes. It then follows that:
$$X=V_1\cup V_2$$ Where the decompositions is given by $$V_1=V(z^2-xy)\cap V(z)$$ $$V_2=V(z^2-xy)\cap V(x-1)$$
By thinking about what a general element of $V_1$ looks like, it should be clear that it decomposes as two components which are "obviously" irreducible. Similarly you can see that $V_2$ is irreducible. It might help to try and draw $X$.
One way you can prove that each of the three factors in this decomposition is irreducible is by considering the corresponding ideals and showing that they're prime directly (looking at their quotients is probably easiest). However, there is a more geometric argument that explains why we would look at the components and think that they were irreducible. Any hyperplane is isomorphic to $\mathbb{A^2}$, and moreover isomorphisms preserve irreducible components. So each of the varieties $V_i$ are isomorphic to varieties in $\mathbb{A^2}$, which are again obviously irreducible, but now we also have that their ideals are principal and generated by prime elements, so they are trivially prime.
Maybe have a think about the geometric interpretation and if something's unclear, let me know and I'll elaborate.
