TITLE: Computing irreducible components of algebraic set
QUESTION [7 upvotes]: Consider the algebraic set $V(X^2-YZ,X-XZ)$. Find the irreducible  components of this set and show that $I(V)=(X^2-YZ,X-XZ)$.
I reasoned that $X-XZ=0$ iff $X=0$ or $Z=1$. If $X=0$, we get $Y=0$ or $Z=0$ from the other equation. If $Z=1$ we get $X^2-Y=0$. Hence the irreducible components are $V(X,Y), V(X,Z), V(Z-1,X^2-Y)$. Is this correct? 
I'm not sure how to do the next question. Of course automatically we get one inclusion, but how about the other?
Thanks

REPLY [3 votes]: Your proof is correct, the variety that you obtain is the union of the varieties given by
$$X=Y=0, X=Z=0, Z=1,Y=X^2.$$
It is then easy to see that each polynomial that vanishes on this set belongs to $$I=(X^2-YZ,X-XZ).$$ Indeed, modulo $I$ you can replace $YZ$ with $X^2$ and $XZ$ with $X$ so every polynomial is equivalent to
$$P(Z)+Q(X,Y)$$
for some polynomials $P,Q$. Because the polynomial vanishes on $X=Y=0$, we get that $P$ is a constant and can assume $P=0$. Because the polynomial vanishes on $Z=1,Y=X^2$ and on $X=Z=0$, $Q(X,Y)$ is a multiple of $X(Y-X^2)$. It remains to see that $X(Y-X^2)\in I$, which is because 
$$X(Y-X^2)=X(Y-YZ)=Y(X-XZ)=0\pmod I.$$