Chapter 1, Exercise 1.3: A Multi-Component Algebraic Set

Let $Y$ be the algebraic set in $\mathbb A^3$ defined by the two polynomials $x^2-yz$ and $xz-x$. Show that $Y$ is a union of three irreducible components. Describe them and find their prime ideals.

This is a relatively relaxing calculation problem to solve before entering the more abstract part of the first problem set. We are looking for three irreducible components of $Y$. To do this, we first consider two cases: $x=0$ and $x\neq 0$.

For the case $x=0$, the second equation becomes trivial and the first equation becomes $yz=0$. The algebraic set $Z(yz)$ is obviously just $Z(y)\cup Z(z)$, and further imposing the condition $x=0$ gives two irreducible components $Y_1=Z(x,y)$ and $Y_2=Z(x,z)$. These correspond geometrically to the $z$ and $y$ axes respectively, and have prime ideals $I(Y_1)=(x,y)$ and $I(Y_2)=(x,z)$.

Next, we have the case $x\neq 0$. In this case we can divide by $x$ in the second equation to obtain $z-1=0$ or $z=1$. Plugging this into the first equation, we find $y=x^2$. There appears to be a problem here however - how are we supposed to turn the condition $x\neq 0$ into a polynomial equation? The good news is that we don't have to: the point in which $x=0$ is $(0,0,1)$ which is already in $Y_1$. Thus we have the final irreducible component $Y_3=Z(y-x^2,z-1)$ with prime ideal $I(Y_3)=(y-x^2,z-1)$. This corresponds to a parabola lifted up one unit above the $xy$ plane along the $z$ axis. By construction, $Y=Y_1\cup Y_2\cup Y_3$ as desired.