Compute $Z(I)\subset \mathbb C^3$ for $I=(xy-z^2,x^2-y^2)$. I have to compute $Z(I)=\{(p,q,r)\in\mathbb C^3\mid \forall f\in I, f(p,q,r)=0\}$. I found $$Z(I)=\{(p,p,p)\mid p\in\mathbb C\}\cup\{(p,p,-p)\mid p\in\mathbb C\}\cup\{(p,-p,ip)\mid p\in\mathbb C\}\cup\{(p,-p,-ip)\mid p\in\mathbb C\}.$$
Then it's written : You should see that this naturally breaks into smaller algebraic sets. What are the ideal of each pieces ? How do the relate to $I$ ? 
I don't really understand what they want to say. It's written that the ideal of the four pieces are $(x-y,x-z),(x-y,x+z),(x+y,x-iz)$ and $(x+y,x+iz)$, but how do they find them ? Them they say that each is a prime ideal which strictly contain $I$. Why do we have this fact ? Is it a property of $Z(I)$ ?
 A: You have already written a decomposition of $Z(I)$ as the union of four pieces and you can observe that every one of them is exactly of the form $Z(J)$ where $J$ is one of the four ideals that are given to you. Moreover, since for every such $J$ you have $Z(J) \subset Z(I)$ then $\sqrt{J} \supset \sqrt{I}$ because of the association between ideals and algebraic sets. However, for every $J$ written in the text you have $\sqrt{J} = J$ because they are defined by polynomials of homogeneous degree $1$, hence $J \supset \sqrt{I} \supseteq I$.
There is another strategy that can be used to tackle this problem which works without knowing in advance the explicit formulas for the ideals $J$. One can reason as follows:
\begin{align}
Z(I) = Z(xy-z^2,x^2-y^2) = &\; Z(xy-z^2,x-y) \cup Z(xy-z^2,x+y) = \\
= &\; Z(x^2-z^2,x-y) \cup Z(-x^2-z^2,x+y).
\end{align}
Now, the first algebraic set splits again in $Z(x-z,x-y) \cup Z(x+z,x-y)$ while the second one splits in $Z(x-iz,x+y) \cup Z(x+iz,x+y)$. In other words one works directly with the polynomials defining $I$. By the same reasoning as before one then deduces that for every $J$ appearing in one of the four smaller algebraic sets the inclusion $J = \sqrt{J} \supset \sqrt{I} \supseteq I$.
